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NCERT Solutions for Class 12 Maths Chapter 5 - Continuity And Differentiability

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NCERT Class 12 Maths Chapter 5: Complete Resource for Continuity and Differentiability

NCERT Solutions for Class Maths Chapter-5 PDF is available to download now. You can download the PDF file from the official website of Vedantu. The Continuity and Differentiability Class 12 NCERT PDF relating to all the solutions have been provided here. You can check all the answers related to Class 12 Maths Chapter 5 here. The Continuity and differentiability Class 12 CBSE NCERT Solutions are created by subject matter experts who have years of teaching experience. They have the proper expertise to explain the questions thoroughly and efficiently while providing adequate justifications and solutions to the problem. If you have any problems, do visit the website and drop your questions and download the NCERT solutions for Class 12 Maths Chapter 5 PDF file. The online helping staff will get in touch with you. 


Class:

NCERT Solutions for Class 12

Subject:

Class 12 Maths

Chapter Name:

Chapter 5 - Continuity and Differentiability

Content-Type:

Text, Videos, Images and PDF Format

Academic Year:

2023-24

Medium:

English and Hindi

Available Materials:

  • Chapter Wise

  • Exercise Wise

Other Materials

  • Important Questions

  • Revision Notes



Continuity and Differentiability Chapter at a Glance - Class 12 NCERT Solutions

  • A real valued function is continuous at a point in its domain if the limit of the function at that point equals the value of the function at that point. A function is continuous if it is continuous on the whole of its domain.

  • Sum, difference, product and quotient of continuous functions are continuous. i.e, if $f$ and 8 are continuous functions, then

$(f \pm g)(x)=f(x) \pm g(x)$ is continuous.

$(f \cdot g)(x)=f(x) g(x)$ is continuous.

$\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}$ (Wherever $\left.g(x) \neq 0\right)$ is continuous.

  • Every differentiable function is continuous, but the converse is not true

  • Chain rule is a rule to differentiate composites of functions.

If $f=V \circ 1<t=11(x)$ and if both $\frac{d t}{d t}$ and $\frac{d}{d t}$ exist, then $\frac{d f}{d}=\frac{d}{d t} \cdot \frac{d}{d x}$

  • Following are some of the standard derivatives (in appropriate domains):

\[\begin{aligned}& \frac{d}{d x}\left(\sin ^{-1} x\right)=\frac{1}{\sqrt{1-x^2}} \\& \frac{d}{d x}\left(\cos ^{-1} x\right)=-\frac{1}{\sqrt{1-x^2}} \\& \frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{1+x^2} \\& \frac{d}{d x}\left(\cot ^{-1} x\right)=\frac{-1}{1+x^2} \\& \frac{d}{d x}\left(\sec ^{-1} x\right)=\frac{1}{x \sqrt{1-x^2}} \\& \frac{d}{d x}\left(\operatorname{cosec}^{-1} x\right)=\frac{-1}{x \sqrt{1-x^2}} \\& \frac{d}{d x}\left(e^x\right)-e^x \\& \frac{d}{d x}\left(\log ^x x\right)=\frac{1}{x}\end{aligned}\]

  • Logarithmic differentiation is a powerful technique to differentiate functions of the form $f(x)-[11(x)]^{\text {rix }}$. Here both $f(x)$ and $u(x)$ need to be positive for this technique to make sense

  • Rolle's Theorem: If $f:[a, b] \rightarrow R$ is continuous on $[a, b]$ and differentiable on $(a, b)$ such that $f(a)-f(b)$, then there exists some $c$ in $(a, b)$ such that $f^{\prime \prime}(c)=0$.

  • Mean Value Theorem: If $f:[a, b] \rightarrow R$ is continuous on $[a, b]$ and differentiable on $(a, b)$. Then there exists some $c$ in $(a, b)$ such that $f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}$ (Not present in current syllabus)

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Exercises under NCERT Class 12 Maths Chapter 5 Continuity and Differentiability

Chapter 5 of NCERT Class 12 Maths textbook covers Continuity and Differentiability. The chapter is divided into nine exercises which are briefly summarized as follows:


Exercise 5.1: This exercise deals with the concept of continuity of a function, and how to check the continuity of a function using algebraic techniques.

Exercise 5.2: The Intermediate Value Theorem is introduced in this exercise, which is used to prove the existence of roots of equations.

Exercise 5.3: This exercise covers the concept of differentiability of a function, and the differentiability of algebraic and trigonometric functions are explored.

Exercise 5.4: The chain rule is introduced in this exercise, which is used to find the derivative of composite functions.

Exercise 5.5: This exercise covers the derivative of implicit functions and related rates problems.

Exercise 5.6: The derivatives of inverse trigonometric functions and logarithmic functions are explored in this exercise.

Exercise 5.7: This exercise covers the derivatives of functions expressed in parametric form.

Exercise 5.8: This exercise explores the concept of derivatives of functions expressed in polar form.

Miscellaneous Exercise: This exercise contains additional problems related to continuity and differentiability.


Overall, these exercises provide a comprehensive understanding of the concepts of continuity and differentiability, and how to apply them to solve various mathematical problems.


Access NCERT Solutions for Class 12 Maths Chapter 5 – Continuity and Differentiability

Exercise 5.1

1. Prove that the function  $ \text{f}\left( \text{x} \right)\text{=5x-3} $  is a continuous at  $ \text{x=0} $ , at  $ \text{x=-3} $  and at  $ \text{x=5} $ .

Ans: The given function is  $ \text{f}\left( \text{x} \right)\text{=5x-3} $ .

At  $ \text{x=0} $ ,  $ \text{f}\left( \text{0} \right)\text{=5 }\!\!\times\!\!\text{ 0-3=-3} $ .

Taking limit as  $ x\to 0 $  both sides of the function give

 $ \underset{x\to 0}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{x\to 0}{\mathop{\lim }}\,\left( \text{5x-3} \right)\text{=5 }\!\!\times\!\!\text{ 0-3=-3} $ 

 $ \therefore \underset{x\to 0}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{0} \right) $ .

Thus,  $ \text{f} $  satisfies continuity at  $ \text{x=0} $ .

Again, at  $ \text{x = -3,}\;f\left( \text{-3} \right)\text{=5 }\!\!\times\!\!\text{ }\left( \text{-3} \right)\text{-3=-18} $ .

Now, taking limit as  $ x\to 3 $  both sides of the function give

 $ \underset{x\to 3}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{x\to 3}{\mathop{\lim }}\,\text{f}\left( \text{5x-3} \right)\text{=5 }\!\!\times\!\!\text{ }\left( \text{-3} \right)\text{-3=-18} $ 

 $ \therefore \underset{x\to 3}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{-3} \right) $ .

Therefore,  $ \text{f} $  satisfies continuity at  $ \text{x=-3} $ 

Also, at  $ \text{x=5,}\;f\left( \text{x} \right)\text{=f}\left( \text{5} \right)\text{=5 }\!\!\times\!\!\text{ 5-3=25-3=22} $ .

Taking limit as  $ x\to 5 $  both sides of the function give

 $ \underset{x\to 5}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{x\to 5}{\mathop{\lim }}\,\left( \text{5x-3} \right)\text{=5 }\!\!\times\!\!\text{ 5-3=22} $ 

 $ \therefore \underset{x\to 5}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{5} \right) $ .

Hence,  $ \text{f} $  satisfies continuity at  $ \text{x=5} $ .


2. Examine the continuity of the function  $ \text{f}\left( \text{x} \right)\text{=2}{{\text{x}}^{\text{2}}}\text{-1} $  at  $ \text{x=3} $ .

Ans: The given function is  $ \text{f}\left( \text{x} \right)\text{=2}{{\text{x}}^{\text{2}}}\text{-1} $ .

Now, at  $ \text{x=3, f}\left( \text{3} \right)\text{=2 }\!\!\times\!\!\text{ }{{\text{3}}^{\text{2}}}\text{-1=17} $ .

Taking limit as  $ x\to 3 $  both sides of the function give

 $ \underset{x\to 3}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{x\to 3}{\mathop{\lim }}\,\left( \text{2}{{\text{x}}^{\text{2}}}\text{-1} \right)\text{=2 }\!\!\times\!\!\text{ }{{\text{3}}^{\text{2}}}\text{-1=17} $ 

 $ \therefore \underset{x\to 3}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f=}\left( \text{3} \right) $ .

Hence,  $ \text{f} $  satisfies continuity at  $ \text{x=3} $ .


3. Examine the following functions for continuity.

(a)  $ \text{f}\left( \text{x} \right)\text{=x-5} $ 

Ans: The given function is  $ \text{f}\left( \text{x} \right)\text{=x-5} $ .

It is assured that for every real number  $ \text{k} $ ,  $ \text{f} $  is defined and its value at  $ k $  is  $ \text{k-5} $ . Also, it can be noted that 

 $ \underset{x\to k}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{x\to k}{\mathop{\lim }}\,\text{f}\left( \text{x-5} \right)\text{=k=k-5=f}\left( \text{k} \right) $ .

 $ \therefore \underset{x\to k}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{k} \right) $ 

Hence,  $ \text{f} $  satisfies continuity at every real number and so, it is a continuous function.


(b)  $ \text{f}\left( \text{x} \right)=\dfrac{\text{1}}{\text{x-5}}\text{,x}\ne \text{5} $ 

Ans: The given function is

 $ \text{f}\left( \text{x} \right)\text{=}\dfrac{\text{1}}{\text{x-5}} $ .

Let  $ \text{k}\ne \text{5} $  is any real number, then taking limit as  $ \text{x}\to \text{k} $  both sides of the function give

 $ \underset{x\to k}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{x\to k}{\mathop{\lim }}\,\dfrac{\text{1}}{\text{x-5}}\text{=}\dfrac{\text{1}}{\text{k-5}} $ 

Also,  $ \text{f(k)=}\dfrac{\text{1}}{\text{k-5}} $  , since  $ \text{k}\ne \text{5} $ 

 $ \therefore \underset{x\to k}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{k} \right) $ 

Therefore,  $ f $  satisfies continuity at every point in the domain of  $ f $  and so, it is a continuous function.


(c)  $ \text{f}\left( \text{x} \right)\text{=}\dfrac{{{\text{x}}^{\text{2}}}\text{-25}}{\text{x+5}}\text{,x}\ne -\text{5} $ 

Ans: The given function is

 $ \text{f}\left( \text{x} \right)\text{=}\dfrac{{{\text{x}}^{\text{2}}}\text{-25}}{\text{x+5}}\text{, x}\ne \text{5} $ 

Now let  $ \text{c}\ne \text{-5} $  be any real number, then taking limit as  $ \text{x}\to \text{c} $  on both sides of the function give

 $ \underset{x\to c}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{x\to c}{\mathop{\lim }}\,\dfrac{{{\text{x}}^{\text{2}}}\text{-25}}{\text{x+5}}\text{=}\underset{x\to c}{\mathop{\lim }}\,\dfrac{\left( \text{x+5} \right)\left( \text{x-5} \right)}{\text{x+5}}\text{=}\underset{x\to c}{\mathop{\lim }}\,\left( \text{x-5} \right)\text{=}\left( \text{c-5} \right) $ 

Again,  $ \text{f}\left( \text{c} \right)\text{=}\dfrac{\left( \text{c+5} \right)\left( \text{c-5} \right)}{\text{c+5}}\text{=}\left( \text{c-5} \right) $ , since  $ \text{c}\ne \text{5} $ .

Hence,  $ \text{f} $  satisfies continuity at every point in the domain of  $ \text{f} $  and so it is a continuous function.


(d)  $ \mathbf{f}\left( \mathbf{x} \right)=\left| \text{x-5} \right| $ 

Ans: The given function is  $ \text{f}\left( \text{x} \right)\text{=}\left| \text{x-5} \right|=\left\{ \begin{align}& \text{5-x, if x}<\text{5} \\ & \text{x-5, if x}\,\ge \text{5} \\ \end{align} \right. $ .

Note that,  $ \text{f} $  is defined at all points in the real line. So, let assume  $ \text{c} $  be a point on a real line. 

Then, we have  $ \text{c}<\text{5} $  or  $ \text{c}=\text{5} $  or  $ \text{c}>\text{5} $ .

Now, let's discuss these three cases one by one.

Case (i).  $ \text{c}<\text{5} $ 

Then, the function becomes  $ \text{f}\left( \text{c} \right)\text{=5-c} $ .

Now,  $ \underset{x\to c}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{x\to c}{\mathop{\lim }}\,\left( \text{5-x} \right)\text{=5-c} $ .

 $ \therefore \underset{x\to c}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right) $ .

Therefore,  $ \text{f} $  is continuous at all real numbers which are less than  $ 5 $ . 

Case (ii).  $ \text{c=5} $ 

Then,  $ \text{f}\left( \text{c} \right)\text{=f}\left( \text{5} \right)\text{=}\left( \text{5-5} \right)\text{=0} $ .

Now,

 $ \underset{x\to {{5}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)=\underset{x\to 5}{\mathop{\lim }}\,\left( \text{5-x} \right)\text{=}\left( \text{5-5} \right)\text{=0} $  and

 $ \underset{x\to {{5}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{x\to 5}{\mathop{\lim }}\,\left( \text{x-5} \right)\text{=0} $ .

Therefore, we have

 $ \underset{x\to {{c}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{x\to {{c}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right) $ .

Thus,  $ \text{f} $  satisfies continuity at  $ \text{x=5} $ , and so  $ \text{f} $  is continuous at  $ \text{x=5} $ .

Case (iii).  $ \text{c}>\text{5} $ 

Then we have,  $ \text{f}\left( \text{c} \right)\text{=f}\left( \text{5} \right)\text{=c-5} $ .

Now,

 $ \underset{x\to c}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{x\to c}{\mathop{\lim }}\,\left( \text{x-5} \right)\text{=c-5} $ .

Therefore,

 $ \underset{x\to c}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right) $ .

So,  $ \text{f} $  is continuous at all real numbers that are greater than  $ 5 $ .

Thus,  $ \text{f} $  satisfies continuity at every real number and hence, it is a continuous function.


4. Prove that the function  $ \text{f}\left( \text{x} \right)={{\text{x}}^{\text{n}}} $   is continuous at  $ \text{x=n} $ , where  $ \mathbf{n} $  is a positive integer.

Ans: The given function is   $ \text{f}\left( \text{x} \right)\text{=}{{\text{x}}^{\text{n}}} $ .

We noticed that the function  $ \text{f} $  is defined at all positive integers  $ \text{n} $  and also its value at  $ \text{x=n} $  is  $ {{\text{n}}^{\text{n}}} $ .

Therefore,  $ \underset{x\to n}{\mathop{\lim }}\,\text{f}\left( \text{n} \right)=\underset{x\to n}{\mathop{\lim }}\,\text{f}\left( {{\text{x}}^{\text{n}}} \right)\text{=}{{\text{n}}^{\text{n}}} $ .

So,  $ \underset{\text{x}\to \text{n}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{n} \right) $ .

Thus, the function  $ \text{f}\left( \text{x} \right)\mathbf{=}{{\text{x}}^{\text{n}}} $   is continuous at  $ \text{x=n} $  , where  $ \text{n} $  is a positive integer.


5. Is the function  $ \mathbf{f} $  defined by 

$ \text{f}\left( \text{x} \right)\text{=}\left\{ \begin{align}& \text{x,}\,\text{if x}\le \text{1} \\ & \text{5, }\,\text{if x}\,>\text{1} \\ \end{align} \right. $ 

continuous at  $ \mathbf{x}=\mathbf{0} $ ,  $ \mathbf{x}=\mathbf{1} $ ?  At  $ \mathbf{x}=\mathbf{2} $ ?

Ans: The given function is$\text{f}\left( \text{x} \right)\text{=}\left\{ \begin{align}& \text{x,}\,\text{ if x}\le \text{1} \\ & \text{5, }\,\text{if x}\,>\text{1} \\ \end{align} \right.$

It is obvious that the function  $ \text{f} $  is defined at  $ \text{x=}0 $  and its value at  $ \text{x=}0 $  is  $ 0 $ .

Now,  $ \underset{x\to 0}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{x\to 0}{\mathop{\lim }}\,\text{x=0} $ .

So,  $ \underset{x\to 0}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{0} \right) $ .

Hence, the function  $ \text{f} $  satisfies continuity at  $ \text{x=0} $ .


It can be observed that  $ \text{f} $  is defined at  $ \text{x=1} $  and its value at this point is  $ \text{1} $ .

Now, the left-hand limit of the function  $ \text{f} $  at  $ \text{x=1} $  is

 $ \underset{x\to {{1}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{x\to {{1}^{-}}}{\mathop{\lim }}\,\text{x=1} $ .

Also, the right-hand limit of the function  $ \text{f} $  at  $ \text{x=1} $  is

 $ \underset{x\to {{1}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{5} \right) $ 

Therefore,  $ \underset{x\to {{1}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\ne \underset{x\to {{1}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right) $ .

Thus,  $ \text{f} $  is not continuous at  $ \text{x=1} $ 


It can be found that  $ \text{f} $  is defined at  $ \text{x=2} $  and its value at this point is  $ 5 $ .

That is,  $ \underset{x\to 2}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{x\to 2}{\mathop{\lim }}\,\text{f}\left( \text{5} \right)\text{=5} $ .

Therefore,  $ \underset{x\to 2}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{2} \right) $ 

Hence,  $ \text{f} $  satisfies continuity at  $ \text{x=2} $ .


Find all points of discontinuity of   $ \mathbf{f} $ , where  $ \mathbf{f} $  is defined by

6.  $ \text{f}\left( \text{x} \right)\text{=}\left\{ \begin{align}& \text{2x+3, if x}\le \text{2} \\ & \text{2x-3, if x}\,>\,\text{2} \\ \end{align} \right. $ .

Ans: The given function is $ \text{f}\left( \text{x} \right)\text{=}\left\{ \begin{align} & \text{2x+3, if x}\le \text{2} \\ & \text{2x-3, if x}\,>\,\text{2} \\ \end{align} \right. $

It can be observed that the function  $ \text{f} $  is defined at all the points in the real line.

Let consider  $ \text{c} $  be a point on the real line. Then, three cases may arise.

I. $ \text{c}\,<\,\text{2} $ 

II. $ \text{c}\,>\,\text{2} $ 

III. $ \text{c}=\text{2} $ 

Case (i). When  $ \text{c}\,<\text{2} $ 

Then, we have  $ \underset{x\to c}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{x\to \infty }{\mathop{\lim }}\,\left( \text{2x+3} \right)\text{=2c+3} $ .

Therefore,

 $ \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right) $ .

Hence,  $ \text{f} $  attains continuity at all points  $ \text{x} $ , where  $ \text{x}<\text{2} $ .

Case (ii). When  $ \text{c}>\text{2} $ 

Then, we have   $ \text{f}\left( \text{c} \right)\text{=2c-3} $ .

So,

 $ \underset{\text{x}\to c}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to \infty }{\mathop{\lim }}\,\left( \text{2x-3} \right)\text{=2c-3} $ .

Therefore,  $ \underset{\text{x}\to c}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right) $ .

Hence,  $ \text{f} $  satisfies continuity at all points  $ \text{x} $  , where  $ \text{x}>\text{2} $ .

Case(iii). When  $ \text{c=2} $ 

Then, the left-hand limit of the function  $ \text{f} $  at  $ \text{x=2} $  is

 $ \underset{x\to {{2}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{2}^{-}}}{\mathop{\lim }}\,\left( \text{2x+3} \right)\text{=2 }\!\!\times\!\!\text{ 2+3=7} $  and 

the right-hand limit of the function  $ \text{f} $  at  $ \text{x=2} $  is,

 $ \underset{\text{x}\to {{2}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to 2+}{\mathop{\lim }}\,\left( \text{2x+3} \right)\text{=2 }\!\!\times\!\!\text{ 2-3=1} $ .

Thus, at  $ \text{x=2} $ ,  $ \underset{x\to {{2}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\ne \underset{\text{x}\to {{2}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right) $ .

So, the function   $ \text{f} $  does not satisfy continuity at  $ \text{x=2} $ .

Hence,  $ \text{x=2} $  is the only point of discontinuity of the function  $ \text{f}\left( \text{x} \right) $ .


7.  $ \mathbf{f}\left( \mathbf{x} \right)\mathbf{=}\left\{ \begin{align}& \left| \mathbf{x} \right|\mathbf{+3, if x}\le \mathbf{-3} \\ & \mathbf{-2x, if -3}\,<\,\mathbf{x}\,<\,\mathbf{3} \\ & \mathbf{6x+2, if x}\ge \mathbf{3} \\ \end{align} \right. $ 

Ans: The given function is$ \text{f}\left( \text{x} \right)\text{=}\left\{ \begin{align} & \left| \text{x} \right|\text{+3, if x}\le \text{-3} \\ & \text{-2x, if -3}<\text{x}<\text{3} \\ & \text{6x+2, if x}\ge \text{3} \\ \end{align} \right. $

Observe that,  $ \text{f} $  is defined at all the points in the real line.

Now, let assume  $ \text{c} $  as a point on the real line.

Then five cases may arise. Either  $ \text{c}<\text{-3} $ , or  $ \text{c}=\text{-3} $  or  $ \text{-3}<\text{c}<\text{3} $ , or  $ \text{c}=\text{3} $ , or  $ \text{c}>\text{3} $ .

Let's discuss the five cases one by one.

Case I. When  $ \text{c}<\text{-3} $ 

Then,  $ \text{f}\left( \text{c} \right)\text{=-c+3} $  and

 $ \underset{\text{x}\to c}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{x\to c}{\mathop{\lim }}\,\left( \text{-x+3} \right)\text{=-c+3} $ .

Therefore,  $ \underset{\text{x}\to c}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right) $ .

Hence,  $ \text{f} $  satisfies continuity at all points  $ \text{x} $ , where  $ \text{x}<\text{-3} $ .

Case II. When  $ \text{c=-3} $ 

Then,  $ \text{f}\left( \text{-3} \right)\text{=-}\left( \text{-3} \right)\text{+3=6} $ .

Also, the left-hand limit

 $ \underset{\text{x}\to -{{3}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to -{{3}^{-}}}{\mathop{\lim }}\,\left( \text{-x+3} \right)\text{=-}\left( \text{-3} \right)\text{+3=6} $ .

and the right-hand limit

 $ \underset{\text{x}\to -{{3}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)=\underset{\text{x}\to -{{3}^{+}}}{\mathop{\lim }}\,\text{f}\left( -2\text{x} \right)\text{=-2}\times \left( \text{-3} \right)\text{=6} $ .

Therefore,  $ \underset{\text{x}\to -3}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{-3} \right) $ .

Hence,  $ \text{f} $  satisfies continuity at  $ \text{x=-3} $ .

Case III. When   $ \text{-3}<\text{c}<\text{3} $ .

Then,  $ f\left( c \right)=-2c $  and also

  $ \underset{x\to c}{\mathop{\lim }}\,\text{f}\left( x \right)\text{=}\underset{x\to c}{\mathop{\lim }}\,\left( -2x \right)=-2c $ .

Therefore,  $ \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right) $ .

Hence,  $ \text{f} $  satisfies continuity at  $ \text{x} $ , where  $ \text{-3}<\text{x}<\text{3} $ .

Case IV. When  $ \text{c=3} $ 

Then, the left-hand limit of the function  $ \text{f} $  at  $ \text{x=3} $  is

 $ \underset{\text{x}\to {{3}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{3}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{-2x} \right)\text{=-2 }\!\!\times\!\!\text{ 3=-6} $  and

the right-hand limit of the function  $ \text{f} $  at  $ \text{x=3} $  is

 $ \underset{\text{x}\to {{3}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{3}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{6x+2} \right)\text{=6 }\!\!\times\!\!\text{ 3+2=20} $ .

Thus, at  $ \text{x=3} $ ,  $ \underset{\text{x}\to {{3}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\ne \underset{\text{x}\to {{3}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right) $ .

Hence,  $ \text{f} $  does not satisfy continuity at  $ \text{x=3} $ .

Case V. When  $ \text{c}>\text{3} $ .

 Then  $ \text{f}\left( \text{c} \right)\text{=6c+2} $  and also

  $ \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\left( \text{6x+2} \right)\text{=6c+2} $ .

Therefore,  $ \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right) $ .

So,  $ \text{f} $  satisfies continuity at all points  $ \text{x} $ , when  $ \text{x}>\text{3} $ .

Thus,  $ \text{x=3} $  is the only point of discontinuity of the function  $ \text{f} $ .


8.  $ \text{f}\left( \text{x} \right)\text{=}\left\{ \begin{align}& \dfrac{\left| \text{x} \right|}{\text{x}}\text{, if x}\ne \text{0} \\ & \text{0,   if x}=\text{0} \\ \end{align} \right. $ .

Ans: 

The given function is  $ \text{f}\left( \text{x} \right)\text{=}\left\{ \begin{align} & \dfrac{\left| \text{x} \right|}{\text{x}}\text{, if x}\ne \text{0} \\ & \text{0, if x}=\text{0} \\ \end{align} \right. $ .

Now,  $ \text{f}\left( \text{x} \right) $  can be rewritten as

 $ \text{f}\left( \text{x} \right)\text{=}\left\{ \begin{align} & \dfrac{\left| \text{x} \right|}{\text{x}}=\dfrac{-\text{x}}{\text{x}}=-\text{1 if x}<\text{0} \\ & \text{0, if x}=\text{0} \\ & \dfrac{\left| \text{x} \right|}{\text{x}}=\dfrac{\text{x}}{\text{x}}=\text{1 if x}>\text{0} \\ \end{align} \right. $

It can be noted that the function  $ \text{f} $  is defined at all points of the real line.

Now, let assume  $ \text{c} $  as a point on the real line.

Then three cases may arise, either  $ \text{c}<\text{0} $ , or  $ \text{c}=\text{0} $ , or  $ \text{c}>\text{0} $ .

Let's discuss three cases one by one.

Case I. When  $ \text{c}<\text{0} $ .

Then,  $ \text{f}\left( \text{c} \right)\text{=-1} $  and

 $ \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\left( \text{-1} \right)\text{=-1} $ .

Therefore,  $ \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right) $ .

Hence,  $ \text{f} $  satisfies continuity at all the points  $ \text{x} $  where  $ \text{x}<\text{0} $ .

Case II. When  $ \text{c=0} $ .

Then, the left-hand limit of the function  $ \text{f} $  at  $ \text{x=0} $  is

 $ \underset{\text{x}\to {{0}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{0}^{-}}}{\mathop{\lim }}\,\left( \text{-1} \right)\text{=-1} $  and

the right-hand limit of the function  $ \text{f} $  at  $ \text{x=0} $  is

 $ \underset{\text{x}\to 0+}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{0}^{+}}}{\mathop{\lim }}\,\left( \text{1} \right)\text{=1} $ .

At  $ \text{x=0} $ ,  $ \underset{\text{x}\to {{0}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\ne \underset{\text{x}\to {{0}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right) $ .

Hence, the function  $ \text{f} $  does not satisfy continuity at  $ \text{x=0} $ .

Case III. When  $ \text{c}>\text{0} $ .

Then  $ \text{f}\left( \text{c} \right)\text{=1} $  and also

 $ \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\left( \text{1} \right)\text{=1} $ .

Therefore,  $ \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right) $ .

So, the function  $ \text{f} $  is continuous at all the points  $ \text{x} $ , for  $ \text{x}>\text{0} $ .

Thus,  $ \text{x=0} $  is the only point of discontinuity for the function  $ \text{f} $ .


9.  $ \text{f}\left( \text{x} \right)\text{=}\left\{ \begin{align}& \dfrac{\text{x}}{\left| \text{x} \right|}\text{, if x}<\text{0} \\ & \text{-1, if x}\ge \text{0} \\ \end{align} \right. $ 

Ans: The given function is $ \text{f}\left( \text{x} \right)\text{=}\left\{ \begin{align} & \dfrac{\text{x}}{\left| \text{x} \right|}\text{, if x}\,<\,\text{0} \\ & \text{-1, if x}\ge \text{0} \\ \end{align} \right. $

Now, we know that, if  $ \text{x}<\text{0} $ , then   $ \left| \text{x} \right|\text{=-x} $ .

Therefore, the  $ \text{f}\left( \text{x} \right) $  can be written as

 $ \text{f}\left( \text{x} \right)\text{=}\left\{ \begin{align} & \dfrac{x}{\left| \text{x} \right|}=\dfrac{\text{x}}{\text{-x}}=\text{-1 if x}<\text{0} \\ & -\text{1, if x}=\text{0} \\ & -\text{1, if x}>\text{0} \\ \end{align} \right. $

 $ \Rightarrow \text{f}\left( \text{x} \right)\text{=-1} $  for all positive real numbers.

Now, let's assume  $ c $  as any real number.

Then three cases may arise, either  $ \text{c}<\text{0} $ , or  $ \text{c}=\text{0} $ , or  $ \text{c}>\text{0} $ .

Let's discuss three cases one by one.

Case I. When  $ \text{c}<\text{0} $ .

Then,  $ \text{f}\left( \text{c} \right)\text{=-1} $  and

 $ \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\left( \text{-1} \right)\text{=-1} $ .

Therefore,  $ \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right) $ .

Hence,  $ \text{f} $  satisfies continuity at all the points  $ \text{x} $  where  $ \text{x}<\text{0} $ .

Case II. When  $ \text{c=0} $ .

Then, the left-hand limit of the function  $ \text{f} $  at  $ \text{x=0} $  is

 $ \underset{\text{x}\to {{0}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{0}^{-}}}{\mathop{\lim }}\,\left( \text{-1} \right)\text{=-1} $  and

the right-hand limit of the function  $ \text{f} $  at  $ \text{x=0} $  is

 $ \underset{\text{x}\to 0+}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{0}^{+}}}{\mathop{\lim }}\,\left( \text{-1} \right)\text{=-1} $ .

At  $ \text{x=0} $ ,  $ \underset{\text{x}\to {{0}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)=\underset{\text{x}\to {{0}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right) $ .

Hence, the function  $ \text{f} $  satisfies continuity at  $ \text{x=0} $ .

Case III. When  $ \text{c}>\text{0} $ .

Then  $ \text{f}\left( \text{c} \right)\text{=-1} $  and also

 $ \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\left( \text{-1} \right)\text{=-1} $ .

Therefore,  $ \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right) $ .

So, the function  $ \text{f} $  is continuous at all the points  $ \text{x} $ , for  $ \text{x}>\text{0} $ .

Then, we have  $ \underset{\text{x}\to c}{\mathop{\lim }}\,\text{f(x)}=\underset{\text{x}\to c}{\mathop{\lim }}\,(-1)=-1 $   and  $ \text{f}\left( \text{c} \right)\text{=-1}=\underset{\text{x}\to e}{\mathop{\lim }}\,\text{f}\left( \text{x} \right) $ .

Therefore, the function  $ \text{f}\left( \text{x} \right) $  is a continuous function.

Thus, there does not exist any point of discontinuity.


10.$ \text{f(x)=}\left\{ \begin{align} & \text{x+1, if x}\ge \text{1} \\ & {{\text{x}}^{\text{2}}}\text{+1, if x}\,<\,\text{1} \\ \end{align} \right. $

Ans: The given function is $ \text{f(x)=}\left\{ \begin{align} & \text{x+1, if x}\ge \text{1} \\ & {{\text{x}}^{\text{2}}}\text{+1, if x}\,<\,\text{1} \\ \end{align} \right. $ .

Note that,  $ \text{f}\left( \text{x} \right) $  is defined at all the points of the real line.

Now, let's assume  $ c $  as a point on the real line.

Then three cases may arise, either either  $ \text{c}<\text{1} $ , or  $ \text{c}=\text{1} $ , or  $ \text{c}>\text{1} $ .

Let's discuss the three cases one by one.

Case I. When  $ \text{c}<\text{1} $ .

 Then,  $ \text{f}\left( \text{c} \right)\text{=}{{\text{c}}^{\text{2}}}\text{+1} $  and also

 $ \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)=\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( {{\text{x}}^{\text{2}}}\text{+1} \right)\text{=}{{\text{c}}^{\text{2}}}\text{+1} $ .

Therefore,  $ \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right) $ .

Hence,  $ \text{f} $  satisfies continuity at all the points  $ \text{x} $ , where  $ \text{x}<\text{1} $ .

Case II. When  $ \text{c=1} $ .

Then, we have  $ \text{f}\left( \text{c} \right)\text{=f}\left( \text{1} \right)\text{=1+1=2} $ .

Now, the left-hand limit of  $ \text{f} $  at  $ \text{x=1} $  is

 $ \underset{\text{x}\to {{\text{1}}^{-}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{\text{1}}^{-}}}{\mathop{\lim }}\,\text{(}{{\text{x}}^{\text{2}}}\text{+1)=}{{\text{1}}^{\text{2}}}\text{+1=2} $  and the right-hand limit of  $ \text{f} $  at  $ \text{x=1} $  is,  $ \underset{\text{x}\to {{1}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{1}^{+}}}{\mathop{\lim }}\,\left( \text{x+1} \right)\text{=1+1=2} $ .

Therefore,  $ \underset{\text{x}\to 1}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right) $ .

Hence,  $ \text{f} $  satisfies continuity at  $ \text{x=1} $ .

Case III. When  $ \text{c}>\text{1} $ .

Then, we have   $ \text{f}\left( \text{c} \right)\text{=c+1} $  and 

 $ \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\left( \text{x+1} \right)\text{=c+1} $ .

Therefore,

 $ \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right) $ .

So,  $ \text{f} $  satisfies continuity at all the points  $ \text{x} $ , where  $ \text{x}>\text{1} $ .

Hence, there does not exist any discontinuity points.


11. $ \mathbf{f}\left( \mathbf{x} \right)=\left\{ \begin{align} & {{\mathbf{x}}^{\mathbf{3}}}-\mathbf{3},\mathbf{if}\,\,\mathbf{x}\le \mathbf{2} \\ & {{\mathbf{x}}^{\mathbf{2}}}+\mathbf{1},\mathbf{if}\,\,\mathbf{x}>\mathbf{2} \\ \end{align} \right. $

Ans: The given function is  $ \text{f}\left( \text{x} \right)\text{=}\left\{ \begin{align}& {{\text{x}}^{\text{3}}}\text{-3, if x}\le \text{2} \\ & {{\text{x}}^{\text{2}}}\text{+1, if x}>\text{2} \\ \end{align} \right. $ 

Observe that, the function  $ \text{f} $  is defined at all points in the real line.

Now, let assume  $ \text{c} $  as a point on the real line.

Case I. When  $ \text{c}<\text{2} $ .

Then, we have   $ \text{f}\left( \text{c} \right)\text{=}{{\text{c}}^{\text{3}}}\text{-3} $  and also  $ \underset{\text{x}\to\text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\left( {{\text{x}}^{\text{3}}}\text{-3} \right)\text{=}{{\text{c}}^{\text{3}}}\text{-3} $ .

Therefore, the function  $ \text{f} $  attains continuity at all the points  $ \text{x} $ , where  $ \text{x}<\text{2} $ .

Case II. When  $ \text{c=2} $ .

Then, we have  $ \text{f}\left( \text{c} \right)\text{=f}\left( \text{2} \right)\text{=}{{\text{2}}^{\text{3}}}\text{-3=5} $ .

Now the left-hand limit of the function is

 $ \underset{\text{x}\to {{2}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{2}^{-}}}{\mathop{\lim }}\,\left( {{\text{x}}^{3}}-3 \right)\text{=}{{\text{2}}^{3}}\text{-3=5} $  and the right-hand limit is

 $ \underset{\text{x}\to {{2}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{2}^{+}}}{\mathop{\lim }}\,\left( {{\text{x}}^{\text{2}}}\text{+1} \right)\text{=}{{\text{2}}^{\text{2}}}\text{+1=5} $ .

Therefore,  $ \underset{\text{x}\to 2}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{2} \right) $ .

Hence, the function  $ \text{f} $  is continuous at  $ \text{x=2} $ .

Case III. When  $ \text{c}>\text{2} $ .

Then,  $ \text{f}\left( \text{c} \right)\text{=}{{\text{c}}^{2}}+1 $  and

 $ \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\left( {{\text{x}}^{\text{2}}}\text{+1} \right)\text{=}{{\text{c}}^{\text{2}}}\text{+1} $ .

Therefore,  $ \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right) $ .

So,  $ \text{f} $  attains continuity at all the points  $ \text{x} $ , where  $ \text{x}>\text{2} $ .

Thus, the function  $ \text{f} $  is continuous at all the points on the real line.

Hence,  $ \text{f} $  does not have any point of discontinuity.


12.  $ \text{f}\left( \text{x} \right)=\left\{ \begin{align}& {{\text{x}}^{\text{10}}}\text{-1, if x}\le \text{1} \\ & {{\text{x}}^{\text{2}}}\text{,       if x}\,>\,\text{1} \\ \end{align} \right. $ . 

Ans: The given function is  $ \text{f}\left( \text{x} \right)=\left\{ \begin{align} & {{\text{x}}^{\text{10}}}\text{-1, if x}\le \text{1} \\ & {{\text{x}}^{\text{2}}}\text{, if x}\,>\text{1} \\ \end{align} \right. $

Observe that, the function  $ \text{f} $  is defined at every point of the real line.

Now, let assume  $ \text{c} $  as a point on the real number line.

Case I. When  $ \text{c}<\text{1} $ .

Then  $ \text{f}\left( \text{c} \right)\text{=}{{\text{c}}^{\text{10}}}\text{-1} $ .

Also,   $ \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\left( {{\text{x}}^{\text{10}}}\text{-1} \right)\text{=}{{\text{c}}^{\text{10}}}\text{-1} $ 

Therefore,  $ \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right) $ .

Hence, the function  $ \text{f} $  attains continuity at every point  $ \text{x} $ , for  $ \text{x}<\text{1} $ .

Case II. When  $ \text{c=1} $ .

Then the left-hand limit of the function  $ \text{f}\left( \text{x} \right) $  at  $ \text{x=1} $  is

 $ \underset{\text{x}\to {{1}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)=\underset{\text{x}\to {{1}^{-}}}{\mathop{\lim }}\,\left( {{\text{x}}^{10}}-1 \right)={{1}^{10}}-1=1-1=0 $  and

the right-hand limit of the function  $ \text{f} $  at  $ \text{x=1} $  is

 $ \underset{\text{x}\to {{1}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{1}^{+}}}{\mathop{\lim }}\,\left( {{\text{x}}^{\text{2}}} \right)\text{=}{{\text{1}}^{\text{2}}}\text{=1} $ .

So, we can notice that,  $ \underset{\text{x}\to {{\text{1}}^{\text{-}}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\ne \underset{\text{x}\to {{\text{1}}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right) $ .

Hence, the function  $ \text{f} $  does not satisfy continuity at  $ \text{x=1} $ .

Case III. When  $ \text{c}>\text{1} $ .

Then,  $ \text{f}\left( \text{c} \right)\text{=}{{\text{c}}^{\text{2}}} $ .

Also,  $ \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\left( {{\text{x}}^{\text{2}}} \right)\text{=}{{\text{c}}^{\text{2}}} $ .

Therefore,  $ \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right) $ .

Thus, the function  $ \text{f} $  attains continuity at every point  $ \text{x} $ , for  $ \text{x}>\text{1} $ .

Hence, we can conclude that  $ \text{x=1} $  is the only point of discontinuity for the function  $ \text{f} $ .


13. Is the function defined by  $ \text{f}\left( \text{x} \right)\text{=}\left\{ \begin{align} & \text{x+5, if x}\le \text{1} \\ & \text{x-5, if x}\,>\,\text{1} \\ \end{align} \right. $  a continuous function? 

Ans: The given function is  $ \text{f}\left( \text{x} \right)\text{=}\left\{ \begin{align}& \text{x+5, if x}\le \text{1} \\ & \text{x-5, if x}\,>\text{1} \\ \end{align} \right. $  .

It can be noted that the function  $ \text{f} $  is defined at every point on the real line. 

Now, let assume  $ \text{c} $  as a point on the real line.

Case I. When  $ \text{c}<\text{1} $ .

Then,  $ \text{f}\left( \text{c} \right)\text{=c+5} $ .

Also,  $ \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\left( \text{x+5} \right)\text{=c+5} $ .

Hence,  $ \text{f} $  satisfies continuity at every point  $ \text{x} $ , for  $ \text{x}<\text{1} $ .

Case II. When  $ \text{c=1} $ .

Then,  $ \text{f}\left( \text{1} \right)\text{=1+5=6} $ .

Now, the left-hand limit of the function  $ \text{f} $  at  $ \text{x=1} $  is

 $ \underset{\text{x}\to {{1}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{1}^{-}}}{\mathop{\lim }}\,\left( \text{x+5} \right)\text{=1+5=6} $  and the right-hand limit of the function at  $ \text{x=1} $  is  $ \underset{\text{x}\to {{1}^{+}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{1}^{+}}}{\mathop{\lim }}\,\text{(x-5)=1-5=4} $ .

Thus, it is seen that,  $ \underset{\text{x}\to {{1}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\ne \underset{\text{x}\to {{1}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right) $ .

Hence,  $ \text{f} $  does not attain continuity at  $ \text{x=1} $ .

Case III. When  $ \text{c}>\text{1} $ .

Then  $ \text{f}\left( \text{c} \right)\text{=c-5} $ .

Also,  $ \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\left( \text{x-5} \right)\text{=c-5} $ .

Therefore,  $ \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right) $ .

Thus, the function  $ \text{f} $  is continuous at every point  $ \text{x} $ , for  $ \text{x}>\text{1} $ .

Hence, we can conclude that  $ \text{x=1} $  is the only point of discontinuity for the function  $ \text{f} $ .


14. Discuss the continuity of the function  $ \mathbf{f} $ , where  $ \mathbf{f} $  is defined by 

 $ \text{f}\left( \text{x} \right)\text{=}\left\{ \begin{align}& \text{3, if 0}\le \text{x}\le \text{1} \\ & \text{4, if 1}\,<\,\text{x}\,<\,\text{3} \\ & \text{5, if 3}\le \text{x}\le \text{10} \\ \end{align} \right. $ .

Ans: The given function is $ \text{f}\left( \text{x} \right)\text{=}\left\{ \begin{align} & \text{3, if 0}\le \text{x}\le \text{1} \\ & \text{4, if 1}\,<\,\text{x}\,<\,\text{3} \\ & \text{5, if 3}\le \text{x}\le \text{10} \\ \end{align} \right. $ .

Therefore,  $ \text{f} $  is defined in the interval  $ \text{ }\!\![\!\!\text{ 0,10 }\!\!]\!\!\text{ } $ .

Now let's assume  $ \text{c} $  as a point in the interval  $ \text{ }\!\![\!\!\text{ 0,10 }\!\!]\!\!\text{ } $ .

Then there may arise five cases.

Case I. When  $ \text{0}\le \text{c}<\text{1} $ .

Then  $ \text{f}\left( \text{c} \right)\text{=3} $ .

Also,  $ \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\left( \text{3} \right)\text{=3} $ .

Therefore,  $ \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right) $ .

Hence, the function  $ \text{f} $  attains continuity at the interval  $ \text{ }\!\![\!\!\text{ 0,1 }\!\!]\!\!\text{ } $ .

Case II. When  $ \text{c=1} $ .

Then  $ \text{f}\left( \text{3} \right)\text{=3} $ .

Also, the left-hand-limit of the function at  $ \text{x=1} $  is

 $ \underset{\text{x}\to {{\text{1}}^{\text{-}}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{\text{1}}^{\text{-}}}}{\mathop{\lim }}\,\left( \text{3} \right)\text{=3} $  and the right-hand-limit of the function at  $ \text{x=1} $  is

 $ \underset{\text{x}\to {{1}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{1}^{+}}}{\mathop{\lim }}\,\left( \text{4} \right)\text{=4} $ .

Thus, it is noticed that  $ \underset{\text{x}\to {{\text{1}}^{\text{-}}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\ne \underset{\text{x}\to {{\text{1}}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right) $ .

Hence, the function  $ \text{f} $  does not satisfy continuity at  $ \text{x=1} $ .

Case III. When  $ \text{1}<\text{c}<\text{3} $ .

Then  $ \text{f}\left( \text{c} \right)\text{=4} $ .

Also,  $ \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\left( \text{4} \right)\text{=4} $ .

Thus,  $ \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right) $ .

Hence, the function  $ \text{f} $  attains continuity at every point in the interval  $ \text{ }\!\![\!\!\text{ 1,3 }\!\!]\!\!\text{ } $ .

Case IV. When  $ \text{c=3} $ .

Then  $ \text{f}\left( \text{c} \right)\text{=5} $ .

Now, the left-hand-limit of the function  $ \text{f} $  at  $ \text{x=3} $  is

 $ \underset{\text{x}\to {{\text{3}}^{\text{-}}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{\text{3}}^{\text{-}}}}{\mathop{\lim }}\,\left( \text{4} \right)\text{=4} $  and the right-hand-limit of the function  $ \text{f} $  at  $ \text{x=3} $  is

 $ \underset{\text{x}\to {{\text{3}}^{\text{+}}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{3}^{+}}}{\mathop{\lim }}\,\left( \text{5} \right)\text{=5} $ .

Therefore, it is noted that  $ \underset{\text{x}\to {{\text{3}}^{\text{-}}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\ne \underset{\text{x}\to {{\text{3}}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right) $  .

Hence, the function  $ \text{f} $  is not continuous at  $ \text{x=3} $ .

Case V. When  $ \text{3}<\text{c}\le \text{10} $ .

Then  $ \text{f}\left( \text{c} \right)\text{=5} $ .

Also,   $ \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\left( \text{5} \right)\text{=5} $ .

Therefore,  $ \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right) $ .

So, the function  $ \text{f} $  attains continuity at every point in the interval  $ \text{ }\!\![\!\!\text{ 3,10 }\!\!]\!\!\text{ } $ .

Hence, the function  $ \text{f} $  is not continuous at  $ \text{x=1} $  and  $ \text{x=3} $ .


15.  $ \text{f}\left( \text{x} \right)\text{=}\left\{ \begin{align}& \text{2x, if x}\,<\,\text{0} \\ & \text{0, }\,\,\,\text{if 0}\le \text{x}\le \text{1} \\ & \text{4x, if x}\,>\,\text{1} \\ \end{align} \right. $  .

 Ans:  The given function is 

 $ \text{f}\left( \text{x} \right)\text{=}\left\{ \begin{align} & \text{2x, if x}<\text{0} \\ & \text{0, if 0}\le \text{x}\le \text{1} \\ & \text{4x, if x}>\text{1} \\ \end{align} \right. $ 

Now, let consider  $ \text{c} $  be a point on the real number line. 

Then, five cases may arrive.

Case I. When  $ \text{c}<\text{0} $ .

Then,  $ \text{f}\left( \text{c} \right)\text{=2c} $ .

Also,   $ \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\left( \text{2x} \right)\text{=2c} $ .

Therefore,  $ \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{c} \right) $ .

Hence, the function  $ \text{f} $  attains continuity at every point  $ \text{x} $  whenever  $ \text{x}<\text{0} $ .

Case II. When  $ \text{c = 0} $ .

Then,  $ \text{f}\left( \text{c} \right)\text{=f}\left( \text{0} \right)\text{=0} $ .

Now, the left-hand-limit of the function  $ \text{f} $ at  $ \text{x = 0} $  is

 $ \underset{\text{x}\to {{0}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{0}^{-}}}{\mathop{\lim }}\,\left( \text{2x} \right)=0 $  and the right-hand limit of the function  $ \text{f} $ at  $ \text{x = 0} $  is, 

 $ \underset{\text{x}\to {{0}^{+}}}{\mathop{\lim }}\,\left( \text{x} \right)\text{=}\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\left( \text{0} \right)\text{=0} $ .

Therefore,  $ \underset{\text{x}\to 0}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=f}\left( \text{0} \right) $ .

Thus, the function  $ \text{f} $  attains continuity at  $ \text{x = 0} $ .

Case III. When  $ \text{0}<\text{c}<\text{1} $  

Then,  $ \text{f(x)=0} $ .

Also,  $ \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{(0)=0} $ .

Therefore,  $ \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f(x)=f(c)} $ .

Hence,  $ \text{f} $  attains continuity at every point in the interval  $ \text{(0,1)} $ .

Case IV. When  $ \text{c =1} $ .

Then,  $ \text{f(c)=f(1)=0} $ .

Now, the left-hand-limit at  $ \text{x = 1} $  is

 $ \underset{\text{x}\to {{\text{1}}^{\text{-}}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{\text{1}}^{\text{-}}}}{\mathop{\lim }}\,\text{(0)=0} $  and the right-hand-limit at  $ \text{x = 1} $  is

 $ \underset{\text{x}\to {{1}^{+}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{1}^{+}}}{\mathop{\lim }}\,\text{(4x)=4 }\!\!\times\!\!\text{ 1=4} $ .

Thus, it is noticed that,  $ \underset{\text{x}\to {{\text{1}}^{\text{-}}}}{\mathop{\lim }}\,\text{f(x)}\ne \underset{\text{x}\to {{\text{1}}^{+}}}{\mathop{\lim }}\,\text{f(x)} $ .

Hence, the function  $ \text{f} $  is not continuous at  $ \text{x = 1} $ .

Case V. When  $ \text{c}>\text{1} $ .

Then,  $ \text{f(c)=f(1)=0} $ .

Also,  $ \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{(4x)=4c} $ 

Therefore,  $ \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f(x)=f(c)} $ .

So, the function  $ \text{f} $  attains continuity at every point  $ \text{x} $ , for  $ \text{c}>\text{1} $ .

Hence, the function  $ \text{f} $  is discontinuous only at  $ \text{x = 1} $ .


16.  $ \text{f(x)=}\left\{ \begin{align} & \text{-2, if x}\le \text{-1} \\ & \text{2x, if -1}\,<\,\text{x}\le \text{1} \\ & \text{2, }\,\,\,\,\text{if x}\,>\,\text{1} \\ \end{align} \right. $ .

Ans:  The given function is  $ \text{f(x)=}\left\{ \begin{align} & \text{-2, if x}\le \text{-1} \\ & \text{2x, if -1}<\text{x}\le \text{1} \\ & \text{2, }\,\,\,\,\text{if x}>\text{1} \\ \end{align} \right. $

Note that,  $ \text{f} $  is defined at every point in the interval  $ \left[ -1,\infty  \right) $ .

Now, let assume  $ \text{c} $  is a point on the real number line.

Case I. When  $ \text{c}<-\text{1} $ .

Then,  $ \text{f(c)=-2} $ .

Also,  $ \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{(-2)=-2} $ .

Therefore,  $ \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f(x)=f(c)} $ .

Hence, the function  $ \text{f} $  attains continuity at every point  $ \text{x} $  , for  $ \text{x}<-\text{1} $ .

Case II. When  $ \text{c=-1} $ .

Then,  $ \text{f(c)=f(-1)=-2} $ .

Now, the left-hand-limit of the function at  $ \text{x=-1} $  is

 $ \underset{\text{x}\to {{\text{1}}^{\text{-}}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{\text{1}}^{\text{-}}}}{\mathop{\lim }}\,\text{(-2)=-2} $  and the right-hand-limit at  $ \text{x=-1} $  is

 $ \underset{\text{x}\to {{1}^{+}}}{\mathop{\lim }}\,\text{(x)=}\underset{\text{x}\to {{1}^{+}}}{\mathop{\lim }}\,\text{=2 }\!\!\times\!\!\text{ (-1)=-2} $ .

Therefore,  $ \underset{\text{x}\to -1}{\mathop{\lim }}\,\text{f(x)=f(-1)} $ .

Hence, the function  $ \text{f} $  satisfies continuity at  $ \text{x=-1} $ .

Case III. When  $ \text{-1}<\text{c}<\text{1} $ .

Then,  $ \text{f(c)=2c} $  and  $ \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{(2x)=2c} $ 

Therefore,  $ \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f(x)=f(c)} $ .

Hence, the function  $ \text{f} $  attains continuity at every point in the interval  $ \text{(-1,1)} $ .

Case IV. When  $ \text{c=1} $ .

Then,  $ \text{f(c)=f(1)=2 }\!\!\times\!\!\text{ 1=2} $ 

Now, the left-hand-limit of the function at  $ \text{x = 1} $  is 

 $ \underset{\text{x}\to {{\text{1}}^{\text{-}}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{\text{1}}^{\text{-}}}}{\mathop{\lim }}\,\text{(2x)=2 }\!\!\times\!\!\text{ 1=2} $  and the right-hand-limit at  $ \text{x = 1} $  is

 $ \underset{\text{x}\to {{1}^{+}}}{\mathop{\lim }}\,\text{f(x)=}\underset{x\to {{1}^{+}}}{\mathop{\lim }}\,\text{2=2} $ .

Therefore,  $ \underset{\text{x}\to 1}{\mathop{\lim }}\,\text{f(x)=f(c)} $ .

Thus, the function  $ \text{f} $  attains continuity at  $ \text{x=2} $ .

Case V. When  $ \text{c}>\text{1} $ .

Then  $ \text{f(c)=2} $ .

Also,  $ \underset{\text{x}\to 2}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to 2}{\mathop{\lim }}\,\text{(2)=2} $ .

Therefore,  $ \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f(x)=f(c)} $ .

Hence, the function  $ \text{f} $  is continuous at every point  $ \text{x} $ , for  $ \text{x}>\text{1} $ .

Thus, it can be concluded that the function  $ \text{f} $  is continuous for all the points.


17. Find the relationship between  $ \mathbf{a} $ and $ \mathbf{b} $ so that the function $ \mathbf{f} $ defined by $ \text{f(x)=}\left\{ \begin{align} & \text{ax+1, if x}\le \text{3} \\ & \text{bx+3, if x}\,>\,\text{3} \\ \end{align} \right. $ is continuous at $ \text{x=3} $ .

Ans:  The given function is$ \text{f(x)=}\left\{ \begin{align} & \text{ax+1, if x}\le \text{3} \\ & \text{bx+3, if x}\,>\text{3} \\ \end{align} \right. $ .

The function  $ \text{f} $  will be continuous at  $ \text{x = 3} $  if

 $ \underset{\text{x}\to {{\text{3}}^{\text{-}}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{\text{3}}^{\text{-}}}}{\mathop{\lim }}\,\text{f(x)=f(3)} $ , ..….. (1)

 $ \underset{x\to {{\text{3}}^{\text{-}}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{\text{3}}^{\text{-}}}}{\mathop{\lim }}\,\text{f(ax+1)=3a+1} $ ,

 $ \underset{\text{x}\to {{3}^{+}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{3}^{+}}}{\mathop{\lim }}\,\text{f(bx+1)=3b+3} $ ,                    …… (2)

 and

 $ \text{f(3)=3a+1} $ .                                                     …… (3)

Therefore, from the equation (1), (2), and (3) gives

 $ \Rightarrow \text{3a+1=3b+3} $ 

 $ \Rightarrow \text{3a=3b+2} $ 

 $ \Rightarrow \text{a=b+}\dfrac{\text{2}}{\text{3}} $ 

Hence, the required relationship between  $ \text{a} $  and  $ \text{b} $  is given by  $ \text{a=b+}\dfrac{\text{2}}{\text{3}} $ .


18. For what value of  $ \mathbf{\lambda } $  is the function defined by  $ \mathbf{f(x)=}\left\{ \begin{align} & \mathbf{\lambda (}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{-2x), }\,\,\,\mathbf{if x}\le \mathbf{0} \\ & \mathbf{ 4x+1, if x}>\mathbf{0} \\ \end{align} \right. $ is continuous at $ \mathbf{x=0} $ . What about continuity at $ \mathbf{x = 1} $ ?

Ans: The given function is  $ \text{f(x)=}\left\{ \begin{align} & \text{ }\!\!\lambda\!\!\text{ (}{{\text{x}}^{\text{2}}}\text{-2x), if x}\le \text{0} \\ & \text{ 4x+1, if x}\,>\,\text{0} \\ \end{align} \right. $

Now the function will be continuous at  $ \text{x = 0} $  if

 $ \underset{\text{x}\to {{\text{0}}^{\text{-}}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{\text{0}}^{+}}}{\mathop{\lim }}\,\text{f(x)=f(0)} $ .

Also, the R.H.L and L.H.L are given by,

 $ \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\left( 4x+1 \right)=4\left( 0 \right)+1=1 $ ,

 $ \underset{\text{x}\to {{\text{0}}^{\text{-}}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{\text{0}}^{\text{-}}}}{\mathop{\lim }}\,\text{ }\!\!\lambda\!\!\text{ (}{{\text{x}}^{\text{2}}}\text{-2x)}\,\text{=}\,\text{ }\!\!\lambda\!\!\text{ (}{{\text{0}}^{\text{2}}}\text{-2 }\!\!\times\!\!\text{ 0)=0} $ .

So,  $ \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)\ne \underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right) $ .

Thus, there does not exist any value of  $ \text{ }\!\!\lambda\!\!\text{ } $  for which  $ \text{f} $  is continuous at  $ \text{x = 0} $ .

Now, at  $ \text{x = 1} $ ,

 $ \text{f(1)=4x+1=4 }\!\!\times\!\!\text{ 1+1=5} $  and

 $ \underset{\text{x}\to 1}{\mathop{\lim }}\,\text{(4x+1)=4 }\!\!\times\!\!\text{ 1+1=5} $ .

Therefore,  $ \underset{\text{x}\to 1}{\mathop{\lim }}\,\text{f(x)=f(1)} $ .

Hence, the function  $ \text{f} $  is continuous at  $ \text{x = 1} $ , for all values of  $ \text{ }\!\!\lambda\!\!\text{ } $ .


19. Show that the function defined by  $ \text{g(x)=x- }\!\![\!\!\text{ x }\!\!]\!\!\text{ } $  is discontinuous at all integral points. Here  $ \mathbf{[x]} $  denotes the greatest integer less than or equal to  $ \mathbf{x} $ . 

Ans:  The given function is  $ \text{g(x)=x-}\left[ \text{x} \right] $ .

Note that the function is defined at every integral point.

Now, let assume that  $ \text{n} $  is an integer.

Then,  $ \text{g(n)=n-}\left[ \text{n} \right]\text{=n-n=0} $ .

Now taking left-hand-limit as  $ \text{x}\to \text{n} $  to the function  $ \text{g} $  gives 

 $ \underset{\text{x}\to {{\text{n}}^{\text{-}}}}{\mathop{\lim }}\,\text{g(x)=}\underset{\text{x}\to {{\text{n}}^{\text{-}}}}{\mathop{\lim }}\,\left[ \text{x-}\left[ \text{x} \right] \right]\text{=}\underset{\text{x}\to {{\text{n}}^{\text{-}}}}{\mathop{\lim }}\,\left( \text{x} \right)\text{-}\underset{\text{x}\to \text{n-}}{\mathop{\lim }}\,\left[ \text{x} \right]\text{=n-}\left( \text{n-1} \right)\text{=1} $ .

Again, the right-hand-limit on the function at  $ \text{x=n} $  is

 $ \underset{\text{x}\to {{\text{n}}^{+}}}{\mathop{\lim }}\,\text{g(x)=}\underset{\text{x}\to {{\text{n}}^{+}}}{\mathop{\lim }}\,\left[ \text{x-}\left[ \text{x} \right] \right]\text{=}\underset{\text{x}\to {{\text{n}}^{\text{+}}}}{\mathop{\lim }}\,\left( \text{x} \right)\text{-}\underset{\text{x}\to {{\text{n}}^{+}}}{\mathop{\lim }}\,\left[ \text{x} \right]\text{=n-n=0} $ .

Note that,  $ \underset{\text{x}\to {{\text{n}}^{\text{-}}}}{\mathop{\lim }}\,\text{g(x)}\ne \underset{\text{x}\to {{\text{n}}^{+}}}{\mathop{\lim }}\,\text{g(x)} $ .

Thus, the function  $ \text{f} $  is cannot be continuous at  $ \text{x=n,} $  

Hence, the function  $ \text{g} $  is not continuous at any integral point. 


20. Is the function defined by  $ \text{f(x)=}{{\text{x}}^{\text{2}}}\text{-sinx+5} $  continuous at  $ \text{x= }\!\!\pi\!\!\text{ } $  ?

Ans:  The given function is  $ \text{f(x)=}{{\text{x}}^{\text{2}}}\text{-sinx+5} $ .

Now, at  $ \text{x= }\!\!\pi\!\!\text{ } $ , 

 $ \text{f(x)}\,\text{=}\,\text{f( }\!\!\pi\!\!\text{ )}\,\text{=}\,{{\text{ }\!\!\pi\!\!\text{ }}^{\text{2}}}\text{-sin }\!\!\pi\!\!\text{ }\,\text{+}\,\text{5}\,\text{=}\,{{\text{ }\!\!\pi\!\!\text{ }}^{\text{2}}}\text{-0+5}\,\text{=}\,{{\text{ }\!\!\pi\!\!\text{ }}^{\text{2}}}\text{+5} $ 

Taking limit as  $ \text{x}\to \text{ }\!\!\pi\!\!\text{ } $  on the function  $ \text{f}\left( \text{x} \right) $  gives

 $ \underset{\text{x}\to \text{ }\!\!\pi\!\!\text{ }}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to \text{ }\!\!\pi\!\!\text{ }}{\mathop{\lim }}\,\text{(}{{\text{x}}^{\text{2}}}\text{-sinx+5)} $ .

Now substitute  $ \text{x= }\!\!\pi\!\!\text{ +h} $  into the function  $ \text{f}\left( \text{x} \right) $ .

When  $ \text{x}\to \text{ }\!\!\pi\!\!\text{ } $ , then  $ \text{h}\to 0 $ .

Therefore, 

 $ \underset{\text{x}\to \text{ }\!\!\pi\!\!\text{ }}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to \text{ }\!\!\pi\!\!\text{ }}{\mathop{\lim }}\,\left( {{\text{x}}^{\text{2}}}\text{-sinx} \right)\text{+5} $ .

 $ =\underset{\text{h}\to 0}{\mathop{\lim }}\,\left[ \left( \text{ }\!\!\pi\!\!\text{ +}{{\text{h}}^{\text{2}}} \right)\text{-sin}\left( \text{ }\!\!\pi\!\!\text{ +h} \right)\text{+5} \right] $ 

 $ =\underset{\text{h}\to 0}{\mathop{\lim }}\,{{\left( \text{ }\!\!\pi\!\!\text{ +h} \right)}^{\text{2}}}-\underset{\text{h}\to 0}{\mathop{\lim }}\,\text{sin}\left( \text{ }\!\!\pi\!\!\text{ +h} \right)\text{+}\underset{\text{h}\to 0}{\mathop{\lim }}\,5 $ 

 $ \text{=}{{\left( \text{ }\!\!\pi\!\!\text{ +0} \right)}^{\text{2}}}-\underset{\text{h}\to 0}{\mathop{\lim }}\,\text{ }\!\![\!\!\text{ sin }\!\!\pi\!\!\text{ }\cdot \text{cosh+cos }\!\!\pi\!\!\text{ }\cdot \text{sinh }\!\!]\!\!\text{ +5} $ 

 $ \text{=}{{\text{ }\!\!\pi\!\!\text{ }}^{\text{2}}}-\underset{\text{h}\to 0}{\mathop{\lim }}\,\left( \text{sin }\!\!\pi\!\!\text{ }\cdot \text{cosh} \right)\text{-}\underset{\text{h}\to 0}{\mathop{\lim }}\,\left( \text{cos }\!\!\pi\!\!\text{ }\cdot \text{sinh} \right)\text{+5} $ 

 $ \text{=}{{\text{ }\!\!\pi\!\!\text{ }}^{\text{2}}}-\text{sin }\!\!\pi\!\!\text{ }\cdot \text{cos0}-\text{cos }\!\!\pi\!\!\text{ }\cdot \text{sin0+5} $ 

 $ \text{=}{{\text{ }\!\!\pi\!\!\text{ }}^{\text{2}}}-\text{0 }\!\!\times\!\!\text{ 1}-\text{(}-\text{1) }\!\!\times\!\!\text{ 0+5=}{{\text{ }\!\!\pi\!\!\text{ }}^{\text{2}}}\text{+5} $ .

So,  $ \underset{\text{x}\to \text{x}}{\mathop{\lim }}\,\text{f(x)=f( }\!\!\pi\!\!\text{ )} $ .

Hence, it is concluded that the function  $ \text{f} $  is continuous at  $ \text{x=n} $ .


21. Discuss the continuity of the following functions:

(a)  $ \mathbf{f(x)}=\mathbf{sinx+cosx} $   

Ans. Suppose that, ‘ $ a $ ’ is any real number. Therefore,

 $ \underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,f\left( a+h \right) $ 

Now,

 $ \begin{align} & \underset{h\to 0}{\mathop{\lim }}\,f\left( a+h \right)=\underset{h\to 0}{\mathop{\lim }}\,\left[ \sin \left( a+h \right)+\cos \left( a+h \right) \right] \\ & =\underset{h\to 0}{\mathop{\lim }}\,\left[ \sin a\cdot \cosh +\cos a\cdot \sinh +\cos a\cdot \cosh -\sin a\cdot \sinh \right] \\ & =\sin a\cdot \cos 0+\cos a\cdot \sin 0+\cos a\cdot \cos 0-\sin a\cdot \sin 0 \\ \end{align} $

 $ =\sin a+\cos a $  ( since,  $ \cos 0=1 $  and  $ \sin 0=0 $ )

 $ =f\left( a \right) $ .

In a similar way, it can be shown that,

 $ \underset{x\to {{a}^{-}}}{\mathop{\lim }}\,=f\left( a \right) $ 

Thus,  $ \underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right)=\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f\left( x \right) $ .

Hence,  $ f\left( x \right) $  is continuous at  $ x=a $ ,  $ \forall a\in \mathbb{R} $ , that is,

the function  $ \text{f(x)=sinx+cosx} $ is continuous.


(b)  $ \mathbf{f(x)=sinx-cosx} $   

Ans. Suppose that, ‘ $ a $ ’ is any real number. Therefore,

 $ \underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,f\left( a+h \right) $ 

Now,

  $ \begin{align} & \underset{h\to 0}{\mathop{\lim }}\,f\left( a+h \right)=\underset{h\to 0}{\mathop{\lim }}\,\left[ \sin \left( a+h \right)-\cos \left( a-h \right) \right] \\ & =\underset{h\to 0}{\mathop{\lim }}\,\left[ \sin a\cdot \cosh +\cos a\cdot \sinh -\cos a\cdot \cosh -\sin a\cdot \sinh \right] \\ & =\sin a\cdot \cos 0+\cos a\cdot \sin 0-\cos a\cdot \cos 0-\sin a\cdot \sin 0 \\ \end{align} $

 $ =\sin a-\cos a $  [ since,  $ \cos 0=1 $  and  $ \sin 0=0 $ ]

 $ =f\left( a \right) $ .

In a similar way, it can be shown that,

 $ \underset{x\to {{a}^{-}}}{\mathop{\lim }}\,=f\left( a \right) $ 

Thus,  $ \underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right)=\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f\left( x \right) $ .

Hence,  $ f\left( x \right) $  is continuous at  $ x=a $ ,  $ \forall a\in \mathbb{R} $ , that is,

the function  $ \text{f(x)=sinx}-\text{cosx} $ is continuous.


(c)  $ \mathbf{f(x)=sinx}\cdot \mathbf{cosx} $ .

Ans. Suppose that, ‘ $ a $ ’ is any real number. Therefore,

 $ \underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{h\to 0}{\mathop{\lim }}\,f\left( a+h \right) $ .

Now,

 $ \begin{align} & \underset{h\to 0}{\mathop{\lim }}\,f\left( a+h \right)=\underset{h\to 0}{\mathop{\lim }}\,\left[ \sin \left( a+h \right)\cdot \cos \left( a+h \right) \right] \\ & =\underset{h\to 0}{\mathop{\lim }}\,\left( \sin a\cdot \cosh +\cos a\cdot \sinh \right)\left( \cos a\cdot \cosh -\sin a\cdot \sinh \right) \\ & =\left( \sin a\cdot \cos 0+\cos a\cdot \sin 0 \right)\left( \cos a\cdot \cos a-\sin a\cdot \sin 0 \right) \\ & =\left( \sin a+0 \right)\left( \cos a-0 \right) \\ & =\sin a\cdot \cos a \\ \end{align} $

 $ =f\left( a \right) $ .

In a similar way, it can be shown that,

 $ \underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right) $ 

Thus,  $ \underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right)=\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f\left( x \right) $ .

Hence,  $ f\left( x \right) $  is continuous at  $ x=a $ ,  $ \forall a\in \mathbb{R} $ , that is, the function  $ \text{f(x)=sinx}\cdot \text{cosx} $  is continuous.


22. Discuss the continuity of the cosine, cosecant, secant and cotangent functions.

Ans:  

Sine function:

It can be observed that the function  $ \text{g(x)=sinx} $  is defined for all real numbers.

Now, let's consider  $ \text{c} $  to be a real number and substitute  $ \text{x=c+h} $  into the function  $ \text{g} $ .

When,  $ \text{x}\to \text{c} $ , then  $ \text{h}\to 0 $ .

So,  $ \text{g(c)=sinc} $  and

 $ \begin{align} & \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{g(x)=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{sinx} \\ & =\underset{\text{h}\to 0}{\mathop{\lim }}\,\text{sin}\left( \text{c+h} \right) \\ & =\underset{\text{h}\to 0}{\mathop{\lim }}\,\left[ \text{sinc}\cdot \text{cosh+cosc}\cdot \text{sinh} \right] \\ & =\underset{\text{h}\to 0}{\mathop{\lim }}\,\left( \text{sinc}\cdot \text{cosh} \right)\text{+}\underset{\text{h}\to 0}{\mathop{\lim }}\,\left( \text{cosc}\cdot \text{sinh} \right) \\ & \text{=sinc}\cdot \text{cos0+cosc}\cdot \text{sin0} \\ & \text{=sinc+0} \\ & \text{=sinc} \\ \end{align} $ 

Therefore,  $ \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{g(x)=g(c)} $ .

Thus, the function  $ \text{g}\left( \text{x} \right)\text{=sinx} $  is continuous.


Cosine function:

let  $ \text{h(x)=cosx} $ .

It can be noted that  $ \text{h(x)=cosx} $  is defined for all real numbers.

Now, let's consider  $ \text{c} $  to be a real number and substitute  $ \text{x=c+h} $  into the function  $ \text{h} $ .

Then,  $ \text{h(c)=cosc} $  and

 $ \begin{align} & \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{h(x)=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{cosx} \\ & =\underset{\text{h}\to 0}{\mathop{\lim }}\,\text{cos}\left( \text{c+h} \right) \\ & =\underset{\text{h}\to 0}{\mathop{\lim }}\,\left[ \text{cosc}\cdot \text{cosh-sinc}\cdot \text{sinh} \right] \\ & =\underset{\text{h}\to 0}{\mathop{\lim }}\,\left( \text{cosc}\cdot \text{cosh} \right)\text{-}\underset{\text{h}\to 0}{\mathop{\lim }}\,\left( \text{sinc}\cdot \text{sinh} \right) \\ & \text{=cosc}\cdot \text{cos0}-\text{sinc}\cdot \text{sin0} \\ & \text{=cosc }\!\!\times\!\!\text{ 1}-\text{sinc }\!\!\times\!\!\text{ 0} \\ & \text{=cosc} \\ \end{align} $ 

Therefore,  $ \underset{\text{h}\to 0}{\mathop{\lim }}\,\text{h(x)=h(c)} $ .

Thus, the function  $ \text{h(x)=cosx} $  is continuous.


Cosecant function:

We know that if a function  $ g\left( x \right) $  is continuous, then

 $ \dfrac{\text{1}}{\text{g(x)}}\text{,g(x)}\ne \text{0} $  is continuous. 

Now, let  $ g\left( x \right)=\sin x $ . 

Then,  $ \text{cosec x=}\dfrac{\text{1}}{\text{sinx}}\text{,} $  and  $ \text{sinx}\ne \text{0} $ .

Therefore,  $ \cos ecx=\dfrac{1}{\sin x} $ , when  $ x\ne n\pi  $ ,  $ n\in \mathbb{N} $ .

Since, the function  $ \sin x $  is continuous, so  $ \text{f}\left( \text{x} \right)=\text{cosec x} $  is continuous for all  $ x $ , except at  $ x=n\pi ,\text{  n}\in \mathbb{N} $ .


Secant function:

We know that if a function  $ h\left( x \right) $  is continuous, then

 $ \dfrac{\text{1}}{\text{h(x)}}\text{,h(x)}\ne \text{0} $  is continuous.

Now, let  $ h\left( x \right)=\cos x $ .

Again,  $ \text{secx}=\dfrac{\text{1}}{\text{cosx}}\text{,} $  and  $ \text{cosx}\ne \text{0} $ .

Therefore,  $ \sec x=\dfrac{1}{\cos x} $  and  $ x\ne \left( 2n+1 \right)\dfrac{\pi }{2} $ 

Since, the function  $ \cos x $  is continuous, so the function  $ f\left( x \right)=\sec x $  is continuous for all  $ x $ , except at  $ \text{x}=\text{(2n+1)}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{,  n}\in \mathbb{N} $ .


Cotangent function:

We know that if two functions say  $ \text{g} $  and  $ \text{h} $  are continuous, then

 $ \dfrac{\text{h(x)}}{\text{g(x)}}\text{,g(x)}\ne \text{0} $  is continuous.

Now, let  $ h\left( x \right)=\cos x $  and  $ g\left( x \right)=\sin x $ .

Then,  $ \text{cotx}=\dfrac{\text{cosx}}{\text{sinx}} $  and  $ \text{sinx}\ne 0 $ .

That is,  $ \cot x=\dfrac{\cos x}{\sin x} $  and  $ x\ne n\pi ,\text{  n}\in \mathbb{N} $ .

Since,  $ \cos x $  and  $ \sin x $  are both continuous, so the function  $ f\left( x \right)=\cot x $  is continuous except at  $ x=n\pi ,\text{  n}\in \mathbb{N} $ .


23.Find all points of discontinuity of  $ \mathbf{f} $ , where  $ \text{f(x)=}\left\{ \begin{align}& \dfrac{\text{sinx}}{\text{x}}\text{, if x}\,<\,\text{0} \\& \text{x+1, }\,\,\text{if x}\ge \text{0} \\ \end{align} \right. $ .

Ans: The given function is $ \text{f(x)=}\left\{ \begin{align} & \dfrac{\text{sinx}}{\text{x}}\text{, if x}\,<\,\text{0} \\ & \text{x+1, }\,\text{if x}\ge \text{0} \\ \end{align} \right. $ .

Note that, the function  $ \text{f} $  is defined at every point on the real number line.

Now, let's consider  $ \text{c} $  be a real number.

Then there may arise three cases, either  $ \text{c}<\text{0} $ , or  $ \text{c}>\text{0} $ , or  $ \text{c}=\text{0} $ .

Let us discuss one after another.

Case I. When  $ \text{c}<\text{0} $ .

Then,  $ \text{f(c)=}\dfrac{\text{sinc}}{\text{c}} $ .

Also,  $ \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f(x)}\left( \dfrac{\text{sinx}}{\text{x}} \right)\text{=}\dfrac{\text{sinc}}{\text{c}} $ .

Therefore,  $ \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f(x)=f(c)} $ .

Hence, the function  $ \text{f} $  is continuous at every point  $ \text{x} $ , for  $ \text{x0} $ .

Case II. When  $ \text{c}>\text{0} $ .

Then  $ \text{f(c)=c+1} $ .

Also,  $ \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{(x+1)=c+1} $ .

Therefore,  $ \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f(x)=f(c)} $ .

Hence, the function  $ \text{f} $  is continuous at every point, where  $ \text{x0} $ .

Case III. When  $ \text{c = 0} $ .

Then  $ \text{f(c)=f(0)=0+1=1} $ .

Now, the left-hand-limit of the function  $ \text{f} $  at  $ \text{x=0} $  is

 $ \underset{\text{x}\to {{\text{0}}^{\text{-}}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{\text{0}}^{\text{-}}}}{\mathop{\lim }}\,\dfrac{\text{sinx}}{\text{x}}\text{=1} $  and the right-hand-limit is

 $ \underset{\text{x}\to {{\text{0}}^{\text{-}}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{\text{0}}^{\text{-}}}}{\mathop{\lim }}\,\text{(x+1)=1} $ 

Therefore,  $ \underset{\text{x}\to {{\text{0}}^{\text{-}}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{\text{0}}^{\text{-}}}}{\mathop{\lim }}\,\text{f(x)=f(0)} $ .

So, the function  $ \text{f} $  is continuous at  $ \text{x = 0} $ .

Thus, the function  $ \text{f} $  is continuous at every real point.

Hence, the function  $ \text{f} $  does not have any point of discontinuity.


24. Determine if  $ \text{f(x)}=\left\{ \begin{align}& {{\text{x}}^{\text{2}}}\text{sin}\dfrac{\text{1}}{\text{x}}\text{,  if x}\ne \text{0} \\  & \text{0, if x}=\text{0} \\ \end{align} \right. $  

 is a continuous function?

Ans: 

The given function is $ \text{f(x)}=\left\{ \begin{align} & {{\text{x}}^{\text{2}}}\text{sin}\dfrac{\text{1}}{\text{x}}\text{, if x}\ne \text{0} \\ & \text{0, if x}=\text{0} \\ \end{align} \right. $

We can observe that the function  $ \text{f} $  is defined at every point on the real number line.

Now, let's consider  $ \text{c} $  to be a real number.

Then, there may arise two cases, either  $ \text{c}\ne \text{0} $  or  $ \text{c=0} $ .

Let us discuss the cases one after another.

Case I. When  $ \text{c}\ne 0 $ .

Then  $ \text{f(c)=}{{\text{c}}^{\text{2}}}\text{sin}\dfrac{\text{1}}{\text{c}} $ .

Also,

 $ \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\left( {{\text{x}}^{\text{2}}}\text{sin}\dfrac{\text{1}}{\text{x}} \right)=\left( \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,{{\text{x}}^{\text{2}}} \right)\left( \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{sin}\dfrac{\text{1}}{\text{x}} \right)={{\text{c}}^{\text{2}}}\text{sin}\dfrac{\text{1}}{\text{c}} $ .

Therefore,  $ \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f(x)=f(c)} $ .

Hence, the function  $ \text{f} $  is continuous at every point  $ \text{x}\ne 0 $ .

Case II. When  $ \text{c = 0} $ .

Then  $ \text{f(0)=0} $  and also

 $ \underset{\text{x}\to {{0}^{\text{-}}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{\text{0}}^{\text{-}}}}{\mathop{\lim }}\,\left( {{\text{x}}^{\text{2}}}\text{sin}\dfrac{\text{1}}{\text{x}} \right)\text{=}\underset{\text{x}\to 0}{\mathop{\lim }}\,\left( {{\text{x}}^{\text{2}}}\text{sin}\dfrac{\text{1}}{\text{x}} \right) $ 

Now, we know that,

  $ \text{-1}\le \text{sin}\dfrac{\text{1}}{\text{x}}\le 1,\text{ x}\ne 0 $ .

 $ \begin{align} & \Rightarrow \text{-}{{\text{x}}^{\text{2}}}\le \text{sin}\dfrac{\text{1}}{\text{x}}\le {{\text{x}}^{\text{2}}} \\ & \Rightarrow \underset{\text{x}\to 0}{\mathop{\lim }}\,\text{(-}{{\text{x}}^{\text{2}}}\text{)}\le \underset{\text{x}\to 0}{\mathop{\lim }}\,\left( {{\text{x}}^{\text{2}}}\text{sin}\dfrac{\text{1}}{\text{x}} \right)\le 0 \\ & \Rightarrow 0\le \underset{\text{x}\to 0}{\mathop{\lim }}\,\left( {{\text{x}}^{\text{2}}}\text{sin}\dfrac{\text{1}}{\text{x}} \right)\le 0 \\ & \Rightarrow \underset{\text{x}\to 0}{\mathop{\lim }}\,\left( {{\text{x}}^{\text{2}}}\text{sin}\dfrac{\text{1}}{\text{x}} \right)=0 \\ \end{align} $ 

Therefore,  $ \underset{\text{x}\to {{0}^{\text{-}}}}{\mathop{\lim }}\,\text{f(x)=0} $ .

Similarly, we have, 

 $ \underset{\text{x}\to {{0}^{+}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{0}^{+}}}{\mathop{\lim }}\,\left( {{\text{x}}^{\text{2}}}\text{sin}\dfrac{\text{1}}{\text{x}} \right)=\underset{\text{x}\to 0}{\mathop{\lim }}\,\left( {{\text{x}}^{\text{2}}}\text{sin}\dfrac{\text{1}}{\text{x}} \right)\text{=0} $ 

Therefore,  $ \underset{\text{x}\to {{0}^{\text{-}}}}{\mathop{\lim }}\,\text{f(x)=f(0)=}\underset{\text{x}\to {{0}^{+}}}{\mathop{\lim }}\,\text{f(x)} $ .

Thus, the function  $ \text{f} $  is continuous at the point  $ \text{x = 0} $ .

So, the function  $ \text{f} $  is continuous at all real points.

Hence, the function  $ \text{f} $  is continuous.


 25. Examine the continuity of  $ \mathbf{f} $ , where  $ \mathbf{f} $  is defined by  $ \text{f(x)=}\left\{ \begin{align}& \text{sinx-cosx, if x}\ne \text{0} \\ & \text{ 1  if x=0} \\ \end{align} \right. $ .

Ans: The given function is` $ \text{f(x)=}\left\{ \begin{align} & \text{sinx-cosx, if x}\ne \text{0} \\ & \text{ 1 if x=0} \\ \end{align} \right. $ .

It can be observed that the function  $ \text{f} $  is defined at every point on the real number line.

Now, let's consider  $ \text{c} $  to be a real number.

Then, there may arise two cases, either  $ \text{c}\ne \text{0} $  or  $ \text{c=0} $ .

Let us discuss the cases one after another.

Case I. When  $ \text{c}\ne 0 $ .

Then,  $ \text{f(c)=sinc-cosc} $ .

Also,  $ \underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{f(x)=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{(sinx-cosx)=sinc-cosc} $ .

Therefore,  $ \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{f(x)=f(c)} $ .

Hence, the function  $ \text{f} $  is continuous at every point  $ \text{x} $  for  $ \text{x}\ne 0 $ .

Case II. When  $ \text{c =0} $ .

Then,  $ \text{f(0)=-1} $ .

Now the left-hand-limit of the function  $ \text{f} $  at  $ \text{x=0} $  is

 $ \underset{\text{x}\to {{0}^{\text{-}}}}{\mathop{\lim }}\,\text{f(x)}=\underset{\text{x}\to {{0}^{\text{-}}}}{\mathop{\lim }}\,\text{(sinx-cosx)=sin0-cos0=0-1=-1} $  and the right-hand-limit is 

 $ \underset{\text{x}\to {{0}^{+}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{0}^{+}}}{\mathop{\lim }}\,\text{(sinx-cosx)=sin0-cos0=0-1=-1} $ .

Therefore,  $ \underset{\text{x}\to {{\text{0}}^{\text{-}}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{0}^{+}}}{\mathop{\lim }}\,\text{f(x)=f(0)} $ .

So, the function  $ \text{f} $  is continuous at  $ \text{x = 0} $ .

Thus, the function  $ \text{f} $  is continuous at all real points.

Hence, the function  $ \text{f} $  is continuous.


Find the values of  $ \mathbf{k} $  so that the function  $ \mathbf{f} $  is continuous at the indicated point in Exercises 26 to 29.

26.  $ \text{f(x)}\mathbf{=}\left\{ \begin{align} & \dfrac{\text{kcosx}}{\text{ }\!\!\pi\!\!\text{ -2x}}\text{, if x}\ne \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}} \\ & \text{    3,      if x=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}} \\ \end{align} \right. $ 

at  $ \mathbf{x}=\dfrac{\mathbf{\pi }}{\mathbf{2}} $ .

Ans: The given function is  $ \text{f(x)}=\left\{ \begin{align} & \dfrac{\text{kcosx}}{\text{ }\!\!\pi\!\!\text{ -2x}}\text{, if x}\ne \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}} \\ & \text{ 3, if x=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}} \\ \end{align} \right. $ .

Observe that,  $ \text{f} $  is defined and continuous at  $ \text{x=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}} $  , since the value of the  $ \text{f} $  at  $ \text{x=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}} $  is equal with the limiting value of  $ \text{f} $  at  $ \text{x=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}} $ .

Since,  $ \text{f} $  is defined at  $ \text{x=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}} $  and  $ \text{f}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}} \right)\text{=3} $ , so

 $ \underset{\text{x}\to \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}}{\mathop{\lim }}\,\text{f(x)}=\underset{\text{x}\to \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}}{\mathop{\lim }}\,\dfrac{\text{kcosx}}{\text{ }\!\!\pi\!\!\text{ -2x}} $ .

Substitute  $ \text{x=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{+h} $  into the function  $ \text{f}\left( \text{x} \right) $ .

So, we have,  $ \text{x}\to \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\Rightarrow \text{h}\to 0 $ .

Then,

 $ \underset{\text{x}\to \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}}{\mathop{\lim }}\,\dfrac{\text{kcosx}}{\text{ }\!\!\pi\!\!\text{ -2x}}=\underset{\text{h}\to 0}{\mathop{\lim }}\,\dfrac{\text{kcos}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{+h} \right)}{\text{ }\!\!\pi\!\!\text{ -2}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{+h} \right)} $ .

  $ \Rightarrow \text{k}\underset{\text{h}\to 0}{\mathop{\lim }}\,\dfrac{\text{-sinh}}{\text{-2h}}\text{=}\dfrac{\text{k}}{\text{2}}\underset{\text{h}\to 0}{\mathop{\lim }}\,\dfrac{\text{sinh}}{\text{h}}\text{=}\dfrac{\text{k}}{\text{2}}\text{.1}=\dfrac{\text{k}}{\text{2}} $ 

Therefore,  $ \underset{\text{x}\to \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}}{\mathop{\lim }}\,\text{f(x)=f}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}} \right) $ 

  $ \begin{align} & \Rightarrow \dfrac{\text{k}}{\text{2}}\text{=3} \\ & \Rightarrow \text{k=6} \\ \end{align} $

Hence, the value of  $ \text{k} $  is  $ \text{6} $  for which the function  $ \text{f} $  is continuous.


27.  $ \text{f(x)=}\left\{ \begin{align}& \text{k}{{\text{x}}^{\text{2}}}\text{, if x}\le \text{2} \\ & \text{3,    if x}\,>\,\text{2} \\ \end{align} \right. $  at  $ \mathbf{x=2} $ .

Ans: The given function is  $ \text{f(x)=}\left\{ \begin{align} & \text{k}{{\text{x}}^{\text{2}}}\text{, if x}\le \text{2} \\ & \text{3, if x}>\text{2} \\ \end{align} \right. $

It is known that,  $ \text{f} $  is continuous at  $ \text{x = 2} $  only if  $ \text{f} $  is defined at  $ \text{x=2} $  and if the value of  $ \text{f} $  at  $ \text{x = 2} $  is equal with the limiting value of  $ \text{f} $  at  $ \text{x = 2} $ .

So, at  $ x=2 $ ,

Now, the left-hand-limit and right-hand-limit of the function  $ \text{f}\left( \text{x} \right) $  at  $ \text{x}=\text{2} $ respectively are,

 $ \underset{\text{x}\to {{2}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)=\underset{\text{x}\to {{2}^{-}}}{\mathop{\lim }}\,\left( \text{k}{{\text{x}}^{\text{2}}} \right)=\text{k}{{\left( \text{2} \right)}^{\text{2}}}=\text{4k} $ , 

and  $ \underset{\text{x}\to {{2}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)=\underset{\text{x}\to {{2}^{+}}}{\mathop{\lim }}\,\left( 3 \right)=3 $ .

Since, the function is continuous at  $ \text{x}=\text{2} $ , so

 $ \begin{align} & \underset{\text{x}\to {{2}^{\text{-}}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{2}^{+}}}{\mathop{\lim }}\,\text{f(x)=f(2)} \\ & \Rightarrow \text{4k=3} \\ & \Rightarrow \text{k=}\dfrac{\text{3}}{\text{4}} \\ \end{align} $

Hence, the value of  $ \text{k} $  is  $ \dfrac{\text{3}}{\text{4}} $  for which the function  $ \text{f} $  is continuous.


28.  $ \text{f(x)=}\left\{ \begin{align} & \text{kx+1, if x}\le \mathbf{\pi } \\ & \mathbf{cosx}\text{,  if x}\,>\,\mathbf{\pi } \\ \end{align} \right. $  at  $ \mathbf{x=\pi } $ .

Ans: The given function is $ \text{f(x)=}\left\{ \begin{align} & \text{kx+1, if x}\le \text{ }\!\!\pi\!\!\text{ } \\ & \text{cosx, if x}\,>\,\text{ }\!\!\pi\!\!\text{ } \\ \end{align} \right. $ .

It is known that,  $ \text{f} $  is continuous at  $ \text{x= }\!\!\pi\!\!\text{ } $  only if the value of  $ \text{f} $  at  $ \text{x= }\!\!\pi\!\!\text{ } $  is equal with the limiting value of  $ \text{f} $  at  $ \text{x= }\!\!\pi\!\!\text{ } $ .

It is provided that the function  $ \text{f} $  is defined at  $ \text{x= }\!\!\pi\!\!\text{ } $ .

Also,  $ \text{f( }\!\!\pi\!\!\text{ )=k }\!\!\pi\!\!\text{ +1} $ .

Now, the left-hand-limit,

 $ \underset{\text{x}\to {{\text{ }\!\!\pi\!\!\text{ }}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{\text{ }\!\!\pi\!\!\text{ }}^{-}}}{\mathop{\lim }}\,\text{(kx+1)=k }\!\!\pi\!\!\text{ +1} $ .

Also, the right-hand-limit,

 $ \underset{\text{x}\to {{\text{ }\!\!\pi\!\!\text{ }}^{\text{+}}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)\text{=}\underset{\text{x}\to {{\text{ }\!\!\pi\!\!\text{ }}^{\text{+}}}}{\mathop{\lim }}\,\text{cosx=cos }\!\!\pi\!\!\text{ =-1} $ .

Since, the function  $ \text{f} $  is continuous, so

 $ \underset{\text{x}\to {{\text{ }\!\!\pi\!\!\text{ }}^{-}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)=\underset{\text{x}\to {{\text{ }\!\!\pi\!\!\text{ }}^{\text{+}}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)=\text{f}\left( \text{ }\!\!\pi\!\!\text{ } \right) $ 

 $ \begin{align} & \Rightarrow \text{k }\!\!\pi\!\!\text{ +1=-1} \\ & \Rightarrow \text{k }\!\!\pi\!\!\text{ =-2} \\ & \Rightarrow \text{k=-}\dfrac{\text{2}}{\text{ }\!\!\pi\!\!\text{ }} \\ \end{align} $ 

Hence, the value of  $ \text{k} $  is  $ \text{-}\dfrac{\text{2}}{\text{ }\!\!\pi\!\!\text{ }} $  for which the function  $ \text{f} $  is continuous at  $ \text{x= }\!\!\pi\!\!\text{ } $ . 


29.  $ \text{f(x)=}\left\{ \begin{align} & \text{kx+1, if x}\le \text{5} \\ & \text{3x-5, }\,\text{if x}\,>\,\text{5} \\ \end{align} \right. $  at  $ \mathbf{x = 5} $ 

Ans: The given function is   $ \text{f(x)=}\left\{ \begin{align} & \text{kx+1, if x}\le \text{5} \\ & \text{3x-5, }\,\,\text{if x}>\text{5} \\ \end{align} \right. $ .

Recall that, the function  $ \text{f} $  is continuous at  $ \text{x = 5} $  only if the value of  $ \text{f} $  at  $ \text{x = 5} $  is equal to the limiting value of  $ \text{f} $  at  $ \text{x = 5} $ .

Note that, the function  $ \text{f} $  is defined at  $ \text{x = 5} $ .

Also,  $ \text{f(5)=kx+1=5k+1} $ .

Then, the left-hand-limit of the function,

 $ \underset{\text{x}\to {{\text{5}}^{-}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{\text{5}}^{-}}}{\mathop{\lim }}\,\left( \text{kx+1} \right)\text{=5k+1} $ .

The right-hand-limit of the function,

 $ \underset{\text{x}\to {{\text{5}}^{+}}}{\mathop{\lim }}\,\text{f}\left( \text{x} \right)=\underset{\text{x}\to {{\text{5}}^{+}}}{\mathop{\lim }}\,\left( \text{3x-5} \right)\text{=3}\left( 5 \right)-5=15-5=10 $ .

Since, the function  $ \text{f} $  is continuous, so

 $ \begin{align} & \underset{\text{x}\to {{\text{5}}^{\text{-}}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{5}^{+}}}{\mathop{\lim }}\,\text{(3x-5)=5k+1} \\  & \Rightarrow \text{5k+1=10} \\ & \Rightarrow \text{5k=9} \\ & \Rightarrow \text{k=}\dfrac{\text{9}}{\text{5}} \\ \end{align} $ 

Hence, the value of  $ \text{k} $  is  $ \dfrac{9}{5} $  for which the function  $ \text{f} $  is continuous at  $ \text{x=5} $ .


30. Find the values of  $ \mathbf{a} $  and  $ \mathbf{b} $  such that the function defined by 

$ \mathbf{f} $  such that  $ \text{f(x)=}\left\{ \begin{align} & \text{5,     if    x}\le \text{2} \\ & \text{ax+b, if 2}\,<\,\text{x}\,<\,\text{10} \\  & \text{   21,  if    x}\ge \text{10} \\ \end{align} \right. $  is a continuous function.

Ans: The given function is $ \text{f(x)=}\left\{ \begin{align} & \text{5, }\,\,\,\text{if x}\le \text{2} \\ & \text{ax+b,}\,\text{ if }\,\,\text{2}<\text{x}<\text{10} \\ & \text{ 21, if x}\ge \text{10} \\ \end{align} \right. $ .

Note that,  $ \text{f} $  is defined at every point on the real number line.

Now, realise that if the function  $ \text{f} $  is continuous then  $ \text{f} $  is continuous at every real number.

So, let  $ \text{f} $  satisfies continuity at  $ \text{x=2} $  and  $ \text{x=10} $ .

Then, since  $ \text{f} $  is continuous at  $ \text{x=2} $ , so

 $ \begin{align}& \underset{\text{x}\to {{\text{2}}^{\text{-}}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{2}^{+}}}{\mathop{\lim }}\,\text{f(x)=f(2)} \\  & \Rightarrow \underset{\text{x}\to {{\text{2}}^{\text{-}}}}{\mathop{\lim }}\,\text{(5)=}\underset{\text{x}\to {{2}^{+}}}{\mathop{\lim }}\,\text{(ax+b)=5} \\  & \Rightarrow \text{5=2a+b} \\ \end{align} $ 

 $ \Rightarrow \text{2a+b=5} $           …… (1)

Again, since  $ \text{f} $  attains continuity at  $ \text{x=10} $ , so

 $ \begin{align}& \underset{\text{x}\to 1{{\text{0}}^{\text{-}}}}{\mathop{\lim }}\,\text{f(x)=}\underset{\text{x}\to {{10}^{+}}}{\mathop{\lim }}\,\text{f(x)=f(10)} \\ & \Rightarrow \underset{\text{x}\to 1{{\text{0}}^{\text{-}}}}{\mathop{\lim }}\,\text{(ax+b)=}\underset{\text{x}\to {{10}^{+}}}{\mathop{\lim }}\,\text{(21)=21} \\ \end{align} $

 $ \Rightarrow 1\text{0a+b=21} $ …… (2)

Subtracting the equation (1) from the equation (2), gives

 $ \text{8a=16}\Rightarrow \text{a=2} $ 

Substituting  $ \text{a=2} $  in the equation (1), gives

 $ \begin{align} & \text{2 }\!\!\times\!\!\text{ 2+b=5} \\ & \Rightarrow \text{4+b=5}\Rightarrow \text{b=1} \\ \end{align} $ 

Hence, the values of  $ \text{a} $  and  $ \text{b} $  are  $ 2 $  and  $ 1 $  respectively for which  $ \text{f} $  is a continuous function.


31. Show that the function defined by  $ \text{f(x)=cos(}{{\text{x}}^{\text{2}}}\text{)} $  is a continuous function.

Ans: The given function is  $ \text{f(x)=cos(}{{\text{x}}^{\text{2}}}\text{)} $ .

Note that,  $ \text{f} $  is defined for all real numbers and so  $ \text{f} $  can be expressed as the composition of two functions as,  $ \text{f=g}\circ \text{h} $ , where  $ \text{g(x)=cosx} $  and  $ \text{h(x)=}{{\text{x}}^{\text{2}}} $ .

 $ \text{ }\!\![\!\!\text{ }\therefore \text{(goh)(x)=g(h(x))=g(}{{\text{x}}^{\text{2}}}\text{)=cos(}{{\text{x}}^{\text{2}}}\text{)=f(x) }\!\!]\!\!\text{ } $ 

Now, it is to be Proven that, the functions  $ \text{g(x)=cosx} $  and  $ \text{h(x)=}{{\text{x}}^{\text{2}}} $  are continuous.

Since the function  $ \text{g} $  is defined for all the real numbers, let 's consider  $ \text{c} $  be a real number.

Then,  $ \text{g(c)=cosc} $ .

Substitute  $ \text{x=c+h} $  into the function  $ \text{g} $ .

When,  $ \text{x}\to \text{c} $ , then  $ \text{h}\to 0 $ .

Then we have,

 $ \begin{align} & \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{g(x)=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{cosx} \\ & =\underset{\text{h}\to 0}{\mathop{\lim }}\,\text{cos}\left( \text{c+h} \right) \\ & =\underset{\text{h}\to 0}{\mathop{\lim }}\,\left[ \text{cosc}\cdot \text{cosh-sinc}\cdot \text{sinh} \right] \\ & =\underset{\text{h}\to 0}{\mathop{\lim }}\,\left( \text{cosc}\cdot \text{cosh} \right)-\underset{\text{h}\to 0}{\mathop{\lim }}\,\left( \text{sinc}\cdot \text{sinh} \right) \\ & \text{=cosc}\cdot \text{cos0}-\text{sinc}\cdot \text{sin0} \\ & \text{=cosc }\!\!\times\!\!\text{ 1}-\text{sinc }\!\!\times\!\!\text{ 0} \\ & \text{=cosc} \\ \end{align} $

Therefore,  $ \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{g(x)=g(c)} $ .

Hence, the function  $ \text{g(x)=cosx} $  is continuous.

Again,  $ \text{h(x)=}{{\text{x}}^{\text{2}}} $  is defined for every real point.

So, let consider  $ \text{k} $  be a real number, then  $ \text{h(k)=}{{\text{k}}^{\text{2}}} $  and

 $ \underset{\text{x}\to \text{k}}{\mathop{\lim }}\,\text{h(x)=}\underset{\text{x}\to \text{k}}{\mathop{\lim }}\,{{\text{x}}^{\text{2}}}\text{=}{{\text{k}}^{\text{2}}} $ .

Therefore,  $ \underset{\text{x}\to \text{k}}{\mathop{\lim }}\,\text{h(x)=h(k)} $ .

Hence, the function  $ \text{h} $  is continuous.

Now, remember that for real valued functions  $ \text{g} $  and  $ \text{h} $  , such that  $ \text{(g }\circ \text{ h)} $  is defined at  $ \text{c} $  , if  $ \text{g} $  is continuous at  $ \text{c} $  and  $ \text{f} $  is continuous at  $ \text{g(c)} $ , then  $ \text{(f }\circ \text{ h)} $ is continuous at  $ \text{c} $ .

Hence, the function  $ \text{f(x)=(g }\circ \text{ h)(x)=cos(}{{\text{x}}^{2}}\text{)} $  is continuous.


32. Show that the function defined by  $ \text{f(x)=}\left| \text{cosx} \right| $  is a continuous function.

Ans: The given function is  $ \text{f(x)=}\left| \text{cosx} \right| $ .

Note that, the function  $ \text{f} $  is defined for all real numbers. So, the function  $ \text{f} $  can be expressed as the composition of two functions as,  $ \text{f=g}\circ \text{h} $ , where  $ \text{g(x)=}\left| \text{x} \right| $  and  $ \text{h(x)=cosx} $ .

 $ [\because (\text{goh)(x)=g(h(x))=g(cosx)=}\left| \text{cosx} \right|\text{=f(x) }\!\!]\!\!\text{ } $ 

Now, it is to be proved that the functions  $ \text{g(x)=}\left| \text{x} \right| $  and  $ \text{h(x)=cosx} $  are continuous.

Remember that,  $ \text{g(x)=}\left| \text{x} \right| $ , can be written as

 $ \text{g(x)=}\left\{ \begin{align} & \text{-x, if x}\,<\,\text{0} \\ & \text{ x, if x}\ge \text{0} \\ \end{align} \right. $ .

Now, since the function  $ \text{g} $  is defined for every real number, let consider  $ \text{c} $  be a real number.

Then there may arise three cases, either  $ \text{c}<\text{0} $ , or  $ \text{c}>\text{0} $ , or  $ \text{c=0} $ .

Let's discuss the cases one after another.

Case I. When  $ \text{c}<\text{0} $ .

Then,  $ \text{g(c)=-c} $ .

Also,  $ \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{g(x)=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{(-x)=-c} $ .

Therefore,  $ \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{g(x)=g(c)} $ .

Hence, the function  $ \text{g} $  is continuous at every point  $ \text{x} $ , for  $ \text{x}<\text{0} $ .

Case II. When  $ \text{c}>\text{0} $ .

Then,  $ \text{g(c)=c} $ .

Also,  $ \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{g(x)=}\underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{x=c} $ .

Therefore,  $ \underset{\text{x}\to \text{c}}{\mathop{\lim }}\,\text{g(x)=g(c)} $ .

Hence, the function  $ \text{g} $  is continuous at every point  $ \text{x} $  for  $ \text{x}>\text{0} $ .

Case III. When  $ \text{c=0} $ .

Then,  $ \text{g(c)=g(0)=0} $ .

Now, the left-hand-limit of the function  $ \text{g} $  at  $ \text{x=0} $  is

 $ \underset{\text{x}\to {{0}^{-}}}{\mathop{\lim }}\,\text{g(x)=}\underset{\text{x}\to {{0}^{\text{-}}}}{\mathop{\lim }}\,\text{(-x)=0} $  and the right-hand-limit is

 $ \underset{\text{x}\to {{0}^{+}}}{\mathop{\lim }}\,\text{g(x)=}\underset{\text{x}\to {{0}^{+}}}{\mathop{\lim }}\,\text{(x)=0} $ .

Therefore,  $ \underset{\text{x}\to {{0}^{-}}}{\mathop{\lim }}\,\text{g(x)=}\underset{\text{x}\to {{0}^{+}}}{\mathop{\lim }}\,\text{g(x)=g(0)} $ .

Hence, the function  $ \text{g} $  is continuous at  $ \text{x=0} $ .

By observing the above three discussions, we can conclude that the function  $ \text{g} $  is continuous at every real point.

Now, since the function  $ \text{h(x)=cosx} $  is defined for all real numbers, let 's consider  $ c $  be a real number. Then, substitute  $ \text{x=c+h} $  into the function  $ \text{h} $ .

So, when  $ \text{x}\to \text{c} $ , then  $ \text{h}\to 0 $ .

Then, we have

 $ \text{h}\left( \text{c} \right)=\text{cosc} $  and

$\underset{\text{x}\to\text{c}}{\mathop{\text{lim}}}\,\text{h(x)=}\underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{cosx} $ 

 $ \text{=}\underset{\text{h}\to \text{0}}{\mathop{\text{lim}}}\,\text{cos}\left( \text{c+h} \right) $ 

 $ \text{=}\underset{\text{h}\to \text{0}}{\mathop{\text{lim}}}\,\left[ \text{cosc}\cdot \text{cosh-sinc}\cdot \text{sinh} \right] $ 

 $ \text{=}\underset{\text{h}\to \text{0}}{\mathop{\text{lim}}}\,\left( \text{cosc}\cdot \text{cosh} \right)-\underset{\text{h}\to \text{0}}{\mathop{\text{lim}}}\,\left( \text{sinc}\cdot \text{sinh} \right) $ 

 $ \begin{align} & \text{=cosc}\cdot \text{cos0}-\text{sinc}\cdot \text{sin0} \\ & \text{=cosc }\!\!\times\!\!\text{ 1-sinc }\!\!\times\!\!\text{ 0} \\ & \text{=cosc} \\ \end{align} $

Therefore,  $ \underset{x\to \text{c}}{\mathop{\lim }}\,\text{h}\left( \text{x} \right)=\text{h}\left( \text{c} \right) $ .

Hence, the function  $ \text{h(x)=cosx} $  is continuous.

Now remember that, for real valued functions  $ \text{g} $  and  $ \text{h} $ , such that  $ \text{(g}\circ \text{h)} $  is defined at  $ \text{x=c} $  only if  $ \text{g} $  is continuous at  $ \text{c} $  and  $ \text{f} $  is continuous at  $ \text{g(c)} $ , then the composition functions  $ \text{(f }\circ \text{ g)} $  is continuous at  $ \text{x=c} $ .

Thus, the function  $ \text{f(x)=(goh)(x)=g(h(x))=g(cosx)=}\left| \text{cosx} \right| $  is continuous.


33. Examine that  $ \mathbf{sin}\left| \mathbf{x} \right| $  is a continuous function. 

Ans: First suppose that,  $ f\left( \text{x} \right)=\sin \left| \text{x} \right| $ .

Now, note that the function  $ \text{f} $  is defined for all real numbers and so  $ \text{f} $  can be expressed as the composition of  functions as,  $ \text{f=g}\circ \text{h,} $  where  $ \text{g(x)=sinx} $  and  $ \text{h(x)=}\left| \text{x} \right| $ .

$ \left[ \text{(g}\circ \text{h)(x)=g(h(x))=g(}\left| \text{x}\right|\text{)=sin}\left| \text{x} \right|\text{=f(x)} \right] $ 

So, it is to be proved that the functions  $ \text{g(x)=sinx} $  and  $ \text{h(x)=}\left| \text{x} \right| $  are continuous.

Now, remember that, the function  $ \text{h(x)=}\left| \text{x} \right| $  can be written as 

 $ \text{h(x)=}\left\{ \begin{align}& \text{-x, if x}<\text{0} \\ & \text{x,  if x}\ge \text{0} \\ \end{align} \right. $ 

Note that, the function  $ \text{h} $  is defined for every real number, and so let consider  $ \text{c} $  be a real number. 

Then, there may arise three cases, either  $ \text{c}<\text{0} $ , or  $ \text{c}>\text{0} $ , or  $ \text{c=0} $ .

Let us discuss the cases one after another.

Case I. When  $ \text{c}<\text{0} $ .

Then  $ \text{h(c)=-c} $ .

Also,  $ \underset{\text{x}\to\text{c}}{\mathop{\text{lim}}}\,\text{(-x)=}\underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{x=-c} $ .

Therefore,  $ \underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{h(x)=h(c)} $ .

Hence, the function  $ \text{h} $  is continuous at every point  $ \text{x} $   for  $ \text{x}<\text{0} $ .

Case II. When  $ \text{c}>\text{0} $ .

Then,  $ \text{h(c)=c} $ 

Also,  $ \underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{(-x)=}\underset{\text{x}\to\text{c}}{\mathop{\text{lim}}}\,\text{x=c} $ .

Therefore,  $ \underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{h(x)=h(c)} $ .

Thus, the function  $ \text{h} $  is continuous at every point  $ \text{x} $   for  $ \text{x}>\text{0} $ .

Case III. When  $ \text{c = 0} $ .

 Then,  $ \text{h(c)=h(0)=0} $ .

Also, the left-hand-limit of the function  $ \text{h} $  at  $ \text{x=0} $  is

 $ \underset{\text{x}\to {{\text{0}}^{\text{-}}}}{\mathop{\text{lim}}}\,\text{h(x)=}\underset{\text{x}\to {{\text{0}}^{\text{-}}}}{\mathop{\text{lim}}}\,\text{(-x)=0} $  and the right-hand -limit is 

 $ \underset{\text{x}\to {{\text{0}}^{\text{+}}}}{\mathop{\text{lim}}}\,\text{h(x)=}\underset{\text{x}\to {{\text{0}}^{\text{+}}}}{\mathop{\text{lim}}}\,\text{(x)=0} $ .

Therefore,  $ \underset{\text{x}\to {{\text{0}}^{\text{-}}}}{\mathop{\text{lim}}}\,\text{h(x)=}\underset{\text{x}\to {{\text{0}}^{\text{+}}}}{\mathop{\text{lim}}}\,\text{(x)=h(0)} $ .

Thus, the function  $ \text{h} $  is continuous at  $ \text{x = 0} $ .

By observing the above three discussions, we can conclude that the function  $ \text{h} $  is continuous at every point. 

Again, since the function  $ \text{g(x)=sinx} $  is defined for all real numbers, so let consider  $ \text{c} $  be a real number and substitute  $ \text{x=c+k} $  into the function.

Now, when  $ \text{x}\to \text{c} $  then  $ \text{k }\to \text{ 0} $ .

Then, we have

 $ \text{g(c)=sinc} $ .

Also,

 $ \begin{align} & \underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{g(x)=}\underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{sinx} \\ & \text{=}\underset{\text{k}\to \text{0}}{\mathop{\text{lim}}}\,\text{sin}\left( \text{c+k} \right) \\ & \text{=}\underset{\text{k}\to \text{0}}{\mathop{\text{lim}}}\,\left[ \text{sinc}\cdot \text{cosk+cosc}\cdot \text{sink} \right] \\ & \text{=}\underset{\text{k}\to \text{0}}{\mathop{\text{lim}}}\,\left( \text{sinc}\cdot \text{cosk} \right)\text{+}\underset{\text{h}\to \text{0}}{\mathop{\text{lim}}}\,\left( \text{cosc}\cdot \text{sink} \right) \\ & \text{=sinc}\cdot \text{cos0+cosc}\cdot \text{sin0} \\ & \text{=sinc+0} \\ & \text{=sinc} \\ \end{align} $ 

Therefore,  $ \underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{g(x)=g(c)} $ .

Hence, the function  $ \text{g} $  is continuous.

Now, remember that, for any two real valued functions  $ \text{g} $  and  $ \text{h} $ , such that the composition of functions  $ \text{g}\circ \text{h} $  is defined at  $ \text{c} $ , if  $ \text{g} $  is continuous at  $ \text{c} $  and  $ \text{f} $  is continuous at  $ \text{g(c)} $ , then the composition function  $ \text{g}\circ \text{h} $  is continuous at  $ \text{c} $ .

Thus, the function  $ \text{(g}\circ \text{h)(x)=g(h(x))=g(}\left| \text{x} \right|\text{)=sin}\left| \text{x} \right|\text{=f(x)} $  is continuous.


34.Find all the points of discontinuity of  $ \mathbf{f} $  defined by  $ \mathbf{f(x)=}\left| \mathbf{x} \right|\mathbf{-}\left| \mathbf{x}+\mathbf{1} \right| $ . 

Ans: The given function is  $ \text{f(x)=}\left| \text{x} \right|\text{-}\left| \text{x+1} \right| $ .

Let consider two functions

 $ \text{g(x)=}\left| \text{x} \right| $  and  $ \text{h(x)=}\left| \text{x+1} \right| $ .

Then we get,   $ \text{f=g-h} $ .

Now, the function  $ \text{g(x)=}\left| \text{x} \right| $  can be written as

 $ \text{g(x)=}\left\{ \begin{align} & \text{-x, if x}<\text{0} \\ & \text{x, if x}\ge \text{0} \\ \end{align} \right. $ .

Note that, the function  $ \text{g} $  is defined for every real number and so let consider  $ \text{c} $  be a real number.

Then there may arise three cases, either  $ \text{c}<\text{0} $ , or  $ \text{c}>\text{0} $ , or  $ \text{c=0} $ .

Let us discuss the cases one after another.

Case I. When  $ \text{c}<\text{0} $ .

Then,  $ \text{g(c)=g(0)=-c} $ .

Also,  $ \underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{g(x)=}\underset{\text{x}\to\text{c}}{\mathop{\text{lim}}}\,\text{(-x)=-c} $ .

Therefore,  $ \underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{g(x)=g(c)} $ .

Hence, the function  $ \text{g} $  is continuous at every point  $ \text{x} $  for  $ \text{x}<\text{0} $ .

Case II. When  $ \text{c}>\text{0} $ .

Then  $ \text{g(c)=c} $ .

Also,  $ \underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{g(x)=}\underset{\text{x}\to\text{c}}{\mathop{\text{lim}}}\,\text{x=c} $ .

Therefore,  $ \underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{g(x)=g(c)} $ .

Hence, the function  $ \text{g} $  is continuous at every point  $ \text{x} $ , where  $ \text{x}>\text{0} $ .

Case III. When  $ \text{c = 0} $ .

Then  $ \text{g(c)=g(0)=0} $ .

Also, the left-hand-limit of the function  $ \text{g} $  at  $ \text{x=0} $  is

 $ \underset{\text{x}\to {{\text{0}}^{\text{-}}}}{\mathop{\text{lim}}}\,\text{g(x)=}\underset{\text{x}\to {{\text{0}}^{\text{-}}}}{\mathop{\text{lim}}}\,\text{(-x)=0} $  and the right-hand-limit is

$ \underset{\text{x}\to {{\text{0}}^{\text{+}}}}{\mathop{\text{lim}}}\,\text{g(x)=}\underset{\text{x}\to {{\text{0}}^{\text{+}}}}{\mathop{\text{lim}}}\,\text{(x)=0} $ .

Therefore,  $ \underset{\text{x}\to {{\text{0}}^{\text{-}}}}{\mathop{\text{lim}}}\,\text{g(x)=}\underset{\text{x}\to {{\text{0}}^{\text{+}}}}{\mathop{\text{lim}}}\,\text{(x)=g(0)} $ .

Hence, the function  $ \text{g} $  is continuous at  $ \text{x = 0} $ .

Thus, we can conclude by observing the above three discussions that  $ \text{g} $  is continuous at every real point.

Now, remember that, the function  $ \text{h(x)=}\left| \text{x+1} \right| $  can be written as

 $ \text{h(x)=}\left\{ \begin{align} & \text{-x(x+1), if, x}<\text{-1} \\ & \text{x+1, }\,\,\,\,\,\,\,\,\text{if, x}\ge \text{-1} \\ \end{align} \right. $ .

Note that, the function  $ \text{h} $  is defined for all real numbers, and so let consider  $ \text{c} $  be a real number.

Case I. When  $ \text{c}<\text{-1} $ .

Then  $ \text{h(c)=-(c+1)} $ .

Also,  $ \underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\left[ \text{-(x+1)} \right]\text{=-(c+1)} $ .

Therefore,  $ \underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{h(x)=h(c)} $ .

Hence, the function  $ \text{h} $  attains continuity at every real point  $ \text{x} $ , where  $ \text{x}<\text{-1} $ .

Case II. When  $ \text{c}>\text{-1} $ .

Then,  $ \text{h(c)=c+1} $ .

Also,  $ \underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{h(x)=}\underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{(x+1)=(c+1)} $ .

Therefore,  $ \underset{\text{x}\to \text{c}}{\mathop{\text{lim}}}\,\text{h(x)=h(c)} $ .

Hence, the function  $ \text{h} $  satisfies continuity at every real point  $ \text{x} $  for  $ \text{x}>\text{-1} $ .

Case III. When  $ \text{c =-1} $ .

Then,  $ \text{h(c)=h(-1)=-1+1=0} $ .

Also, the left-hand-limit of the function  $ \text{h} $  at  $ \text{x=1} $  is

 $ \underset{\text{x}\to {{\text{1}}^{\text{-}}}}{\mathop{\text{lim}}}\,\text{h(x)=}\underset{\text{x}\to {{\text{1}}^{\text{-}}}}{\mathop{\text{lim}}}\,\left[ \text{-(x+1)} \right]\text{=-(-1+1)=0} $  and the right-hand-limit is

 $ \underset{\text{x}\to {{\text{1}}^{\text{+}}}}{\mathop{\text{lim}}}\,\text{h(x)=}\underset{\text{x}\to {{\text{1}}^{\text{+}}}}{\mathop{\text{lim}}}\,\text{(x+1)=(-1+1)=0} $ .

Therefore,  $ \underset{\text{x}\to {{\text{1}}^{\text{-}}}}{\mathop{\text{lim}}}\,\text{h=}\underset{\text{x}\to {{\text{1}}^{\text{+}}}}{\mathop{\text{lim}}}\,\text{h(x)=h(-1)} $ .

Thus, the function  $ \text{h} $  satisfies continuity at  $ \text{x=-1} $ .

Hence, by observing the above three discussions, we can conclude that the function  $ \text{h} $  is continuous for every real point.

Now, since the functions  $ \text{g} $  and  $ \text{h} $  are both continuous, so the function  $ \text{f=g-h} $  is also continuous.

Hence, the function  $ \text{f} $  does not have any discontinuity points.


Exercise 5.2

Differentiate the function with respect to  $ \mathbf{x} $  in Exercises 1 to 8.

1.  $ \text{f}\left( \mathbf{x} \right)\text{=sin(}{{\text{x}}^{\text{2}}}\text{+5)} $  

Ans: Let  $ \text{f(x)=sin(}{{\text{x}}^{\text{2}}}\text{+5), u(x)=}{{\text{x}}^{\text{2}}}\text{+5,} $  and  $ \text{v(t)=sint} $ 

Then,  $ \text{(v}\circ \text{u)(x)=v(u(x))=v(}{{\text{x}}^{\text{2}}}\text{+5)=sin(}{{\text{x}}^{\text{2}}}\text{+5)=f(x)} $ 

Therefore,  $ \text{f} $  is a composition of two functions  $ \text{u} $  and  $ \text{v} $ .

Substitute  $ \text{t=u(x)=}{{\text{x}}^{\text{2}}}\text{+5} $ .

Then, it gives 

 $ \begin{align} & \dfrac{\text{dv}}{\text{dt}}\text{=}\dfrac{\text{d}}{\text{dt}}\text{(sint)=cost=cos(}{{\text{x}}^{\text{2}}}\text{+5)} \\ & \dfrac{\text{dt}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{x}}^{\text{2}}}\text{+5)=}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{x}}^{\text{2}}}\text{)+}\dfrac{\text{d}}{\text{dx}}\text{(5)=2x+0=2x} \\ \end{align} $ 

Applying the chain rule of derivatives gives 

$\dfrac{\text{df}}{\text{dx}}\text{=}\dfrac{\text{dv}}{\text{dt}}\text{.}\dfrac{\text{dt}}{\text{dx}}\text{=cos(}{{\text{x}}^{\text{2}}}\text{+5) }\!\!\times\!\!\text{ 2x=2xcos(}{{\text{x}}^{\text{2}}}\text{+5)} $ 

An alternate method.

 $ \begin{align}& \dfrac{\text{df}}{\text{dx}}\text{=}\dfrac{\text{dv}}{\text{dt}}\text{.}\dfrac{\text{dt}}{\text{dx}}\text{=cos(}{{\text{x}}^{\text{2}}}\text{+5)}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{x}}^{\text{2}}}\text{+5)} \\ & \text{=cos(}{{\text{x}}^{\text{2}}}\text{+5)}\text{.}\left[\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{x}}^{\text{2}}}\text{)+}\dfrac{\text{d}}{\text{dx}}\text{(5)} \right] \\ & \text{=cos(}{{\text{x}}^{\text{2}}}\text{+5)}\text{. }\!\![\!\!\text{ 2x+0 }\!\!]\!\!\text{ } \\ & \text{=2xcos(}{{\text{x}}^{\text{2}}}\text{+5)} \\ \end{align} $ 

Hence, the derivative of the function  $ \text{f}\left( \text{x} \right)=\text{sin}\left( {{\text{x}}^{2}}+5 \right) $  is  $ \text{2xcos(}{{\text{x}}^{\text{2}}}\text{+5)} $ .


2.  $ \text{f}\left( \mathbf{x} \right)\text{=cos(sinx)} $  

Ans: Let suppose that,  $ \text{f(x)=cos(sinx), u(x)=sinx,} $  and  $ \text{v(t)=cost} $ 

Then,  $ \text{(v}\circ \text{u)(x)=v(u(x))=v(sinx)=cos(sinx)=f(x)} $ 

Therefore, it is observed that  $ \text{f} $  is the composition of two functions  $ \text{u} $  and  $ \text{v} $ .

Now, substitute  $ \text{t=u(x)=sinx} $ .

Then,

 $ \dfrac{\text{dv}}{\text{dt}}\text{=}\dfrac{\text{d}}{\text{dt}}\text{(cost)=-sint=-sin(sinx)} $  and

 $ \dfrac{\text{dt}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{(sinx)=cosx} $ .

Applying the chain rule of derivatives gives

 $ \dfrac{\text{df}}{\text{dx}}\text{=}\dfrac{\text{dv}}{\text{dt}}\text{.}\dfrac{\text{dt}}{\text{dx}}\text{= -sin(sinx)}\text{.cosx=-cosxsin(sinx)} $ 

An alternate method. 

 $ \dfrac{\text{d}}{\text{dx}}\left[ \text{cos(sinx)} \right]\text{=-sin(sinx)}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(sinx)=-sin(sinx)cosx=-cosxsin(sinx)} $ .

Hence, the derivative of the function  $ \text{f}\left( \text{x} \right)=\text{cos}\left( \text{sinx} \right) $  is  $ \text{-cosxsin(sinx)} $ .


3.  $ \text{f}\left( \mathbf{x} \right)\text{=sin(ax+b)} $  .

Ans: Let suppose that,  $ \text{f(x)=sin(ax+b), u(x)=ax+b,} $ , and  $ \text{v(t)=sint} $  

Then we get,  $ \text{(v}\circ \text{u)(x)=v(u(x))=v(ax+b)=sin(ax+b)=f(x)} $ .

It is observed that the function  $ \text{f} $  is the composition of two functions  $ \text{u} $  and  $ \text{v} $  . 

Now, substitute  $ \text{t=u(x)=ax+b} $ .

Therefore,

 $ \dfrac{\text{dv}}{\text{dt}}\text{=}\dfrac{\text{d}}{\text{dt}}\text{(sint)=cost=cos(ax+b)} $  and

 $ \dfrac{\text{dt}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{(ax+b)=}\dfrac{\text{dt}}{\text{dx}}\text{(ax)+}\dfrac{\text{d}}{\text{dx}}\text{(b)=a+0=a} $ .

Applying the chain rule derivatives, gives 

 $ \dfrac{\text{df}}{\text{dx}}\text{=}\dfrac{\text{dv}}{\text{dt}}\text{.}\dfrac{\text{dt}}{\text{dx}}\text{=cos(ax+b)}\text{.a=acos(ax+b)} $ .

Alternate method.

 $ \begin{align} & \dfrac{\text{d}}{\text{dx}}\left[ \text{sin(ax+b)} \right]\text{=cos(ax+b)}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(ax+b)} \\ & \text{=cos(ax+b) }\!\!\times\!\!\text{ }\left[ \dfrac{\text{d}}{\text{dx}}\text{(ax)+}\dfrac{\text{d}}{\text{dt}}\text{(b)} \right] \\  & \text{=cos(ax+b) }\!\!\times\!\!\text{ (a+0)} \\  & \text{=acos(ax+b)} \\ \end{align} $ 

Hence, the derivative of the function  $ \text{f}\left( \text{x} \right)=\sin \left( \text{ax+b} \right) $  is  $ \text{acos(ax+b)} $ .


4.  $ \mathbf{f}\left( \mathbf{x} \right)=\mathbf{sec(tan(}\sqrt{\mathbf{x}}\mathbf{))} $  with respect to  $ \mathbf{x} $ .

Ans: Let suppose that,  $ \text{f(x)=sec(tan(}\sqrt{\text{x}}\text{)), u(x)=}\sqrt{\text{x}}\text{,v(t)=tant,} $  and  $ \text{w(s)=secs} $ 

Then, we get,  $ \text{(w}\circ \text{v}\circ \text{u)(x)=w }\!\![\!\!\text{ v(u(x)) }\!\!]\!\!\text{ =w }\!\![\!\!\text{ v(}\sqrt{\text{x}}\text{) }\!\!]\!\!\text{ =w(tan}\sqrt{\text{x}}\text{)=f(x)} $ .

It is observed that the function  $ \text{g} $  is the composition of three functions  $ \text{u} $  ,  $ \text{v} $  and  $ \text{w} $ . 

Now, substitute  $ \text{s=v(t)} $  and  $ \text{t=u(x)=}\sqrt{\text{x}} $ .

Then, we get

 $ \begin{align}& \dfrac{\text{dw}}{\text{ds}}\text{=}\dfrac{\text{d}}{\text{ds}}\text{(secs)=secs=sec(tant) }\!\!\times\!\!\text{ tan(tant)}\ \text{  }\!\![\!\!\text{ s=tant }\!\!]\!\!\text{ } \\  \text{=sec(tan}\sqrt{\text{x}}\text{) }\!\!\times\!\!\text{ tan(tan}\sqrt{\text{x}}\text{)}\ \text{   }\!\![\!\!\text{ t=}\sqrt{\text{x}}\text{ }\!\!]\!\!\text{ }  \end{align} $  

Thus, applying the chain rule of derivatives gives

 $ \begin{align}& \dfrac{\text{dt}}{\text{dx}}\text{=}\dfrac{\text{dw}}{\text{ds}}\text{.}\dfrac{\text{ds}}{\text{dt}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{dt}}{\text{dx}} \\ & \text{=sec(tan(}\sqrt{\text{x}}\text{) }\!\!\times\!\!\text{ (tan(}\sqrt{\text{x}}\text{)xse}{{\text{c}}^{\text{2}}}\sqrt{\text{x}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{2}\sqrt{\text{x}}} \\ & =\dfrac{\text{1}}{\text{2}\sqrt{\text{x}}}\text{se}{{\text{c}}^{\text{2}}}\sqrt{\text{x}}\text{sec(tan}\sqrt{\text{x}}\text{)tan(tan}\sqrt{\text{x}}\text{)} \\  & =\dfrac{\text{se}{{\text{c}}^{\text{2}}}\sqrt{\text{x}}\text{sec(tan}\sqrt{\text{x}}\text{)tan(tan}\sqrt{\text{x}}\text{)}}{\text{2}\sqrt{\text{x}}}  \end{align} $ 

An alternate method.

 $ \begin{align}& \dfrac{\text{d}}{\text{dx}}\left[ \text{sec(tan(}\sqrt{\text{x}}\text{))}\right]\text{=sec(tan(}\sqrt{\text{x}}\text{)}\text{.(tan(}\sqrt{\text{x}}\text{)}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(tan(}\sqrt{\text{x}}\text{)} \\ & \text{=sec(tan}\sqrt{\text{x}}\text{) }\!\!\times\!\!\text{ tan(tan}\sqrt{\text{x}}\text{) }\!\!\times\!\!\text{ se}{{\text{c}}^{\text{2}}}\text{(}\sqrt{\text{x}}\text{) }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(}\sqrt{\text{x}}\text{)} \\ & \text{=sec(tan}\sqrt{\text{x}}\text{) }\!\!\times\!\!\text{ tan(tan}\sqrt{\text{x}}\text{) }\!\!\times\!\!\text{ se}{{\text{c}}^{\text{2}}}\text{(}\sqrt{\text{x}}\text{) }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{2}\sqrt{\text{x}}} \\ & =\dfrac{\text{se}{{\text{c}}^{\text{2}}}\sqrt{\text{x}}\text{sec(tan}\sqrt{\text{x}}\text{)tan(tan}\sqrt{\text{x}}\text{)}}{\text{2}\sqrt{x}}  \end{align} $ 

Hence, the derivative of the function  $ \text{f}\left( \text{x} \right)=\text{sec}\left( \text{tan}\left( \sqrt{\text{x}} \right) \right) $  is  $ \dfrac{\text{se}{{\text{c}}^{\text{2}}}\sqrt{\text{x}}\text{sec(tan}\sqrt{\text{x}}\text{)tan(tan}\sqrt{\text{x}}\text{)}}{\text{2}\sqrt{x}} $ .

5.  $ \text{f}\left( \mathbf{x} \right)\mathbf{=}\dfrac{\text{sin(ax+b)}}{\text{cos(cx+d)}} $ .

Ans: The given function is  $ \text{f(x)=}\dfrac{\text{sin(ax+b)}}{\text{cos(cx+d)}} $ .

Now, let  $ \text{g(x)=sin(ax+b)} $  and  $ \text{h(x)=cos(cx+d)} $ .

Here we will use the divide formula of derivatives  $ \text{{f}'}\,\text{=}\dfrac{\text{g }\!\!'\!\!\text{ h-gh }\!\!'\!\!\text{ }}{{{\text{h}}^{\text{2}}}} $ .    …… (1)

First, consider the function  $ \text{g(x)=sin(ax+b)} $ .

Let's assume   $ \text{u(x)=ax+b,} $  and  $ \text{v(t)=sint} $ .

Then, we get  $ \text{(v}\circ \text{u)(x)=v(u(x))=v(ax+b)=sin(ax+b)=g(x)} $ .

Therefore, we observe that the function  $ \text{g} $  is the composition of two functions,  $ \text{u} $  and  $ \text{v} $  .

So, substitute  $ \text{t=u(x)=ax+b} $ .

Then,

 $ \dfrac{\text{dv}}{\text{dt}}\text{=}\dfrac{\text{d}}{\text{dt}}\text{(sint)=cost=cos(ax+b)} $  and

 $ \dfrac{\text{dt}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{(ax+b)=}\dfrac{\text{dt}}{\text{dx}}\text{(ax)+}\dfrac{\text{d}}{\text{dx}}\text{(b)=a+0=a} $ .

Therefore, applying the chain rule of derivatives gives

 $ \text{{g}'=}\dfrac{\text{dg}}{\text{dx}}\text{=}\dfrac{\text{dv}}{\text{dt}}\text{.}\dfrac{\text{dt}}{\text{dx}}\text{=cos(ax+b)}\text{.a=acos(ax+b)} $ .

Now, consider the function  $ \text{h(x)=cos(cx+d)} $ .

Let's suppose  $ \text{p(x)=cx+d,} $  and  $ \text{q(t)=cosy} $ .

Then, we have  $ \text{(q}\circ \text{p)(x)=q(p(x))=q(cx+d)=cos(cx+d)=h(x)} $ .

Therefore, the function  $ \text{h} $  is the composition of two functions  $ \text{p} $  and  $ \text{q} $ . 

Now, substitute  $ \text{y=p(x)=cx+d} $ .

Then we have,

 $ \dfrac{\text{dq}}{\text{dy}}\text{=}\dfrac{\text{d}}{\text{dy}}\text{(cosy)=-siny=-sin(cx+d)} $  and

 $ \dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{(cx+d)=}\dfrac{\text{d}}{\text{dx}}\text{(cx)+}\dfrac{\text{d}}{\text{dx}}\text{(d)=c} $ .

Therefore, applying the chain rule of derivatives gives

 $ \text{{h}'=}\dfrac{\text{dh}}{\text{dx}}\text{=}\dfrac{\text{dq}}{\text{dy}}\text{.}\dfrac{\text{dy}}{\text{dx}}\text{=-sin(cx+d) }\!\!\times\!\!\text{ c=-csin(cx+d)} $ .

Now, substituting all the obtained derivatives into the formula (1) gives

 $ \begin{align}& \text{{f}'=}\dfrac{\text{acos(ax+b) }\!\!\times\!\!\text{ cos(cx+d)+sin(ax+b)}\times \text{csin(cx+d)}}{\text{ }\!\![\!\!\text{ cos(cx+d}{{\text{ }\!\!]\!\!\text{ }}^{\text{2}}}} \\ & =\dfrac{\text{acos(ax+b)}}{\text{cos(cx+d)}}\text{+csin(ax+b) }\!\!\times\!\!\text{ }\dfrac{\text{sin(cx+d)}}{\text{cos(cx+d)}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{cos(cx+d)}} \\ & \text{=acos(ax+b)sec(cx+d)+csin(ax+b)tan(cx+d)sec(cx+d)} \end{align} $

Hence, the derivative of the function  $ \text{f(x)=}\dfrac{\text{sin(ax+b)}}{\text{cos(cx+d)}} $  is  $ \text{acos(ax+b)sec(cx+d)+c.sin(ax+b)tan(cx+d)sec(cx+d)} $ .


6.  $ \mathbf{f}\left( \mathbf{x} \right)=\mathbf{cos}\left( {{\mathbf{x}}^{\mathbf{3}}} \right)\mathbf{\times si}{{\mathbf{n}}^{\mathbf{2}}}\mathbf{(}{{\mathbf{x}}^{\mathbf{5}}}\mathbf{)} $ . 

Ans: The given function is  $ \text{f}\left( \text{x} \right)\text{=cos}\left( {{\text{x}}^{\text{3}}} \right)\text{ }\!\!\times\!\!\text{ si}{{\text{n}}^{\text{2}}}\left( {{\text{x}}^{\text{5}}} \right) $ .

Then,

 $ \begin{align}& \dfrac{\text{d}}{\text{dx}}\text{ }\!\![\!\!\text{ cos}{{\text{x}}^{\text{3}}}\text{ }\!\!\times\!\!\text{ si}{{\text{n}}^{\text{2}}}\text{(}{{\text{x}}^{\text{5}}}\text{) }\!\!]\!\!\text{ =si}{{\text{n}}^{\text{2}}}\text{(}{{\text{x}}^{\text{5}}}\text{) }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(cos}{{\text{x}}^{\text{3}}}\text{)+cos}{{\text{x}}^{\text{3}}}\times \dfrac{\text{d}}{\text{dx}}\text{ }\!\![\!\!\text{ si}{{\text{n}}^{\text{2}}}\text{(}{{\text{x}}^{\text{5}}}\text{) }\!\!]\!\!\text{ } \\ & \text{=si}{{\text{n}}^{\text{2}}}\text{(}{{\text{x}}^{\text{5}}}\text{) }\!\!\times\!\!\text{ (-sin}{{\text{x}}^{\text{3}}}\text{) }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{x}}^{\text{3}}}\text{)+cos}{{\text{x}}^{\text{3}}}\text{+2sin(}{{\text{x}}^{\text{5}}}\text{) }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{ }\!\![\!\!\text{ sin}{{\text{x}}^{\text{5}}}\text{ }\!\!]\!\!\text{ } \\ & \text{=-sin}{{\text{x}}^{\text{3}}}\text{si}{{\text{n}}^{\text{2}}}\text{(}{{\text{x}}^{\text{5}}}\text{) }\!\!\times\!\!\text{ 3}{{\text{x}}^{\text{2}}}\text{+2sin}{{\text{x}}^{\text{5}}}\text{cos}{{\text{x}}^{\text{3}}}\text{ }\!\!\times\!\!\text{ cos}{{\text{x}}^{\text{5}}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{x}}^{\text{5}}}\text{)} \\ & \text{=-3}{{\text{x}}^{\text{2}}}\text{sin}{{\text{x}}^{\text{3}}}\text{ }\!\!\times\!\!\text{ si}{{\text{n}}^{2}}\text{(}{{\text{x}}^{\text{5}}}\text{)+2sin}{{\text{x}}^{\text{5}}}\text{cos}{{\text{x}}^{\text{5}}}\text{cos}{{\text{x}}^{\text{3}}}\text{ }\!\!\times\!\!\text{ 5}{{\text{x}}^{\text{4}}} \\ & \text{=10}{{\text{x}}^{\text{4}}}\text{sin}{{\text{x}}^{\text{5}}}\text{cos}{{\text{x}}^{\text{5}}}\text{cos}{{\text{x}}^{\text{3}}}\text{-3}{{\text{x}}^{\text{2}}}\text{sin}{{\text{x}}^{\text{3}}}\text{si}{{\text{n}}^{\text{2}}}\text{(}{{\text{x}}^{\text{5}}}) \\ \end{align} $ 

Hence, the derivative of the function  $ \text{f}\left( \text{x} \right)\text{=cos}\left( {{\text{x}}^{\text{3}}} \right)\text{ }\!\!\times\!\!\text{ si}{{\text{n}}^{\text{2}}}\left( {{\text{x}}^{\text{5}}} \right) $  is $ \text{10}{{\text{x}}^{\text{4}}}\text{sin}{{\text{x}}^{\text{5}}}\text{cos}{{\text{x}}^{\text{5}}}\text{cos}{{\text{x}}^{\text{3}}}\text{-3}{{\text{x}}^{\text{2}}}\text{sin}{{\text{x}}^{\text{3}}}\text{si}{{\text{n}}^{\text{2}}}\text{(}{{\text{x}}^{\text{5}}}) $ .


7.  $ \text{f}\left( \mathbf{x} \right)\text{=}\mathbf{2}\sqrt{\mathbf{cot}\left( {{\mathbf{x}}^{\mathbf{2}}} \right)} $ .

Ans:  The given function is  $ \text{f}\left( \text{x} \right)=2\sqrt{\cot \left( {{x}^{2}} \right)} $ .

Then,

 $ \begin{align} & \dfrac{\text{d}}{\text{dx}}\left[ 2\sqrt{\cot \left( {{x}^{2}} \right)} \right] \\ & \text{=2}\cdot \dfrac{\text{1}}{2\sqrt{\cot \left( {{x}^{2}} \right)}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\left[ \text{cot(}{{\text{x}}^{\text{2}}}\text{)} \right] \\ & =-\dfrac{1}{\sqrt{\cot \left( {{x}^{2}} \right)}}\text{cose}{{\text{c}}^{\text{2}}}\text{(}{{\text{x}}^{\text{2}}}\text{) }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{x}}^{\text{2}}}\text{)} \\ & =-\dfrac{1}{\sqrt{\cot \left( {{x}^{2}} \right)}}\text{cose}{{\text{c}}^{\text{2}}}\text{(}{{\text{x}}^{\text{2}}}\text{)}\times \text{2x} \\ & \text{=-2xcose}{{\text{c}}^{2}}\left( {{\text{x}}^{2}} \right)\sqrt{\tan \left( {{\text{x}}^{2}} \right)}  \end{align} $ 

Hence, the derivative of the function  $ \text{f}\left( \text{x} \right)=2\sqrt{\cot \left( {{x}^{2}} \right)} $  is  $ \text{-2xcose}{{\text{c}}^{2}}\left( {{\text{x}}^{2}} \right)\sqrt{\tan \left( {{\text{x}}^{2}} \right)} $ .


8.  $ \text{f}\left( \mathbf{x} \right)\text{=cos(}\sqrt{\text{x}}\text{)} $ .

Ans:  The given function is  $ \text{f(x)=cos(}\sqrt{\text{x}}\text{)} $ 

Now, let  $ \text{u(x)=}\sqrt{\text{x}} $  and  $ \text{v(t)=cost} $ .

Then, we have,  $ \text{(v}\circ\text{u)(x)=v(u(x))=v(}\sqrt{\text{x}}\text{)=cos}\sqrt{\text{x}}\text{=f(x)} $ .

It is observed that the function  $ \text{f} $  is the composition of two functions  $ \text{u} $  and  $ \text{v} $ . So, let  $ \text{t=u(x)=}\sqrt{\text{x}} $ .

Then, 

$\dfrac{\text{dt}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{(}\sqrt{\text{x}}\text{)=}\dfrac{\text{d}}{\text{dx}}\left({{\text{x}}^{\dfrac{\text{1}}{\text{2}}}}\right)\text{=}\dfrac{\text{1}}{\text{2}}{{\text{x}}^{\dfrac{\text{1}}{\text{2}}}}=\dfrac{\text{1}}{\text{2}\sqrt{\text{x}}} $  .

Also,  $\dfrac{\text{dv}}{\text{dt}}\text{=}\dfrac{\text{d}}{\text{dt}}\text{(cost)=-sint=sin(}\sqrt{\text{x}}\text{)} $ .

Now, by applying the chain rule of derivatives, gives 

 $ \begin{align}& \dfrac{\text{dt}}{\text{dx}}\text{=}\dfrac{\text{dv}}{\text{dt}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{dt}}{\text{dx}} \\ & \text{=-sin(}\sqrt{\text{x}}\text{) }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{2}\sqrt{\text{x}}} \\ & \text{=-}\dfrac{\text{1}}{\text{2}\sqrt{\text{x}}}\text{sin(}\sqrt{\text{x}}\text{)} \\ & \text{=-}\dfrac{\text{sin(}\sqrt{\text{x}}\text{)}}{\text{2}\sqrt{\text{x}}}  \end{align} $ 

An alternate method.

 $ \begin{align}& \dfrac{\text{d}}{\text{dx}}\left[ \text{cos(}\sqrt{\text{x}}\text{)} \right] \\ & \text{=-sin(}\sqrt{\text{x}}\text{)}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(}\sqrt{\text{x}}\text{)} \\ & \text{=-sin(}\sqrt{\text{x}}\text{) }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\left( {{\text{x}}^{\dfrac{\text{1}}{\text{2}}}} \right) \\  & \text{=-sin}\sqrt{\text{x}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{2}}{{\text{x}}^{\dfrac{\text{1}}{\text{2}}}} \\ & =\dfrac{\text{-sin}\sqrt{\text{x}}}{\text{2}\sqrt{\text{x}}}  \end{align} $ 

Hence, the derivative of the function  $ \text{f(x)=cos(}\sqrt{\text{x}}\text{)} $  is  $ \text{-}\dfrac{\text{sin(}\sqrt{\text{x}}\text{)}}{\text{2}\sqrt{\text{x}}} $ .


9. Prove that the function  $ \mathbf{f} $  given by  $ \text{f(x)=}\left| \text{x-1} \right|\text{, x}\in \text{R} $  is not differentiable at  $ \mathbf{x =1} $ .

Ans:  The given function is  $ \text{f(x)=}\left| \text{x-1} \right|\text{,}\,\text{x}\in \text{R} $ .

We know that a function  $ \text{f} $  is called differentiable at a point  $ \text{x=c} $  in its domain if both the  $ \underset{\text{k}\to {{\text{0}}^{\text{-}}}}{\mathop{\text{lim}}}\,\dfrac{\text{f(c+h)-f(c)}}{\text{h}} $  and  $ \underset{\text{h}\to {{\text{0}}^{\text{+}}}}{\mathop{\text{lim}}}\,\dfrac{\text{f(c+h)-f(c)}}{\text{h}} $  are finite and equal.

Now verify the differentiability for the function  $ \text{f} $  at the point  $ \text{x =1} $ .

First, the left-hand-derivative is

 $ \underset{\text{h}\to {{\text{0}}^{\text{-}}}}{\mathop{\text{lim}}}\,\dfrac{\text{f(1+h)-f(1)}}{\text{h}}\text{=}\underset{\text{h}\to{{\text{0}}^{\text{-}}}}{\mathop{\text{lim}}}\,\dfrac{\text{f}\left| \text{1+h-1} \right|\left| \text{1-1} \right|}{\text{h}} $ 

 $ \underset{\text{h}\to {{\text{0}}^{\text{-}}}}{\mathop{\text{lim}}}\,\dfrac{\text{f}\left| \text{h} \right|\text{-0}}{\text{h}}\text{=}\underset{\text{h}\to{{\text{0}}^{\text{+}}}}{\mathop{\text{lim}}}\,\dfrac{\text{-h}}{\text{h}}=1 $ , since   $ \text{h0}\Rightarrow \left| \text{h} \right|\text{=-h} $ .

Now the right-hand-derivative is

 $ \underset{\text{h}\to {{\text{0}}^{\text{+}}}}{\mathop{\text{lim}}}\,\dfrac{\text{f(1+h)-f(1)}}{\text{h}}\text{=}\underset{\text{h}\to{{\text{0}}^{\text{+}}}}{\mathop{\text{lim}}}\,\dfrac{\text{f}\left| \text{1+h-1} \right|\left| \text{1-1} \right|}{\text{h}} $ 

 $ \underset{\text{h}\to {{\text{0}}^{\text{+}}}}{\mathop{\text{lim}}}\,\dfrac{\text{f}\left| \text{h} \right|\text{-0}}{\text{h}}\text{=}\underset{\text{h}\to{{\text{0}}^{\text{+}}}}{\mathop{\text{lim}}}\,\dfrac{\text{-h}}{\text{h}}=\text{-1} $ , since  $ \text{h0}\Rightarrow \left| \text{h} \right|\text{=h} $ .

From the above, it is noted that  $ \underset{\text{h}\to {{\text{0}}^{-}}}{\mathop{\text{lim}}}\,\dfrac{\text{f(1+h)-f(1)}}{\text{h}}\ne \underset{\text{h}\to {{\text{0}}^{\text{+}}}}{\mathop{\text{lim}}}\,\dfrac{\text{f(1+h)-f(1)}}{\text{h}} $ .

Hence, the function  $ \text{f(x)=}\left| \text{x-1} \right|\text{,}\,\text{x}\in \text{R} $  is not differentiable at the point  $ \text{x=1} $ .


10. Prove that the greatest integer function defined by  $ \text{f(x)= }\!\![\!\!\text{ x }\!\!]\!\!\text{ ,}\,\,\,\text{0x3} $ , is not differentiable at  $ \mathbf{x=1} $  and  $ \mathbf{x=2} $  .

Ans: The given function is  $ \text{f(x)= }\!\![\!\!\text{ x }\!\!]\!\!\text{ ,}\,\,\,\text{0x3} $ .

Remember that a function  $ \text{f} $  is called differentiable at a point  $ \text{x=c} $  in its domain if both the limits,  $ \underset{\text{h}\to {{\text{0}}^{\text{-}}}}{\mathop{\text{lim}}}\,\dfrac{\text{f(c+h)-f(c)}}{\text{h}} $  and  $ \underset{\text{h}\to {{\text{0}}^{\text{+}}}}{\mathop{\text{lim}}}\,\dfrac{\text{f(c+h)-f(c)}}{\text{h}} $  are finite and equal.

First, take the left-hand-derivative of the function  $ \text{f} $  at  $ \text{x=1} $  such that

 $ \underset{\text{h}\to {{\text{0}}^{\text{-}}}}{\mathop{\text{lim}}}\,\dfrac{\text{f(1+h)-f(1)}}{\text{h}}\text{=}\underset{\text{h}\to{{\text{0}}^{\text{-}}}}{\mathop{\text{lim}}}\,\dfrac{\text{ }\!\![\!\!\text{ 1+h }\!\!]\!\!\text{ - }\!\![\!\!\text{ 1 }\!\!]\!\!\text{ }}{\text{h}}=\underset{\text{h}\to{{\text{0}}^{\text{-}}}}{\mathop{\text{lim}}}\,\dfrac{\text{(0-1)}}{\text{h}}\text{=}\underset{\text{h}\to {{\text{0}}^{\text{+}}}}{\mathop{\text{lim}}}\,\dfrac{\text{-h}}{\text{h}}\text{=}\infty  $ .

Now, take the right-hand-derivative of the function  $ \text{f} $  at  $ \text{x=1} $  such that

 $ \underset{\text{h}\to {{\text{0}}^{\text{+}}}}{\mathop{\text{lim}}}\,\dfrac{\text{f(1+h)-f(1)}}{\text{h}}\text{=}\underset{\text{h}\to {{\text{0}}^{\text{+}}}}{\mathop{\text{lim}}}\,\dfrac{\text{ }\!\![\!\!\text{ 1+h }\!\!]\!\!\text{  }\!\![\!\!\text{ 1 }\!\!]\!\!\text{ }}{\text{h}}=\underset{\text{h}\to {{\text{0}}^{\text{+}}}}{\mathop{\text{lim}}}\,\dfrac{\text{1-1}}{\text{h}}\text{=}\underset{\text{h}\to {{\text{0}}^{\text{+}}}}{\mathop{\text{lim}}}\,\text{0=0} $ .

Therefore, it is being noticed that,  $ \underset{\text{h}\to {{\text{0}}^{\text{-}}}}{\mathop{\text{lim}}}\,\dfrac{\text{f(1+h)-f(1)}}{\text{h}}\ne \underset{\text{h}\to {{\text{0}}^{+}}}{\mathop{\text{lim}}}\,\dfrac{\text{f(1+h)-f(1)}}{\text{h}} $ .

Thus, the function  $ \text{f} $  is not differentiable at  $ \text{x=1} $ .

Now, justify the differentiability of the function  $ \text{f} $  at  $ \text{x=2} $ .

First, take the left-hand-derivative of the function  $ \text{f} $  at  $ \text{x=2} $ , such that

$ \underset{\text{h}\to{{\text{0}}^{\text{-}}}}{\mathop{\text{lim}}}\,\dfrac{\text{f(2+h)-f(2)}}{\text{h}}\text{=}\underset{\text{h}\to {{\text{0}}^{\text{-}}}}{\mathop{\text{lim}}}\,\dfrac{\text{ }\!\![\!\!\text{ 2+h }\!\!]\!\!\text{ - }\!\![\!\!\text{ 2 }\!\!]\!\!\text{ }}{\text{h}}\text{=}\underset{\text{h}\to {{\text{0}}^{\text{-}}}}{\mathop{\text{lim}}}\,\dfrac{\text{(1-2)}}{\text{h}}\text{=}\underset{\text{h}\to {{\text{0}}^{\text{+}}}}{\mathop{\text{lim}}}\,\dfrac{\text{-1}}{\text{h}}\text{=}\infty  $ 

Now, take the right-hand-derivative of the function  $ \text{f} $  at  $ \text{x=2} $ , such that

 $ \underset{\text{h}\to {{\text{0}}^{\text{+}}}}{\mathop{\text{lim}}}\,\dfrac{\text{f(2+h)-f(2)}}{\text{h}}\text{=}\underset{\text{h}\to {{\text{0}}^{\text{+}}}}{\mathop{\text{lim}}}\,\dfrac{\text{ }\!\![\!\!\text{ 2+h }\!\!]\!\!\text{  }\!\![\!\!\text{ 2 }\!\!]\!\!\text{ }}{\text{h}}\text{=}\underset{\text{h}\to {{\text{0}}^{\text{+}}}}{\mathop{\text{lim}}}\,\dfrac{\text{1-2}}{\text{h}}\text{=}\underset{\text{h}\to {{\text{0}}^{\text{+}}}}{\mathop{\text{lim}}}\,\text{0=0} $ 

It is observed from the above discussion that,  $ \underset{\text{h}\to {{\text{0}}^{\text{-}}}}{\mathop{\text{lim}}}\,\dfrac{\text{f(2+h)-f(2)}}{\text{h}}\ne \underset{\text{h}\to {{\text{0}}^{+}}}{\mathop{\text{lim}}}\,\dfrac{\text{f(2+h)-f(2)}}{\text{h}} $ .

Thus, the function  $ \text{f} $  is not differentiable at the point  $ \text{x=2} $.

Exercise 5.3

Find  $ \dfrac{\mathbf{dy}}{\mathbf{dx}} $  in the following.

1.  $ \mathbf{2x+3y=sinx} $ . 

Ans:  The given equation is  $ \text{2x+3y=sinx} $ .

Differentiating both sides of the equation with respect to  $ \text{x} $  , gives

  $ \dfrac{\text{d}}{\text{dy}}\text{(2x+3y)=}\dfrac{\text{d}}{\text{dx}}\text{(sinx)} $ 

 $ \Rightarrow \dfrac{\text{d}}{\text{dx}}\text{(2x)+}\dfrac{\text{d}}{\text{dx}}\text{(3y)=cosx} $ ,  applying the addition rule of derivatives.

 $ \Rightarrow \text{2+3}\dfrac{\text{dy}}{\text{dx}}\text{=cosx} $ 

 $ \Rightarrow \text{3}\dfrac{\text{dy}}{\text{dx}}\text{=cosx-2} $ 

Therefore,  $ \dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{cosx-2}}{\text{3}} $ .

2.  $ \mathbf{2x+3y=siny} $ .

Ans:  The given equation is  $ \mathbf{2x+3y=siny} $ .

Differentiating both sides of the equation with respect to  $ \text{x} $ ,  gives  

 $ \dfrac{\text{d}}{\text{dx}}\text{(2x)+}\dfrac{\text{d}}{\text{dx}}\text{(3y)=}\dfrac{\text{d}}{\text{dx}}\text{(siny)} $  

 $ \Rightarrow \text{2+3}\dfrac{\text{dy}}{\text{dx}}\text{=cosy}\dfrac{\text{dy}}{\text{dx}} $    , applying the chain rule of derivatives.

 $ \Rightarrow 2\text{=}\left( \text{cosy-3} \right)\dfrac{\text{dy}}{\text{dx}} $ 

Therefore,  $ \dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{2}}{\text{cosy-3}} $ .


3.  $ \mathbf{ax+b}{{\mathbf{y}}^{\mathbf{2}}}\mathbf{=cosy} $ .

Ans:  The given function is  $ \text{ax+b}{{\text{y}}^{\text{2}}}\text{=cosy} $ .

Differentiating both sides of the equation with respect to  $ \text{x} $ , gives  $ \dfrac{\text{d}}{\text{dx}}\text{(ax)+}\dfrac{\text{d}}{\text{dx}}\text{(b}{{\text{y}}^{\text{2}}}\text{)=}\dfrac{\text{d}}{\text{dx}}\text{(cosy)} $   

 $ \Rightarrow \text{a }\!\!\times\!\!\text{ 1+b}\dfrac{\text{d}}{\text{dy}}\text{(}{{\text{y}}^{\text{2}}}\text{)}\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dy}}\text{(cosy)}\dfrac{\text{dy}}{\text{dx}} $  , applying the chain rule of derivatives.

 $ \Rightarrow \text{a+b }\!\!\times\!\!\text{ 2y}\dfrac{\text{dy}}{\text{dx}}\text{=-siny}\dfrac{\text{dy}}{\text{dx}} $  

 $ \Rightarrow \text{(2by+siny)}\dfrac{\text{dy}}{\text{dx}}\text{=-a} $ 

Therefore,  $ \dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{-a}}{\text{2by+siny}} $ .


4.  $ \mathbf{xy+}{{\mathbf{y}}^{\mathbf{2}}}\mathbf{=tanx+y} $ .

Ans: The given equation is  $ \text{xy+}{{\text{y}}^{\text{2}}}\text{=tanx+y} $ .

Differentiating both sides of the equation with respect to  $ \text{x} $ , gives  $ \begin{align} & \dfrac{\text{d}}{\text{dx}}\text{(xy+}{{\text{y}}^{\text{2}}}\text{)=}\dfrac{\text{d}}{\text{dx}}\text{(tanx+y)} \\ & \Rightarrow \dfrac{\text{d}}{\text{dx}}\text{(xy)+}\dfrac{\text{dy}}{\text{dx}}\text{(}{{\text{y}}^{\text{2}}}\text{)=}\dfrac{\text{d}}{\text{dx}}\text{(tanx)+}\dfrac{\text{dy}}{\text{dx}} \\ \end{align} $

$ \Rightarrow \left[ \text{y }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(x)+x }\!\!\times\!\!\text{ }\dfrac{\text{dy}}{\text{dx}} \right]\text{+2y}\dfrac{\text{dy}}{\text{dx}}\text{=se}{{\text{c}}^{\text{2}}}\text{x+}\dfrac{\text{dy}}{\text{dx}} $ , applying chain rule of derivatives.

 $ \Rightarrow \text{y }\!\!\times\!\!\text{ 1+x}\dfrac{\text{dy}}{\text{dx}}\text{+2y}\dfrac{\text{dy}}{\text{dx}}\text{=se}{{\text{c}}^{\text{2}}}\text{x+}\dfrac{\text{dy}}{\text{dx}} $ 

Therefore,  $ \dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{se}{{\text{c}}^{\text{2}}}\text{x-y}}{\text{(x+2y-1)}} $ .


5. ${{\mathbf{x}}^{\mathbf{2}}}\mathbf{+xy+}{{\mathbf{y}}^{\mathbf{2}}}\mathbf{=100} $ .

Ans: The given equation is  $ {{\text{x}}^{\text{2}}}\text{+xy+}{{\text{y}}^{\text{2}}}\text{=100} $ .

Differentiating both sides of the equation with respect to  $ \text{x} $ , gives  $ \begin{align} & \dfrac{\text{dy}}{\text{dx}}\text{(}{{\text{x}}^{\text{2}}}\text{+xy+}{{\text{y}}^{\text{2}}}\text{)=}\dfrac{\text{d}}{\text{dx}}\text{100} \\ & \Rightarrow \dfrac{\text{dy}}{\text{dx}}\text{(}{{\text{x}}^{\text{2}}}\text{)+}\dfrac{\text{dy}}{\text{dx}}\text{(xy)+}\dfrac{\text{dy}}{\text{dx}}\text{(}{{\text{y}}^{\text{2}}}\text{)=0} \\ \end{align} $  

$ \Rightarrow \text{2x+}\left[ \text{y }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(x)+x }\!\!\times\!\!\text{ }\dfrac{\text{dy}}{\text{dx}} \right]\text{+2y}\dfrac{\text{dy}}{\text{dx}}\text{=0} $  , applying the chain rule of derivatives.

$ \begin{align}& \Rightarrow \text{2x+y }\!\!\times\!\!\text{ 1+x }\!\!\times\!\!\text{}\dfrac{\text{dy}}{\text{dx}}\text{+2y}\dfrac{\text{dy}}{\text{dx}}\text{=0} \\ & \Rightarrow \text{2x+y+(x+2y)}\dfrac{\text{dy}}{\text{dx}}\text{=0} \\ \end{align} $ 

Therefore,  $ \dfrac{\text{dy}}{\text{dx}}\text{=-}\dfrac{\text{2x+y}}{\text{x+2y}} $ .


6. ${{\mathbf{x}}^{\mathbf{3}}}\mathbf{+}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{y+x}{{\mathbf{y}}^{\mathbf{2}}}\mathbf{+}{{\mathbf{y}}^{\mathbf{3}}}\mathbf{=81} $ .

Ans: The given equation is  $ {{\text{x}}^{\text{3}}}\text{+}{{\text{x}}^{\text{2}}}\text{y+x}{{\text{y}}^{\text{2}}}\text{+}{{\text{y}}^{\text{3}}}\text{=81} $ .

Differentiating both sides of the equation with respect to  $ \text{x} $ , gives  $ \begin{align}& \dfrac{\text{dy}}{\text{dx}}\text{(}{{\text{x}}^{\text{3}}}\text{+}{{\text{x}}^{\text{2}}}\text{y+x}{{\text{y}}^{\text{2}}}\text{+}{{\text{y}}^{\text{3}}}\text{)=}\dfrac{\text{d}}{\text{dx}}\left( \text{81} \right) \\ & \Rightarrow \dfrac{\text{d}}{\text{dx}}\text{(}{{\text{x}}^{\text{3}}}\text{)+}\dfrac{\text{dy}}{\text{dx}}\text{(}{{\text{x}}^{\text{2}}}\text{y)+}\dfrac{\text{dy}}{\text{dx}}\text{(x}{{\text{y}}^{\text{2}}}\text{)+}\dfrac{\text{dy}}{\text{dx}}\text{(}{{\text{y}}^{\text{3}}}\text{)=0} \\ & \Rightarrow \text{3x+}\left[ \text{y}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{x}}^{\text{2}}}\text{)+}{{\text{x}}^{\text{2}}}\dfrac{\text{dy}}{\text{dx}} \right]\text{+}\left[ {{\text{y}}^{\text{2}}}\dfrac{\text{d}}{\text{dx}}\text{(x)+x}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{y}}^{2}}\text{)}\right]\text{+3}{{\text{y}}^{\text{2}}}\dfrac{\text{dy}}{\text{dx}}\text{=0} \\ \end{align} $ 

$ \Rightarrow \text{3x+}\left[ \text{y }\!\!\times\!\!\text{ 2x+}{{\text{x}}^{\text{2}}}\dfrac{\text{dy}}{\text{dx}} \right]\text{+}\left[ {{\text{y}}^{\text{2}}}\text{ }\!\!\times\!\!\text{ 1+x }\!\!\times\!\!\text{ 2y }\!\!\times\!\!\text{ }\dfrac{\text{dy}}{\text{dx}} \right]\text{+3}{{\text{y}}^{\text{2}}}\dfrac{\text{dy}}{\text{dx}}\text{=0} $  , applying chain rule.

 $ \Rightarrow\text{(}{{\text{x}}^{\text{3}}}\text{+2xy+3}{{\text{y}}^{\text{2}}}\text{)}\dfrac{\text{dy}}{\text{dx}}\text{+(3x+2xy+}{{\text{y}}^{\text{2}}}\text{)=0} $ 

Therefore,  $\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{-(2x+2xy+}{{\text{y}}^{\text{2}}}\text{)}}{\text{(}{{\text{x}}^{\text{3}}}\text{+2xy+3}{{\text{y}}^{\text{2}}}\text{)}} $ .

 

7.  $ \text{si}{{\text{n}}^{\text{2}}}\text{y+cosxy= }\text{k} $ .

Ans:  The given equation is  $ \text{si}{{\text{n}}^{\text{2}}}\text{y+cosxy= }\text{k} $ .

Differentiating both sides of the equation with respect to  $ \text{x} $ , gives  $ \begin{align}&\dfrac{\text{d}}{\text{dx}}\text{(si}{{\text{n}}^{\text{2}}}\text{y+cosxy)=}\dfrac{\text{d}}{\text{dx}}\left( \text{ k} \right) \\ & \Rightarrow \dfrac{\text{d}}{\text{dx}}\text{(si}{{\text{n}}^{\text{2}}}\text{y)+}\dfrac{\text{d}}{\text{dx}}\text{(cosxy)=0} \\ \end{align} $                                             …. (1)

Applying the chain rule of derivatives gives

$\dfrac{\text{d}}{\text{dx}}\text{(si}{{\text{n}}^{\text{2}}}\text{y)=2siny}\dfrac{\text{d}}{\text{dx}}\text{(siny)=2sinycosy}\dfrac{\text{dy}}{\text{dx}} $ ….. (2)

and

$\dfrac{\text{d}}{\text{dx}}\text{(cosxy)=-sinxy}\dfrac{\text{d}}{\text{dx}}\text{(xy)=-sinxy}\left[\text{y}\dfrac{\text{d}}{\text{dx}}\text{(x)+x}\dfrac{\text{dy}}{\text{dx}} \right]\text{=-ysinxy-xsinxy}\dfrac{\text{dy}}{\text{dx}} $       …. (3)

From (1), (2) and (3), we obtain

 $\begin{align}&\text{2sinycosy}\dfrac{\text{dy}}{\text{dx}}\text{-ysinxy-xsinxy}\dfrac{\text{dy}}{\text{dx}}\text{=0} \\ & \Rightarrow\text{(2sinycosy-xsinxy)}\dfrac{\text{dy}}{\text{dx}}\text{=ysinxy} \\ & \Rightarrow \text{(sin2y-xsinxy)}\dfrac{\text{dy}}{\text{dx}}\text{=ysinxy} \\ \end{align} $ 

Therefore,  $\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{ysinxy}}{\text{sin2y-xsinxy}} $ .


8. $\text{si}{{\text{n}}^{\mathbf{2}}}\text{x+co}{{\text{s}}^{\mathbf{2}}}\text{y=1} $ .

Ans:  The given equation is  $ \text{si}{{\text{n}}^{\text{2}}}\text{x+co}{{\text{s}}^{\text{2}}}\text{y=1} $ .

Differentiating both sides of the equation with respect to  $ \text{x} $ , gives  $ \begin{align} & \dfrac{\text{dy}}{\text{dx}}\text{(si}{{\text{n}}^{\text{2}}}\text{x+co}{{\text{s}}^{\text{2}}}\text{y)=}\dfrac{\text{d}}{\text{dx}}\text{(1)} \\ & \Rightarrow \dfrac{\text{d}}{\text{dx}}\text{(si}{{\text{n}}^{\text{2}}}\text{x)+}\dfrac{\text{d}}{\text{dx}}\text{(co}{{\text{s}}^{\text{2}}}\text{y)=0} \\ & \Rightarrow \text{2sinx }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(sinx)+2cosy }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(cosy)=0} \\ \end{align} $ 

$ \begin{align}& \Rightarrow \text{2sinxcosx+2cosy(-siny) }\!\!\times\!\!\text{ }\dfrac{\text{dy}}{\text{dx}}\text{=0} \\ & \Rightarrow \text{sin2x-sin2y}\dfrac{\text{dy}}{\text{dx}}\text{=0} \\ \end{align} $ 

 $ \therefore \dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{sin2x}}{\text{sin2y}} $ 


9.  $ \mathbf{y=si}{{\mathbf{n}}^{\mathbf{-1}}}\left( \dfrac{\mathbf{2x}}{\mathbf{1+}{{\mathbf{x}}^{\mathbf{2}}}} \right) $ .

Ans:  The given equation is  $ \text{y=si}{{\text{n}}^{\text{-1}}}\left( \dfrac{\text{2x}}{\text{1+}{{\text{x}}^{\text{2}}}} \right) $ .

Now,  $ \text{y=si}{{\text{n}}^{\text{-1}}}\left( \dfrac{\text{2x}}{\text{1+}{{\text{x}}^{\text{2}}}} \right) $ 

 $ \Rightarrow \text{siny=}\dfrac{\text{2x}}{\text{1+}{{\text{x}}^{\text{2}}}} $ .

Differentiating both sides of the equation with respect to  $ \text{x} $ , gives  $ \dfrac{\text{d}}{\text{dx}}\text{(siny)=}\dfrac{\text{d}}{\text{dx}}\left( \dfrac{\text{2x}}{\text{1+}{{\text{x}}^{\text{2}}}} \right) $

 $ \Rightarrow\text{cosy}\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\left( \dfrac{\text{2x}}{\text{1+}{{\text{x}}^{\text{2}}}} \right) $…… (1)

Now, the function  $ \dfrac{\text{2x}}{\text{1+}{{\text{x}}^{\text{2}}}} $  is of the form of  $ \dfrac{\text{u}}{\text{v}} $ . 

Applying the quotient rule, gives

 $ \begin{align} & \dfrac{\text{d}}{\text{dx}}\left( \dfrac{\text{2x}}{\text{1+}{{\text{x}}^{\text{2}}}} \right)\text{=}\dfrac{\text{(1+}{{\text{x}}^{\text{2}}}\text{)}\dfrac{\text{d}}{\text{dx}}\text{(2x)-2x }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(1+}{{\text{x}}^{\text{2}}}\text{)}}{{{\text{(1+}{{\text{x}}^{\text{2}}}\text{)}}^{2}}} \\& \text{=}\dfrac{\text{(1+}{{\text{x}}^{\text{2}}}\text{) }\!\!\times\!\!\text{ 2-2x }\!\!\times\!\!\text{  }\!\![\!\!\text{ 0+2x }\!\!]\!\!\text{ }}{{{\text{(1+}{{\text{x}}^{\text{2}}}\text{)}}^{2}}}\text{=}\dfrac{\text{2+2}{{\text{x}}^{\text{2}}}\text{-4}{{\text{x}}^{2}}}{{{\text{(1+}{{\text{x}}^{\text{2}}}\text{)}}^{\text{2}}}} \\ \end{align} $

Therefore,  $ \dfrac{\text{d}}{\text{dx}}\left(\dfrac{\text{2x}}{\text{1+}{{\text{x}}^{\text{2}}}} \right)\text{=}\dfrac{\text{2(1-}{{\text{x}}^{\text{2}}}\text{)}}{{{\text{(1+}{{\text{x}}^{\text{2}}}\text{)}}^{\text{2}}}} $                                       ……. (2)

It is given that,$ \text{siny=}\dfrac{\text{2x}}{\text{1+}{{\text{x}}^{\text{2}}}} $ $ \Rightarrow \text{cosy=}\sqrt{\text{1-si}{{\text{n}}^{\text{2}}}\text{y}}\text{=}\sqrt{\text{1-}{{\left( \dfrac{\text{2x}}{\text{1+}{{\text{x}}^{\text{2}}}} \right)}^{\text{2}}}}\text{=}\sqrt{\dfrac{{{\left( \text{1+}{{\text{x}}^{\text{2}}} \right)}^{\text{2}}}\text{-4}{{\text{x}}^{\text{2}}}}{{{\text{(1+}{{\text{x}}^{\text{2}}}\text{)}}^{\text{2}}}}} $ 

 $ \Rightarrow \text{cosy=}\sqrt{\dfrac{{{\left( \text{1-}{{\text{x}}^{\text{2}}} \right)}^{\text{2}}}}{{{\left( \text{1-}{{\text{x}}^{\text{2}}} \right)}^{\text{2}}}}}\text{=}\dfrac{\text{1-}{{\text{x}}^{\text{2}}}}{\text{1+}{{\text{x}}^{\text{2}}}} $                               ….. (3)

From the equation (1), (2) and (3), gives

$\dfrac{\text{1-}{{\text{x}}^{\text{2}}}}{\text{1+}{{\text{x}}^{\text{2}}}}\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{2(1-}{{\text{x}}^{\text{2}}}\text{)}}{{{\text{(1+}{{\text{x}}^{\text{2}}}\text{)}}^{\text{2}}}} $ 

Therefore,  $\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{2}}{\text{1+}{{\text{x}}^{\text{2}}}} $ .


10.  $\text{y=ta}{{\text{n}}^{\text{-1}}}\left(\dfrac{\text{3x-}{{\text{x}}^{\text{3}}}}{\text{1-3}{{\text{x}}^{\text{2}}}}\right)\text{,}\,\,\,\text{-}\dfrac{\text{1}}{\sqrt{\text{3}}}<\text{x}<\dfrac{\text{1}}{\sqrt{\text{3}}} $ .

Ans: The given function is  $ \text{y=ta}{{\text{n}}^{\text{-1}}}\left( \dfrac{\text{3x-}{{\text{x}}^{\text{3}}}}{\text{1-3}{{\text{x}}^{\text{2}}}} \right) $ .

Now,  $\text{y=ta}{{\text{n}}^{\text{-1}}}\left(\dfrac{\text{3x-}{{\text{x}}^{\text{3}}}}{\text{1-3}{{\text{x}}^{\text{2}}}} \right) $

$\Rightarrow\text{tany=}\dfrac{\text{3x-}{{\text{x}}^{\text{3}}}}{\text{1-3}{{\text{x}}^{\text{2}}}} $                                         …… (1)

According to the trigonometric formulas,

$\text{tany=}\dfrac{\text{3tan}\dfrac{\text{y}}{\text{3}}\text{-ta}{{\text{n}}^{\text{3}}}\dfrac{\text{y}}{\text{3}}}{\text{1-3ta}{{\text{n}}^{\text{2}}}\dfrac{\text{y}}{\text{3}}} $                                             …... (2)

By comparing the equations (1) and (2), gives

 $ \text{x=tan}\dfrac{\text{y}}{\text{3}} $ .                                                                                …… (3)

Differentiating both sides of the equation (3) with respect to  $ \text{x}\\ $ , gives 

$ \begin{align}& \dfrac{\text{d}}{\text{dx}}\text{(x)=}\dfrac{\text{d}}{\text{dx}}\left( \text{tan}\dfrac{\text{y}}{\text{3}} \right) \\ & \Rightarrow \text{1=se}{{\text{c}}^{\text{2}}}\dfrac{\text{y}}{\text{3}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\left( \dfrac{\text{y}}{\text{3}} \right) \\ & \Rightarrow\text{1=se}{{\text{c}}^{\text{2}}}\dfrac{\text{y}}{\text{3}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{3}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{dy}}{\text{dx}} \\ & \Rightarrow \dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{3}}{\text{se}{{\text{c}}^{\text{2}}}\dfrac{\text{y}}{\text{3}}}\text{=}\dfrac{\text{3}}{\text{1+ta}{{\text{n}}^{\text{2}}}\dfrac{\text{y}}{\text{3}}} \\ \end{align} $ 

Therefore,  $\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{3}}{\text{1+}{{\text{x}}^{\text{2}}}} $ .


11.  $ \text{y=co}{{\text{s}}^{\text{-1}}}\left( \dfrac{\text{1-}{{\text{x}}^{\text{2}}}}{\text{1+}{{\text{x}}^{\text{2}}}} \right) $ ,  0<x<1.

Ans: The given equation is  $ \text{y=co}{{\text{s}}^{\text{-1}}}\left( \dfrac{\text{1-}{{\text{x}}^{\text{2}}}}{\text{1+}{{\text{x}}^{\text{2}}}} \right) $ 

 $ \Rightarrow\text{cosy=}\dfrac{\text{1-}{{\text{x}}^{\text{2}}}}{\text{1+}{{\text{x}}^{\text{2}}}} $ 

$\Rightarrow\dfrac{\text{1-ta}{{\text{n}}^{\text{2}}}\dfrac{\text{y}}{\text{2}}}{\text{1+ta}{{\text{n}}^{\text{2}}}\dfrac{\text{y}}{\text{2}}}\text{=}\dfrac{\text{1-}{{\text{x}}^{\text{2}}}}{\text{1+}{{\text{x}}^{\text{2}}}} $ .                                                                        …… (1)

By comparing both sides of the equation (1) give

 $ \text{tan}\dfrac{\text{y}}{\text{2}}\text{=x} $                                                                                           …… (2)          

Differentiating both sides of the equation (2) with respect to  $ \text{x}\\ $ , gives  $ \begin{align}& \text{se}{{\text{c}}^{\text{2}}}\dfrac{\text{y}}{\text{2}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\left( \dfrac{\text{y}}{\text{2}} \right)\text{=}\dfrac{\text{d}}{\text{dx}}\text{(x)} \\ & \Rightarrow \text{se}{{\text{c}}^{\text{2}}}\dfrac{\text{y}}{\text{2}}\text{ }\!\!\times\!\!\text{}\dfrac{\text{1}}{\text{2}}\dfrac{\text{dy}}{\text{dx}}\text{=1} \\ \end{align} $ 

$ \begin{align}& \Rightarrow \dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{2}}{\text{se}{{\text{c}}^{\text{2}}}\dfrac{\text{y}}{\text{2}}} \\ & \Rightarrow \dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{2}}{\text{1+ta}{{\text{n}}^{\text{2}}}\dfrac{\text{y}}{\text{2}}} \\ \end{align} $ 

Therefore, using the equation (2), gives

$ \dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{2}{\text{1+}{{\text{x}}^{\text{2}}}} $. 


12.  $ \text{y=si}{{\text{n}}^{\text{-1}}}\left( \dfrac{\text{1-}{{\text{x}}^{\text{2}}}}{\text{1+}{{\text{x}}^{\text{2}}}} \right) $ , 0<x<1.

Ans: The given equation is  $ \text{y=si}{{\text{n}}^{\text{-1}}}\left( \dfrac{\text{1-}{{\text{x}}^{\text{2}}}}{\text{1+}{{\text{x}}^{\text{2}}}} \right) $ .

Now,  $ \text{y=si}{{\text{n}}^{\text{-1}}}\left( \dfrac{\text{1-}{{\text{x}}^{\text{2}}}}{\text{1+}{{\text{x}}^{\text{2}}}} \right) $ 

$ \Rightarrow \text{siny=}\dfrac{\text{1-}{{\text{x}}^{\text{2}}}}{\text{1+}{{\text{x}}^{\text{2}}}} $ .                                                         …… (1)

Differentiating both sides of the equation with respect to  $ \text{x} $ , gives  $ \dfrac{\text{d}}{\text{dx}}\text{(siny)=}\dfrac{\text{d}}{\text{dx}}\left( \dfrac{\text{1-}{{\text{x}}^{\text{2}}}}{\text{1+}{{\text{x}}^{\text{2}}}} \right) $                     …... (2)

Using chain rule, we get

$ \dfrac{\text{d}}{\text{dx}}\text{(siny)=cosy }\!\!\times\!\!\text{ }\dfrac{\text{dy}}{\text{dx}} $                                                 …… (3)

$\text{cosy=}\sqrt{\text{1-si}{{\text{n}}^{\text{2}}}\text{y}}\text{=}\sqrt{\text{1-}{{\left( \dfrac{\text{1-}{{\text{x}}^{\text{2}}}}{\text{1+}{{\text{x}}^{\text{2}}}} \right)}^{\text{2}}}}=\sqrt{\dfrac{{{\text{(1+}{{\text{x}}^{\text{2}}}\text{)}}^{\text{2}}}\text{-(1-}{{\text{x}}^{\text{2}}}{{\text{)}}^{\text{2}}}}{\text{1+}{{\text{x}}^{\text{2}}}}}\text{=}\sqrt{\dfrac{\text{4}{{\text{x}}^{\text{2}}}}{{{\text{(1+}{{\text{x}}^{\text{2}}}\text{)}}^{\text{2}}}}} $ , using the equation (1).

 $ \Rightarrow \text{cosy=}\dfrac{\text{2x}}{\text{1+}{{\text{x}}^{\text{2}}}} $                                                         …… (4)

Therefore, from the equation (3) and (4) gives

$\dfrac{\text{d}}{\text{dx}}\text{(siny)=}\dfrac{\text{2x}}{\text{1+}{{\text{x}}^{\text{2}}}}\dfrac{\text{dy}}{\text{dx}} $ ….. (5)

Now,

 $ \dfrac{\text{d}}{\text{dx}}\left( \dfrac{\text{1-}{{\text{x}}^{\text{2}}}}{\text{1+}{{\text{x}}^{\text{2}}}} \right)\text{=}\dfrac{\text{(1+}{{\text{x}}^{\text{2}}}\text{)(1-}{{\text{x}}^{\text{2}}}\text{)-(1-}{{\text{x}}^{\text{2}}}\text{)(1+}{{\text{x}}^{\text{2}}}\text{)}}{{{\text{(1+}{{\text{x}}^{\text{2}}}\text{)}}^{\text{2}}}} $  , applying the quotient rule.

 $ \begin{align}&=\dfrac{\text{(1+}{{\text{x}}^{\text{2}}}\text{)(-2x)-(1-}{{\text{x}}^{\text{2}}}\text{)(2x)}}{{{\text{(1+}{{\text{x}}^{\text{2}}}\text{)}}^{\text{2}}}} \\ &=\dfrac{\text{-2x-2}{{\text{x}}^{\text{3}}}\text{-2x+2}{{\text{x}}^{\text{3}}}}{{{\text{(1-}{{\text{x}}^{\text{2}}}\text{)}}^{\text{2}}}} \\ \end{align} $ 

 $ \Rightarrow \dfrac{\text{d}}{\text{dx}}\left(\dfrac{\text{1-}{{\text{x}}^{\text{2}}}}{\text{1+}{{\text{x}}^{\text{2}}}} \right)=\text{-}\dfrac{\text{4}}{{{\text{(1+}{{\text{x}}^{\text{2}}}\text{)}}^{\text{2}}}} $             …... (6)

Using the equations (2), (5), and (6), gives

 $\dfrac{\text{2x}}{\text{1+}{{\text{x}}^{\text{2}}}}\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{-4x}}{{{\text{(1+}{{\text{x}}^{\text{2}}}\text{)}}^{\text{2}}}} $ 

Therefore,  $\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{-2}}{\text{1+}{{\text{x}}^{\text{2}}}} $ .

An alternate method.

$ \begin{align} & \text{y=si}{{\text{n}}^{\text{-1}}}\left( \dfrac{\text{1-}{{\text{x}}^{\text{2}}}}{\text{1+}{{\text{x}}^{\text{2}}}} \right) \\& \Rightarrow \text{siny=}\dfrac{\text{1-}{{\text{x}}^{\text{2}}}}{\text{1+}{{\text{x}}^{\text{2}}}} \\ & \Rightarrow \text{(1+}{{\text{x}}^{\text{2}}}\text{)siny=1-}{{\text{x}}^{\text{2}}} \\ & \Rightarrow \text{(1+siny)}{{\text{x}}^{\text{2}}}\text{=1-siny} \\ & \Rightarrow {{\text{x}}^{\text{2}}}\text{=}\dfrac{\text{1-siny}}{\text{1+siny}} \\ & \Rightarrow {{\text{x}}^{\text{2}}}=\dfrac{{{\left( \text{cos}\dfrac{\text{y}}{\text{2}}\text{-sin}\dfrac{\text{y}}{\text{2}} \right)}^{\text{2}}}}{{{\left(\text{cos}\dfrac{\text{y}}{\text{2}}\text{+sin}\dfrac{\text{y}}{\text{x}} \right)}^{\text{2}}}} \\ & \Rightarrow \text{x=}\dfrac{\text{cos}\dfrac{\text{y}}{\text{2}}\text{-sin}\dfrac{\text{y}}{\text{2}}}{\text{cos}\dfrac{\text{y}}{\text{2}}\text{+sin}\dfrac{\text{y}}{\text{2}}} \\ \end{align} $ 

$ \Rightarrow \text{x=tan}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}} \right) $ 

Differentiating both sides of the equation with respect to  $ \text{x} $ , gives $ \begin{align}& \dfrac{\text{d}}{\text{dx}}\text{(x)=}\dfrac{\text{d}}{\text{dx}}\left[ \text{tan}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-}\dfrac{\text{y}}{\text{2}} \right) \right] \\ & \Rightarrow \text{1=se}{{\text{c}}^{\text{2}}}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-}\dfrac{\text{y}}{\text{2}} \right)\text{ }\!\!\times\!\!\text{ }\dfrac{\text{dy}}{\text{dx}}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-}\dfrac{\text{y}}{\text{2}} \right) \\ & \Rightarrow \text{1=}\left[ \text{1+ta}{{\text{n}}^{\text{2}}}\left( \dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}\text{-}\dfrac{\text{y}}{\text{2}} \right)\text{ }\!\!\times\!\!\text{ }\left( \text{-}\dfrac{\text{1}}{\text{2}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{dy}}{\text{dx}} \right) \right] \\ & \Rightarrow \text{1=(1+}{{\text{x}}^{\text{2}}}\text{)}\left( \text{-}\dfrac{\text{1}}{\text{2}}\dfrac{\text{dy}}{\text{dx}} \right) \\\end{align} $

Therefore,  $\dfrac{\text{dx}}{\text{dy}}\text{=}\dfrac{\text{-2}}{\text{1+}{{\text{x}}^{\text{2}}}} $ .


13.  $ \text{y=co}{{\text{s}}^{\text{-1}}}\left( \dfrac{\text{2x}}{\text{1+}{{\text{x}}^{\text{2}}}} \right) $ , -1<x<1.

Ans: The given equation is  $ \text{y=co}{{\text{s}}^{\text{-1}}}\left( \dfrac{\text{2x}}{\text{1+}{{\text{x}}^{\text{2}}}} \right) $ .

Now,  $ \text{y=co}{{\text{s}}^{\text{-1}}}\left( \dfrac{\text{2x}}{\text{1+}{{\text{x}}^{\text{2}}}} \right) $ 

 $ \Rightarrow \text{cosy=}\dfrac{\text{2x}}{\text{1+}{{\text{x}}^{\text{2}}}} $ .                                     …… (1)

Differentiating both sides of the equation with respect to  $ \text{x} $ , gives  $ \dfrac{\text{d}}{\text{dx}}\text{(cosy)=}\dfrac{\text{d}}{\text{dx}}\text{ }\!\!\times\!\!\text{ }\left( \dfrac{\text{2x}}{\text{1+}{{\text{x}}^{\text{2}}}} \right) $ 

$ \Rightarrow \text{-siny }\!\!\times\!\!\text{ }\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{(1-}{{\text{x}}^{\text{2}}}\text{) }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(2x)-2x }\!\!\times\!\!\text{}\dfrac{\text{d}}{\text{dx}}\text{(1+}{{\text{x}}^{\text{2}}}\text{)}}{{{\text{(1+}{{\text{x}}^{\text{2}}}\text{)}}^{\text{2}}}} $ , applying the quotient rule.

$ \Rightarrow \text{-}\sqrt{\text{1-co}{{\text{s}}^{\text{2}}}\text{y}}\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{(1+}{{\text{x}}^{\text{2}}}\text{) }\!\!\times\!\!\text{ 2-2x }\!\!\times\!\!\text{ 2x}}{{{\text{(1+}{{\text{x}}^{\text{2}}}\text{)}}^{\text{2}}}} $ 

$ \Rightarrow \left[ \sqrt{\text{1-}{{\left( \dfrac{\text{2x}}{\text{1+}{{\text{x}}^{\text{2}}}} \right)}^{\text{2}}}} \right]\dfrac{\text{dx}}{\text{dy}}\text{=-}\left[ \dfrac{\text{2(1-x}{{\text{)}}^{\text{2}}}}{{{\text{(1+}{{\text{x}}^{\text{2}}}\text{)}}^{\text{2}}}} \right] $ , using the equation (1).

$ \begin{align}& \Rightarrow \sqrt{\dfrac{{{\text{(1-}{{\text{x}}^{\text{2}}}\text{)}}^{\text{2}}}\text{-4}{{\text{x}}^{\text{2}}}}{{{\text{(1+}{{\text{x}}^{\text{2}}}\text{)}}^{\text{2}}}}}\text{=}\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{-2(1-x}{{\text{)}}^{\text{2}}}}{\text{(1+}{{\text{x}}^{\text{2}}}\text{)}} \\ & \Rightarrow \sqrt{\dfrac{{{\text{(1-}{{\text{x}}^{\text{2}}}\text{)}}^{\text{2}}}}{{{\text{(1+}{{\text{x}}^{\text{2}}}\text{)}}^{\text{2}}}}}\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{-2(1-x}{{\text{)}}^{\text{2}}}}{\text{(1+}{{\text{x}}^{\text{2}}}\text{)}} \\ & \Rightarrow \dfrac{\text{1-}{{\text{x}}^{\text{2}}}}{\text{1+}{{\text{x}}^{\text{2}}}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{-2(1-x}{{\text{)}}^{\text{2}}}}{\text{(1+}{{\text{x}}^{\text{2}}}\text{)}} \\ \end{align} $ 

Therefore,  $\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{-2}}{\text{(1+}{{\text{x}}^{\text{2}}}\text{)}} $ .


14.  $ \mathbf{y=si}{{\mathbf{n}}^{\mathbf{-1}}}\left( \mathbf{2x}\sqrt{\mathbf{1-}{{\mathbf{x}}^{\mathbf{2}}}} \right)\mathbf{,}\text{ }\mathbf{-}\dfrac{\mathbf{1}}{\sqrt{\mathbf{2}}}<\mathbf{x}<\dfrac{\mathbf{1}}{\sqrt{\mathbf{2}}} $ .

Ans:  The given equation is  $ \text{y=si}{{\text{n}}^{\text{-1}}}\left( \text{2x}\sqrt{\text{1-}{{\text{x}}^{\text{2}}}} \right) $ .

Now,  $ \text{y=si}{{\text{n}}^{\text{-1}}}\left(\text{2x}\sqrt{\text{1-}{{\text{x}}^{\text{2}}}} \right) $ 

 $ \Rightarrow \text{siny=2x}\sqrt{\text{1-}{{\text{x}}^{\text{2}}}} $ .                                                                                  …… (1) 

Differentiating both sides of the equation with respect to  $ \text{x} $ , gives  $ \text{cosy}\dfrac{\text{dy}}{\text{dx}}\text{=2}\left[ \text{x}\dfrac{\text{d}}{\text{dx}}\left( \sqrt{\text{1-}{{\text{x}}^{\text{2}}}} \right)\text{+}\sqrt{\text{1-}{{\text{x}}^{\text{2}}}}\dfrac{\text{dx}}{\text{dx}} \right] $ 

$\Rightarrow\sqrt{\text{1-si}{{\text{n}}^{\text{2}}}\text{y}}\dfrac{\text{dy}}{\text{dx}}\text{=2}\left[ \dfrac{\text{x}}{\text{2}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{-2}}{\sqrt{\text{1-}{{\text{x}}^{\text{2}}}}}\text{+}\sqrt{\text{1-}{{\text{x}}^{\text{2}}}} \right] $ 

 $\Rightarrow\sqrt{\text{1-(2x}\sqrt{\text{1-}{{\text{x}}^{\text{2}}}}{{\text{)}}^{\text{2}}}}\dfrac{\text{dy}}{\text{dx}}\text{=2}\left[ \dfrac{\text{-}{{\text{x}}^{\text{2}}}\text{+1-}{{\text{x}}^{\text{2}}}}{\sqrt{\text{1-}{{\text{x}}^{\text{2}}}}} \right] $ , using the equation (1).

$ \begin{align}& \Rightarrow \sqrt{\text{1-4}{{\text{x}}^{\text{2}}}\text{(1-}{{\text{x}}^{\text{2}}}\text{)}}\dfrac{\text{dy}}{\text{dx}}\text{=2}\left[ \dfrac{\text{1-2}{{\text{x}}^{\text{2}}}}{\sqrt{\text{1-}{{\text{x}}^{\text{2}}}}} \right] \\ & \Rightarrow \sqrt{{{\text{(1-2x)}}^{\text{2}}}}\dfrac{\text{dy}}{\text{dx}}\text{=2}\left[ \dfrac{\text{1-2}{{\text{x}}^{\text{2}}}}{\sqrt{\text{1-}{{\text{x}}^{\text{2}}}}} \right] \\ & \Rightarrow \left( \text{1-2}{{\text{x}}^{\text{2}}} \right)\dfrac{\text{dy}}{\text{dx}}\text{=2}\left[ \dfrac{\text{1-2}{{\text{x}}^{\text{2}}}}{\sqrt{\text{1-}{{\text{x}}^{\text{2}}}}} \right] \\ \end{align} $ 

Therefore,  $\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{2}}{\sqrt{\text{1-}{{\text{x}}^{\text{2}}}}} $ .


15.  $ \mathbf{y=se}{{\mathbf{c}}^{\mathbf{-1}}}\left( \dfrac{\mathbf{1}}{\mathbf{2}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{-1}} \right)\mathbf{,}\text{  }\mathbf{0}<\mathbf{x}<\dfrac{\mathbf{1}}{\sqrt{\mathbf{2}}} $ .

Ans: The given equation is  $ \text{y=se}{{\text{c}}^{\text{-1}}}\left( \dfrac{\text{1}}{\text{2}{{\text{x}}^{\text{2}}}\text{-1}} \right) $ .

Now,

 $ \begin{align}& \text{y=se}{{\text{c}}^{\text{-1}}}\left( \dfrac{\text{1}}{\text{2}{{\text{x}}^{\text{2}}}\text{-1}} \right) \\ & \Rightarrow \text{secy=}\dfrac{\text{1}}{\text{2}{{\text{x}}^{\text{2}}}\text{-1}} \\ & \Rightarrow \text{cosy=2}{{\text{x}}^{\text{2}}}\text{-1} \\ & \Rightarrow \text{2}{{\text{x}}^{\text{2}}}\text{=1+cosy} \\ & \Rightarrow \text{2}{{\text{x}}^{\text{2}}}\text{=2co}{{\text{s}}^{\text{2}}}\dfrac{\text{y}}{\text{2}} \\ \end{align} $ 

 $ \Rightarrow \text{x=cos}\dfrac{\text{y}}{\text{2}} $                                                                     ……. (1)  

Differentiating both sides of the equation with respect to  $ \text{x} $ , gives 

 $ \begin{align}& \dfrac{\text{d}}{\text{dx}}\text{(x)=}\dfrac{\text{d}}{\text{dx}}\left( \text{cos}\dfrac{\text{y}}{\text{2}} \right) \\ & \Rightarrow \text{1=sin}\dfrac{\text{y}}{\text{2}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\left( \dfrac{\text{y}}{\text{2}} \right) \\ & \Rightarrow\dfrac{\text{-1}}{\text{sin}\dfrac{\text{y}}{\text{2}}}\text{=}\dfrac{\text{1}}{\text{2}}\dfrac{\text{dy}}{\text{dx}} \\ \end{align} $ 

$\Rightarrow \frac{dy}{dx}=\frac{-2}{sin\frac{y}{2}}=\frac{-2}{\sqrt{1-cos^2}\frac{y}{2}}$

 $ =\dfrac{-2}{\sqrt{1-{{\text{x}}^{2}}}} $ , using the equation (1).

Therefore,  $ \dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{-2}}{\sqrt{{1-x^2}}} $ .


Exercise 5.4 

Differentiate the following w.r.t.  $ \mathbf{x} $ .    

1.  $ \text{y=}\dfrac{{{\text{e}}^{\text{x}}}}{\text{sinx}} $ . 

Ans: The given function is   $ \text{y=}\dfrac{{{\text{e}}^{\text{x}}}}{\text{sinx}} $ .

Then, we have

$\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{sinx}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{e}}^{\text{x}}}\text{)-}{{\text{e}}^{\text{x}}}\dfrac{\text{d}}{\text{dx}}\text{(sinx)}}{\text{si}{{\text{n}}^{\text{2}}}\text{x}} $ , by applying the quotient rule of derivatives.

$=\dfrac{\text{sinx}\!\!\times\!\!\text{(}{{\text{e}}^{\text{x}}}\text{)-}{{\text{e}}^{\text{x}}}\text{}\!\!\times\!\!\text{(cosx)}}{\text{si}{{\text{n}}^{\text{2}}}\text{x}} $ 

Therefore, the derivative of the function  $ \text{y} $  is

 $\dfrac{\text{dy}}{\text{dx}}=\dfrac{{{\text{e}}^{\text{x}}}\text{(sinx-cosx)}}{\text{si}{{\text{n}}^{\text{2}}}\text{x}}\text{,}\,\,\,\,\text{x}\ne \text{n }\!\!\pi\!\!\text{ ,}\,\,\,\text{n}\in \text{Z} $ .


2.  $ \text{y=}{{\text{e}}^{\text{si}{{\text{n}}^{\text{-1}}}\text{x}}} $ .

Ans: The given function is  $ \text{y=}{{\text{e}}^{\text{si}{{\text{n}}^{\text{-1}}}\text{x}}} $ .

Then, we have

 $ \begin{align}& \dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{e}}^{\text{si}{{\text{n}}^{\text{-1}}}\text{x}}}\text{)} \\ & \text{=}{{\text{e}}^{\text{si}{{\text{n}}^{\text{-1}}}\text{x}}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(si}{{\text{n}}^{\text{-1}}}\text{x)} \\ & \text{=}{{\text{e}}^{\text{si}{{\text{n}}^{\text{-1}}}\text{x}}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\sqrt{\text{1-}{{\text{x}}^{\text{2}}}}} \\ & =\dfrac{{{\text{e}}^{\text{si}{{\text{n}}^{\text{-1}}}\text{x}}}}{\sqrt{\text{1-}{{\text{x}}^{\text{2}}}}} \end{align} $ 

Therefore, the derivative of the function  $ \text{y} $  is

$\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{{{\text{e}}^{\text{si}{{\text{n}}^{\text{-1}}}\text{x}}}}{\sqrt{\text{1-}{{\text{x}}^{\text{2}}}}}\text{,}\,\,\text{x}\in \text{(-1,1)} $ .


3.  $ \mathbf{y=}{{\mathbf{e}}^{{{\mathbf{x}}^{\mathbf{3}}}}} $ .

Ans: The given function is  $ \text{y=}{{\text{e}}^{{{\text{x}}^{3}}}} $ .

Then by applying the chain rule of derivatives we have,

$\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{e}}^{{{\text{x}}^{3}}}}\text{)=}{{\text{e}}^{{{\text{x}}^{3}}}}\text{}\!\!\times\!\!\text{}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{x}}^{3}}\text{)=}{{\text{e}}^{{{\text{x}}^{\text{3}}}}}\text{ }\!\!\times\!\!\text{ 3}{{\text{x}}^{\text{2}}} $ .

Therefore, the derivative of the function  $ \text{y} $  is

$\dfrac{\text{dy}}{\text{dx}}\text{=3}{{\text{x}}^{\text{2}}}{{\text{e}}^{{{\text{x}}^{\text{3}}}}} $ .


4.$\mathbf{y}=\mathbf{sin(ta}{{\mathbf{n}}^{\mathbf{-1}}}{{\mathbf{e}}^{\mathbf{-x}}}\mathbf{)} $ . 

Ans:  The given function is  $ \text{y=sin(ta}{{\text{n}}^{\text{-1}}}{{\text{e}}^{\text{-x}}}\text{)} $ .

Now, applying the chain rule of derivatives, give 

 $ \begin{align}& \dfrac{\text{dy}}{\text{dx}}=\dfrac{\text{d}}{\text{dx}}\left[ \text{sin(ta}{{\text{n}}^{\text{-1}}}{{\text{e}}^{\text{-x}}}\text{)} \right] \\ & \text{=cos(ta}{{\text{n}}^{\text{-1}}}{{\text{e}}^{\text{-x}}}\text{) }\!\!\times\!\!\text{}\dfrac{\text{d}}{\text{dx}}\text{(ta}{{\text{n}}^{\text{-1}}}{{\text{e}}^{\text{-x}}}\text{)} \\ &\text{=cos(ta}{{\text{n}}^{\text{-1}}}{{\text{e}}^{\text{-x}}}\text{)}\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{1+(}{{\text{e}}^{\text{-x}}}{{\text{)}}^{2}}}\text{ }\!\!\times\!\!\text{}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{e}}^{\text{-x}}}\text{)} \\ &=\text{-}\dfrac{\text{cos(ta}{{\text{n}}^{\text{-1}}}{{\text{e}}^{\text{-x}}}\text{)}}{\text{1+}{{\text{e}}^{\text{-2x}}}}\text{ }\!\!\times\!\!\text{ }{{\text{e}}^{\text{-x}}} \\ \end{align} $ 

Therefore, the derivative of the function  $ \text{y} $  is

$\dfrac{\text{dy}}{\text{dx}}=\dfrac{\text{-}{{\text{e}}^{\text{-x}}}\text{cos(ta}{{\text{n}}^{\text{-1}}}{{\text{e}}^{\text{-x}}}\text{)}}{\text{1+}{{\text{e}}^{\text{-2x}}}} $ .


5.  $ \mathbf{y=log}\left( \mathbf{cos}\left( {{\mathbf{e}}^{\mathbf{x}}} \right) \right) $ 

Ans:  Let  $ \text{y=log}\left( \text{cos}\left( {{\text{e}}^{\text{x}}} \right) \right) $ 

Now, by applying the chain rule of derivatives give

$\begin{align}&\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\left[ \text{log}\left( \text{cos}\left( {{\text{e}}^{\text{x}}} \right) \right) \right] \\ & =\dfrac{\text{1}}{\text{cos}{{\text{e}}^{\text{x}}}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\left( \text{cos}\left( {{\text{e}}^{\text{x}}} \right) \right) \\ & =\dfrac{\text{1}}{\text{cos}{{\text{e}}^{\text{x}}}}\text{ }\!\!\times\!\!\text{ }\left( \text{-sin}\left( {{\text{e}}^{\text{x}}} \right) \right)\text{ }\!\!\times\!\!\text{}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{e}}^{\text{x}}}\text{)} \\ &\text{=}\dfrac{\text{-sin}{{\text{e}}^{\text{x}}}}{\text{cos}{{\text{e}}^{\text{x}}}}\text{ }\!\!\times\!\!\text{ }{{\text{e}}^{\text{x}}}  \end{align} $ 

Therefore, the derivative of the function  $ \text{y} $  is

$ \dfrac{\text{dy}}{\text{dx}}\text{=-}{{\text{e}}^{\text{x}}}\text{tan}\left( {{\text{e}}^{\text{x}}} \right)\text{,}\,\,\,\,\text{x}\ne \text{(2n+1)}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{2}}\text{,}\,\,\text{n}\in N  $ .


6. $\text{y=}{{\text{e}}^{\text{x}}}\text{+}{{\text{e}}^{{{\text{x}}^{\text{2}}}}}\text{+}...\text{+}{{\text{e}}^{{{\text{x}}^{\text{5}}}}} $ .

Ans: The given function is  $\text{y=}{{\text{e}}^{\text{x}}}\text{+}{{\text{e}}^{{{\text{x}}^{\text{2}}}}}\text{+}...\text{+}{{\text{e}}^{{{\text{x}}^{\text{5}}}}} $ .

Then, differentiating with respect to  $ \text{x} $  both sides, give

$\dfrac{\text{dy}}{\text{dx}}=\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{e}}^{\text{x}}}\text{+}{{\text{e}}^{{{\text{x}}^{\text{2}}}}}\text{+}...\text{+}{{\text{e}}^{{{\text{x}}^{\text{5}}}}}\text{)} $ 

$\text{=}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{e}}^{\text{x}}}\text{)+}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{e}}^{\text{x}}}^{^{\text{2}}}\text{)+}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{e}}^{\text{x}}}^{^{\text{4}}}\text{)+}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{e}}^{\text{x}}}^{^{\text{5}}}\text{)} $ , applying the sum rule of derivatives.

$ \begin{align}& \text{=}{{\text{e}}^{\text{x}}}\text{+}\left[ {{\text{e}}^{{{\text{x}}^{\text{2}}}}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\left( {{\text{x}}^{\text{2}}} \right) \right]\text{+}\left[ {{\text{e}}^{{{\text{x}}^{\text{3}}}}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\left( {{\text{x}}^{\text{3}}} \right) \right]\text{+}\left[ {{\text{e}}^{{{\text{x}}^{\text{4}}}}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\left( {{\text{x}}^{\text{4}}} \right) \right]\text{+}\left[ {{\text{e}}^{{{\text{x}}^{\text{5}}}}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\left( {{\text{x}}^{\text{5}}} \right) \right] \\  &\text{=}{{\text{e}}^{\text{x}}}\text{+(}{{\text{e}}^{{{\text{x}}^{\text{2}}}}}\text{ }\!\!\times\!\!\text{ 2x)+(}{{\text{e}}^{{{\text{x}}^{\text{3}}}}}\text{ }\!\!\times\!\!\text{3}{{\text{x}}^{\text{2}}}\text{)+(}{{\text{e}}^{{{\text{x}}^{\text{4}}}}}\text{}\!\!\times\!\!\text{4}{{\text{x}}^{\text{3}}}\text{)+(}{{\text{e}}^{{{\text{x}}^{\text{5}}}}}\text{ }\!\!\times\!\!\text{ 5}{{\text{x}}^{\text{4}}}\text{)} \\ \end{align} $ 

Therefore, the derivative of the function  $ \text{y} $  is

$\dfrac{\text{dy}}{\text{dx}}\text{=}{{\text{e}}^{\text{x}}}\text{+2x}{{\text{e}}^{{{\text{x}}^{\text{2}}}}}\text{+3}{{\text{x}}^{\text{2}}}{{\text{e}}^{{{\text{x}}^{\text{3}}}}}\text{+4}{{\text{x}}^{\text{3}}}{{\text{e}}^{{{\text{x}}^{\text{4}}}}}\text{+5}{{\text{x}}^{\text{4}}}{{\text{e}}^{{{\text{x}}^{\text{5}}}}} $ .


7. $\mathbf{y=}\sqrt{{{\mathbf{e}}^{\sqrt{\mathbf{x}}}}}\mathbf{,}\,\,\,\mathbf{x>0} $  .

Ans: The given function is  $ \text{y=}\sqrt{{{\text{e}}^{\sqrt{\text{x}}}}} $ .

Then squaring both sides both sides of the equation give

$ {{\text{y}}^{\text{2}}}\text{=}{{\text{e}}^{\sqrt{\text{x}}}} $ 

Now, differentiating both sides with respect to  $ \text{x} $  gives

$ \dfrac{\text{d}}{\text{dx}}\left( {{\text{y}}^{2}} \right)=\dfrac{\text{d}}{\text{dx}}\left( {{\text{e}}^{\sqrt{\text{x}}}} \right) $ 

$ \begin{align}& \Rightarrow2\text{y}\dfrac{\text{dy}}{\text{dx}}\text{=}{{\text{e}}^{\sqrt{\text{x}}}}\dfrac{\text{d}}{\text{dx}}\text{(}\sqrt{\text{x}}\text{)} \\ & \Rightarrow 2\text{y}\dfrac{\text{dy}}{\text{dx}}\text{=}{{\text{e}}^{\sqrt{\text{x}}}}\dfrac{\text{1}}{\text{2}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\sqrt{\text{x}}} \\ & \Rightarrow \dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{{{\text{e}}^{\sqrt{\text{x}}}}}{\text{4y}\sqrt{\text{x}}} \\ \end{align} $ 

$\Rightarrow\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{{{\text{e}}^{\sqrt{\text{x}}}}}{\text{4}\sqrt{{{\text{e}}^{\sqrt{\text{x}}}}}\sqrt{\text{x}}} $ , substituting the value of  $ \text{y} $ .

Therefore, 

$\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{{{\text{e}}^{\sqrt{\text{x}}}}}{\text{4}\sqrt{\text{x}{{\text{e}}^{\sqrt{\text{x}}}}}}\text{,}\,\,\text{x0} $ .


8.  $ \mathbf{y=log(logx),}\,\,\mathbf{x>1} $ .

Ans: The given function is  $ \text{y=log(logx)} $ .

Now, differentiating both sides with respect to  $ \text{x} $  gives 

$ \dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{ }\!\![\!\!\text{ log(logx) }\!\!]\!\!\text{ } $ 

$=\dfrac{\text{1}}{\text{logx}}\text{}\!\!\times\!\!\text{}\dfrac{\text{d}}{\text{dx}}\text{(logx)} $ , by applying the chain rule of derivatives.

$=\dfrac{\text{1}}{\text{logx}}\text{}\!\!\times\!\!\text{}\dfrac{\text{1}}{\text{x}} $ 

Therefore,  $ \dfrac{\text{dy}}{\text{dx}}=\dfrac{\text{1}}{\text{xlog}\left( \text{x} \right)}\text{,  x>1} $ .


9.$\mathbf{y=}\dfrac{\mathbf{cosx}}{\mathbf{logx}}\mathbf{,}\,\,\mathbf{x>0} $ .

Ans: The given function is  $ \text{y=}\dfrac{\text{cosx}}{\text{logx}} $ .

Differentiating both sides with respect to  $ \text{x} $  gives

$\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\dfrac{\text{d}}{\text{dx}}\text{(cos) }\!\!\times\!\!\text{ logx-cosx }\!\!\times\!\!\text{}\dfrac{\text{d}}{\text{dx}}\text{(logx)}}{{{\text{(logx)}}^{\text{2}}}} $ , by applying the quotient rule.

$=\dfrac{\text{-sinxlogx-cosx}\!\!\times\!\!\text{}\dfrac{\text{1}}{\text{x}}}{{{\text{(logx)}}^{\text{2}}}} $ 

Therefore,

$ \dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{- }\!\![\!\!\text{ xlogxsinx+cosx}\!\!]\!\!\text{}}{\text{x(logx}{{\text{)}}^{\text{2}}}}\text{,}\,\,\,\text{x>0} $ .


10.$\mathbf{y=cos(logx+}{{\mathbf{e}}^{\mathbf{x}}}\mathbf{),}\,\,\mathbf{x>0} $ . 

Ans: The given function is  $ \text{y=cos(logx+}{{\text{e}}^{\text{x}}}\text{)} $ .

Then differentiating both sides with respect to  $ \text{x} $  gives

 $ \dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\left[ \text{cos}\left( \text{logx+}{{\text{e}}^{\text{x}}} \right) \right] $ .

 $ \Rightarrow \dfrac{\text{dy}}{\text{dx}}\text{=-sin }\!\![\!\!\text{ logx+}{{\text{e}}^{\text{x}}}\text{ }\!\!]\!\!\text{  }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(logx+}{{\text{e}}^{\text{x}}}\text{)} $ , by applying the chain rule of derivatives.

$ \begin{align} & \text{=-sin(logx+}{{\text{e}}^{\text{x}}}\text{) }\!\!\times\!\!\text{ }\left[ \dfrac{\text{d}}{\text{dx}}\text{(logx)+}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{e}}^{\text{x}}}\text{)} \right] \\ & \text{=-sin(logx+}{{\text{e}}^{\text{x}}}\text{) }\!\!\times\!\!\text{ }\left( \dfrac{\text{1}}{\text{x}}\text{+}{{\text{e}}^{\text{x}}} \right) \\ \end{align} $ 

Therefore,  $ \dfrac{\text{dy}}{\text{dx}}=\text{-}\left( \dfrac{\text{1}}{\text{x}}\text{+}{{\text{e}}^{\text{x}}} \right)\text{sin(logx+}{{\text{e}}^{\text{x}}}\text{)}\,\,\,\text{} $ .


Exercise 5.5

Differentiate the functions given in Exercises 1 to 11 w.r.t  $ \mathbf{x} $ .

1.  $ \text{y=cosx }\!\!\times\!\!\text{ cos2x }\!\!\times\!\!\text{ cos3x} $ .

Ans: The given function is  $ \text{y=cosx }\!\!\times\!\!\text{ cos2x }\!\!\times\!\!\text{ cos3x} $ .

First, taking logarithm both sides of the equation give,

 $ \text{logy=log(cosx }\!\!\times\!\!\text{ cos2x }\!\!\times\!\!\text{ cos3x)} $ 

 $ \Rightarrow \text{logy=log(cosx)+log(cos2x)+log(cos3x)} $ , by the property of logarithm.

Now, differentiating both sides of the equation with respect to  $ \text{x} $  gives

$ \begin{align}& \dfrac{\text{1}}{\text{y}}\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{1}}{\text{cosx}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(cosx)+}\dfrac{\text{1}}{\text{cos2x}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(cos2x)+}\dfrac{\text{1}}{\text{cos3x}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(cos3x)} \\ & \Rightarrow \dfrac{\text{dy}}{\text{dx}}\text{=y}\left[ \text{-}\dfrac{\text{sinx}}{\text{cosx}}\text{-}\dfrac{\text{sin2x}}{\text{cos2x}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(2x)-}\dfrac{\text{sin3x}}{\text{cos3x}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(3x)} \right] \\ \end{align} $ 

Therefore,

$ \dfrac{\text{dy}}{\text{dx}}\text{=-cos x }\!\!\times\!\!\text{ cos2x }\!\!\times\!\!\text{ cos3x}\left[ \text{tanx+2tan2x+3tan3x} \right] $ .


2.  $ \mathbf{y=}\sqrt{\dfrac{\mathbf{(x-1)(x-2)}}{\mathbf{(x-3)(x-4)(x-5)}}} $ .

Ans: The given function is  $\text{y=}\sqrt{\dfrac{\text{(x-1)(x-2)}}{\text{(x-3)(x-4)(x-5)}}} $ .

First taking logarithm both sides of the equation give

$ \begin{align}& \text{logy=log}\sqrt{\dfrac{\text{(x-1)(x-2)}}{\text{(x-3)(x-4)(x-5)}}} \\ & \Rightarrow \text{logy=}\dfrac{\text{1}}{\text{2}}\text{log}\left[ \dfrac{\text{(x-1)(x-2)}}{\text{(x-3)(x-4)(x-5)}} \right] \\ & \Rightarrow \text{logy=}\dfrac{\text{1}}{\text{2}}\left[ \text{log }\!\!\{\!\!\text{ (x-1)(x-2) }\!\!\}\!\!\text{ -log }\!\!\{\!\!\text{ (x-3)(x-4)(x-5) }\!\!\}\!\!\text{ } \right] \\ & \Rightarrow \text{logy=}\dfrac{\text{1}}{\text{2}}\text{ }\!\![\!\!\text{ log(x-1)+log(x-2)-log(x-3)-log(x-4)-log(x-5) }\!\!]\!\!\text{ } \\ \end{align} $ 

Now, differentiating both sides of the equation with respect to  $ \text{x} $  give

$\dfrac{\text{dy}}{\text{dx}}=\dfrac{\text{1}}{\text{2}}\dfrac{\text{d}}{\text{dx}}\text{ }\!\![\!\!\text{ log(x-1)+log(x-2)-log(x-3)-log(x-4)-log(x-5) }\!\!]\!\!\text{ } $ .

$\begin{align}&\dfrac{\text{1}}{\text{y}}\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{1}}{\text{2}}\left[ \dfrac{\text{1}}{\text{x-1}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(x-1)+}\dfrac{\text{1}}{\text{x-2}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(x-2)-}\dfrac{\text{1}}{\text{x-3}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(x-3)-}\dfrac{\text{1}}{\text{x-4}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(x-4)} \right. \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left. \text{-}\dfrac{\text{1}}{\text{x-5}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(x-5)} \right] \\ \end{align} $

$ \Rightarrow \dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{y}}{\text{2}}\left( \dfrac{\text{1}}{\text{x-1}}\text{+}\dfrac{\text{1}}{\text{x-2}}\text{-}\dfrac{\text{1}}{\text{x-3}}\text{-}\dfrac{\text{1}}{\text{x-4}}\text{-}\dfrac{\text{1}}{\text{x-5}} \right) $ 

Therefore,

$\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{1}}{\text{2}}\sqrt{\dfrac{\text{(x-1)(x-2)}}{\text{(x-3)(x-4)(x-5)}}}\left[\dfrac{\text{1}}{\text{x-1}}\text{+}\dfrac{\text{1}}{\text{x-2}}\text{-}\dfrac{\text{1}}{\text{x-3}}\text{-}\dfrac{\text{1}}{\text{x-4}}\text{-}\dfrac{\text{1}}{\text{x-5}} \right] $ .


3. Differentiate the function  $ \text{y=(logx}{{\text{)}}^{\text{cosx}}} $  with respect to  $ \mathbf{x} $ .

Ans: The given function is  $ \text{y=(logx}{{\text{)}}^{\text{cosx}}} $ .

First, taking logarithm both sides of the equation give 

 $ \text{logy=cosx}\text{.log(logx)} $ .

Now, differentiating both sides of the equation with respect to  $ \text{x} $  give

$ \dfrac{\text{1}}{\text{y}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{(cosx) }\!\!\times\!\!\text{ log(logx)+cosx }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\left[ \text{log(logx)} \right] $ 

$ \Rightarrow \dfrac{\text{1}}{\text{y}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{dy}}{\text{dx}}\text{=-sinxlog(logx)+cosx }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{logx}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(logx)} $ , by applying the chain rule.

$ \Rightarrow \dfrac{\text{dy}}{\text{dx}}\text{=y}\left[ \text{-sinxlog(logx)+}\dfrac{\text{cosx}}{\text{logx}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{x}} \right] $ 

Therefore,

$ \dfrac{\text{dy}}{\text{dx}}\text{=(logx}{{\text{)}}^{\text{cosx}}}\left[ \dfrac{\text{cosx}}{\text{xlogx}}\text{-sinx }\!\!\times\!\!\text{ log(logx)} \right] $ .


4.  $ \text{y=}{{\text{x}}^{\mathbf{x}}}\text{-}{{\text{2}}^{\text{sinx}}} $ .

Ans: The given function is  $ \text{y=}{{\text{x}}^{\text{x}}}\text{-}{{\text{2}}^{\text{sinx}}} $ .

Now, let  $ {{\text{x}}^{\text{x}}}\text{=u} $                                                                       …… (1)

and  $ {{\text{2}}^{\text{sinx}}}\text{=v} $ .                                                                           …… (2)

Therefore,  $ \text{y=u-v} $ .                                                                 …… (3)

Then differentiating the equation (3) with respect to  $ \text{x} $  gives

 $ \dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{du}}{\text{dx}}-\dfrac{\text{dv}}{\text{dx}} $                                                                        …… (4)

Now, taking logarithm both sides of the equation (1) give

 $ \begin{align} & \text{log}\left( \text{u} \right)=\log \left( {{\text{x}}^{\text{x}}} \right) \\ & \Rightarrow \log \text{u=xlogx} \\ \end{align} $ 

Differentiating both sides of the equation with respect to  $ \text{x} $  gives $\begin{align}&\dfrac{\text{1}}{\text{u}}\dfrac{\text{du}}{\text{dx}}\text{=}\left[ \dfrac{\text{d}}{\text{dx}}\text{(x) }\!\!\times\!\!\text{ logx+x }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(logx)} \right] \\ & \Rightarrow \dfrac{\text{du}}{\text{dx}}\text{=u}\left[ \text{1 }\!\!\times\!\!\text{ logx+x }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{x}} \right] \\  & \Rightarrow \dfrac{\text{du}}{\text{dx}}\text{=}{{\text{x}}^{\text{x}}}\text{(logx+1)} \\ \end{align} $ 

$\Rightarrow\dfrac{\text{du}}{\text{dx}}\text{=}{{\text{x}}^{\text{x}}}\text{(1+logx)} $                                                       …… (5) 

Now, taking logarithm both sides of the equation (2) give

$ \text{log}\left( {{\text{2}}^{\text{sinx}}} \right)\text{=logv} $ 

$ \Rightarrow \text{logv=sinx }\!\!\times\!\!\text{ log2} $ .

Differentiating both sides of the equation with respect to  $ \text{x} $ , give

$ \begin{align} & \dfrac{\text{1}}{\text{v}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{dv}}{\text{dx}}\text{=log2 }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(sinx}) \\ & \Rightarrow \dfrac{\text{dv}}{\text{dx}}\text{=vlog2cosx} \\ \end{align} $ 

$\Rightarrow\dfrac{\text{dv}}{\text{dx}}\text{=}{{\text{2}}^{\text{sinx}}}\text{cosxlog2} $                                                  …… (6)      

Therefore, from the equation (4), (5) and (6) give

$\dfrac{\text{dy}}{\text{dx}}\text{=}{{\text{x}}^{\text{x}}}\text{(1+logx)-}{{\text{2}}^{\text{sinx}}}\text{cosxlog}2 $ .


5.$\mathbf{y=(x+3}{{\mathbf{)}}^{\mathbf{2}}}{{\mathbf{(x+4)}}^{\mathbf{3}}}{{\mathbf{(x+5)}}^{\mathbf{4}}} $ .

Ans:  The given function is  $ \text{y=(x+3}{{\text{)}}^{\text{2}}}{{\text{(x+4)}}^{\text{3}}}{{\text{(x+5)}}^{\text{4}}} $ .

First, taking logarithm both sides of the equation give

$\text{logy=log}\left[{{\text{(x+3)}}^{\text{2}}}{{\text{(x+4)}}^{\text{3}}}{{\text{(x+5)}}^{\text{4}}} \right] $ 

$ \Rightarrow \text{logy=2log(x+3)+3log(x+4)+4log(x+5)} $ 

Now, differentiating both sides of the equation with respect to  $ \text{x} $ , give

$ \begin{align}& \dfrac{\text{1}}{\text{y}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{dy}}{\text{dy}}\text{=2 }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{x-3}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dz}}\text{(x+3)+3 }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{x+4}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(x+4)+4 }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{x+5}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(x+5)} \\ & \Rightarrow \dfrac{\text{dy}}{\text{dx}}\text{=y}\left[ \dfrac{\text{2}}{\text{x+3}}\text{+}\dfrac{\text{3}}{\text{x+4}}\text{+}\dfrac{\text{4}}{\text{x+5}} \right] \\ & \Rightarrow \dfrac{\text{dy}}{\text{dx}}\text{=(x+3}{{\text{)}}^{\text{2}}}{{\text{(x+4)}}^{\text{3}}}{{\text{(x+5)}}^{\text{4}}}\text{ }\!\!\times\!\!\text{ }\left[ \dfrac{\text{2}}{\text{x+3}}\text{+}\dfrac{\text{3}}{\text{x+4}}\text{+}\dfrac{\text{4}}{\text{x+5}} \right] \\ & \Rightarrow \dfrac{\text{dy}}{\text{dx}}\text{=(x+3}{{\text{)}}^{\text{2}}}{{\text{(x+4)}}^{\text{3}}}{{\text{(x+5)}}^{\text{4}}}\text{ }\!\!\times\!\!\text{ }\left[ \dfrac{\text{2(x+4)(x+5)+3(x+3)(x+5)+4(x+3)(x+4)}}{\text{(x+3)(x+4)(x+5)}} \right] \\ & \Rightarrow \dfrac{\text{dy}}{\text{dx}}\text{=(x+3}{{\text{)}}^{\text{2}}}{{\text{(x+4)}}^{\text{2}}}{{\text{(x+5)}}^{\text{2}}}\text{-}\left[ \text{2(}{{\text{x}}^{\text{2}}}\text{+9x+20)+3(}{{\text{x}}^{\text{2}}}\text{+9x+15)+4(}{{\text{x}}^{\text{2}}}\text{+7x+12)} \right] \\ \end{align} $ 

Therefore,

$\dfrac{\text{dy}}{\text{dx}}\text{=(x+3)(x+4}{{\text{)}}^{\text{2}}}{{\text{(x+5)}}^{\text{3}}}\text{(9}{{\text{x}}^{\text{2}}}\text{+70x+133)} $ .


6.$\mathbf{y=}{{\left(\mathbf{x+}\dfrac{\mathbf{1}}{\mathbf{x}}\right)}^{\mathbf{x}}}\mathbf{+}{{\mathbf{x}}^{\left(\mathbf{1+}\dfrac{\mathbf{1}}{\mathbf{x}} \right)}} $ .

Ans: The given function is  $ \text{y=}{{\left( \text{x+}\dfrac{\text{1}}{\text{x}} \right)}^{\text{x}}}\text{+}{{\text{x}}^{\left( \text{1+}\dfrac{\text{1}}{\text{x}} \right)}} $ .

First, let  $ \text{u=}{{\left( \text{x+}\dfrac{\text{1}}{\text{x}} \right)}^{\text{x}}} $ and  $ \text{v=}{{\text{x}}^{\left( \text{1+}\dfrac{\text{1}}{\text{x}} \right)}} $ 

Therefore,  $ \text{y=u+v} $ .                                         …… (1)

Differentiating the equation (1) both sides with respect to  $ \text{x} $  give

 $ \Rightarrow \dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{du}}{\text{dx}}\text{=}\dfrac{\text{dv}}{\text{dx}} $ …... (2)

Now,  $ \text{u=}{{\left( \text{x+}\dfrac{\text{1}}{\text{x}} \right)}^{\text{x}}} $ 

 $ \Rightarrow \text{logu=log}{{\left( \text{x+}\dfrac{\text{1}}{\text{x}} \right)}^{\text{x}}} $ 

 $ \Rightarrow \text{logu=xlog}\left( \text{x+}\dfrac{\text{1}}{\text{x}} \right) $ 

Differentiating both sides of the equation with respect to  $ \text{x} $  gives

$\begin{align}&\dfrac{\text{1}}{\text{u}}\dfrac{\text{du}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{(x) }\!\!\times\!\!\text{ log}\left( \text{x+}\dfrac{\text{1}}{\text{x}} \right)\text{+x }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\left[ \text{log}\left( \text{x+}\dfrac{\text{1}}{\text{x}} \right) \right] \\ & \Rightarrow \dfrac{\text{1}}{\text{u}}\dfrac{\text{du}}{\text{dx}}\text{=1 }\!\!\times\!\!\text{ log}\left( \text{x+}\dfrac{\text{1}}{\text{x}} \right)\text{+x }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\left( \text{x+}\dfrac{\text{1}}{\text{x}} \right)}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\left( \text{x+}\dfrac{\text{1}}{\text{x}} \right) \\ & \Rightarrow \dfrac{\text{du}}{\text{dx}}\text{=u}\left[ \text{log}\left( \text{x+}\dfrac{\text{1}}{\text{x}} \right)\text{+}\dfrac{\text{x}}{\left( \text{x+}\dfrac{\text{1}}{\text{x}} \right)}\text{ }\!\!\times\!\!\text{ }\left( \text{x+}\dfrac{\text{1}}{{{\text{x}}^{\text{2}}}} \right) \right] \\ & \Rightarrow \dfrac{\text{du}}{\text{dx}}\text{=}{{\left( \text{x+}\dfrac{\text{1}}{\text{x}} \right)}^{\text{x}}}\left[ \text{log}\left( \text{x+}\dfrac{\text{1}}{\text{x}} \right)\text{+}\dfrac{\left( \text{x-}\dfrac{\text{1}}{\text{x}} \right)}{\left( \text{x+}\dfrac{\text{1}}{\text{x}} \right)} \right] \\ & \Rightarrow \dfrac{\text{du}}{\text{dx}}\text{=}{{\left( \text{x+}\dfrac{\text{1}}{\text{x}} \right)}^{\text{x}}}\left[ \text{log}\left( \text{x+}\dfrac{\text{1}}{\text{x}} \right)\text{+}\dfrac{{{\text{x}}^{\text{2}}}\text{+1}}{{{\text{x}}^{\text{2}}}\text{-1}} \right] \\ &  \\ \end{align} $ 

$ \Rightarrow \dfrac{\text{du}}{\text{dx}}\text{=}{{\left(\text{x+}\dfrac{\text{1}}{\text{x}}\right)}^{\text{2}}}\left[\dfrac{{{\text{x}}^{\text{2}}}\text{+1}}{{{\text{x}}^{\text{2}}}\text{-1}}\text{+log}\left( \text{x+}\dfrac{\text{1}}{\text{x}} \right) \right] $ …… (3)

Also,  $ \text{v=}{{\text{x}}^{\left( \text{x+}\dfrac{\text{1}}{\text{x}} \right)}} $ 

$ \begin{align} & \Rightarrow \text{logv=log}\left[ {{\text{x}}^{{{\text{x}}^{\left( \text{x+}\dfrac{\text{1}}{\text{x}} \right)}}}} \right] \\ & \Rightarrow \text{logv=}\left( \text{x+}\dfrac{\text{1}}{\text{x}} \right)\text{logx} \\ \end{align} $ 

Differentiating both sides of the equation with respect to  $ \text{x} $  gives

$ \begin{align}& \dfrac{\text{1}}{\text{v}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{dv}}{\text{dx}}\text{=}\left[ \dfrac{\text{d}}{\text{dx}}\left( \text{1+}\dfrac{\text{1}}{\text{x}} \right) \right]\text{ }\!\!\times\!\!\text{ logx+}\left( \text{1+}\dfrac{\text{1}}{\text{x}} \right)\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{logx} \\ & \Rightarrow \dfrac{\text{1}}{\text{v}}\dfrac{\text{dv}}{\text{dx}}\text{=-}\dfrac{\text{logx}}{{{\text{x}}^{\text{2}}}}\text{+}\dfrac{\text{1}}{\text{x}}\text{+}\dfrac{\text{1}}{{{\text{x}}^{\text{2}}}} \\ \end{align} $ 

$\Rightarrow\dfrac{\text{dv}}{\text{dx}}\text{=v}\left[\dfrac{\text{-logx+x+1}}{{{\text{x}}^{\text{2}}}} \right] $           ……. (4)

Hence, from the equations (2), (3) and (4), give

$ \dfrac{\text{dy}}{\text{dx}}\text{=}{{\left( \text{x+}\dfrac{\text{1}}{\text{x}} \right)}^{\text{x}}}\left[\dfrac{{{\text{x}}^{\text{2}}}\text{-1}}{{{\text{x}}^{\text{2}}}\text{+1}}\text{+log}\left( \text{x+}\dfrac{\text{1}}{\text{x}} \right)\right]\text{+}{{\text{x}}^{\left( \text{x+}\dfrac{\text{1}}{\text{x}} \right)}}\left( \dfrac{\text{x+1-logx}}{{{\text{x}}^{\text{2}}}} \right) $ .


7.$\mathbf{y=(logx}{{\mathbf{)}}^{\mathbf{x}}}\mathbf{+}{{\mathbf{x}}^{\mathbf{logx}}} $ .

Ans: The given function is  $ \text{y=(logx}{{\text{)}}^{\text{x}}}\text{+}{{\text{x}}^{\text{logx}}} $ .

Then, let  $ \text{u=(logx}{{\text{)}}^{\text{x}}} $  and  $ \text{v=}{{\text{x}}^{\text{logx}}} $ .

Therefore,  $ \text{y=u+v} $ .

Differentiating both sides of the equation with respect to  $ \text{x} $  gives

$\Rightarrow\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{du}}{\text{dx}}\text{+}\dfrac{\text{dv}}{\text{dx}} $                             ……. (1)

Now,  $ \text{u=(logx}{{\text{)}}^{\text{x}}} $ 

$ \begin{align}& \Rightarrow \text{logu=log}\left[ {{\text{(logx)}}^{\text{x}}} \right] \\ & \Rightarrow \text{logu=xlog(logx)} \\ \end{align} $ 

Differentiating both sides of the equation with respect to  $ \text{x} $  gives

$ \begin{align}& \dfrac{\text{1}}{\text{u}}\dfrac{\text{du}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{(x) }\!\!\times\!\!\text{ log(logx)+x }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\left[ \text{log(logx)} \right] \\ & \Rightarrow \dfrac{\text{du}}{\text{dx}}\text{=u}\left[ \text{1 }\!\!\times\!\!\text{ log(logx)+x }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{logx}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(logx)} \right] \\ & \Rightarrow \dfrac{\text{du}}{\text{dx}}\text{=(logx}{{\text{)}}^{\text{x}}}\left[ \text{log(logx)+}\dfrac{\text{x}}{\text{logx}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{x}} \right] \\ \end{align} $ 

$ \begin{align}& \Rightarrow \dfrac{\text{du}}{\text{dx}}\text{=(logx}{{\text{)}}^{\text{x}}}\left[ \text{log(logx)+}\dfrac{\text{1}}{\text{logx}} \right] \\ & \Rightarrow \dfrac{\text{du}}{\text{dx}}\text{=(logx}{{\text{)}}^{\text{x}}}\text{=}\left[ \dfrac{\text{log(logx) }\!\!\times\!\!\text{ logx+1}}{\text{logx}} \right] \\ \end{align} $ 

$ \dfrac{\text{du}}{\text{dx}}\text{=(logx}{{\text{)}}^{\text{x-1}}}\left[ \text{1+logx }\!\!\times\!\!\text{ log(logx)} \right] $ ……. (2)

Again,  $ \text{v=}{{\text{x}}^{\text{logx}}} $ 

$ \begin{align}& \Rightarrow \log \text{v=log}\left( {{\text{x}}^{\text{logx}}} \right) \\ & \Rightarrow \log \text{v=logxlogx=}{{\left( \log \text{x} \right)}^{2}} \\ \end{align} $

Differentiating both sides of the equation with respect to  $ \text{x} $  gives

$ \begin{align}& \dfrac{\text{1}}{\text{v}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{dx}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\left[ {{\text{(logx)}}^{\text{2}}} \right] \\ & \Rightarrow \dfrac{\text{1}}{\text{v}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{dx}}{\text{dx}}\text{=2(logx) }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(logx)} \\ & \Rightarrow \dfrac{\text{dv}}{\text{dx}}\text{=2}{{\text{x}}^{\text{logx}}}\dfrac{\text{logx}}{\text{x}} \\ \end{align} $ 

$ \Rightarrow \dfrac{\text{dv}}{\text{dx}}\text{=2}{{\text{x}}^{\text{logx - 1}}}\text{ }\!\!\times\!\!\text{ logx} $ ….…. (3)

Hence, from the equations (1), (2), and (3), gives

$ \dfrac{\text{dy}}{\text{dx}}\text{=(logx}{{\text{)}}^{\text{x-1}}}\left[ \text{1+logx}\!\!\times\!\!\text{log(logx)}\right]\text{+2}{{\text{x}}^{\text{logx-1}}}\text{ }\!\!\times\!\!\text{ logx} $ .


8.$\mathbf{y=(sinx}{{\mathbf{)}}^{\mathbf{x}}}\mathbf{+si}{{\mathbf{n}}^{\mathbf{-1}}}\sqrt{\mathbf{x}} $ .

Ans: The given function is  $ \text{y=(sinx}{{\text{)}}^{\text{x}}}\text{+si}{{\text{n}}^{\text{-1}}}\sqrt{\text{x}} $ .

Now, let  $ \text{u=(sinx}{{\text{)}}^{\text{x}}} $  and  $\text{v=si}{{\text{n}}^{\text{-1}}}\sqrt{\text{x}} $ .

Therefore,  $ \text{y=u+v} $ .

Then, differentiating both sides of the equation with respect to  $ \text{x} $  gives

$\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{du}}{\text{dx}}\text{-}\dfrac{\text{dv}}{\text{dx}} $       …….. (1)

Now,  $ \text{u=(sinx}{{\text{)}}^{\text{x}}} $ 

$ \begin{align}& \Rightarrow \text{logu=xlog(sinx}{{\text{)}}^{\text{x}}} \\ & \Rightarrow \text{logu=xlog(sinx)} \\ \end{align} $ 

Differentiating both sides of the equation with respect to  $ \text{x} $  gives  

$ \begin{align} &\dfrac{\text{1}}{\text{u}}\dfrac{\text{du}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{(x) }\!\!\times\!\!\text{ log(sinx)+x }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\left[ \text{log(sinx)} \right] \\ & \Rightarrow \dfrac{\text{du}}{\text{dx}}\text{=u}\left[ \text{1 }\!\!\times\!\!\text{ log(sinx)+x }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{sinx}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(sinx)} \right] \\ & \Rightarrow \dfrac{\text{du}}{\text{dx}}\text{=(sinx}{{\text{)}}^{\text{x}}}\left[ \text{log(sinx)+}\dfrac{\text{x}}{\text{sinx}}\text{ }\!\!\times\!\!\text{ cosx} \right] \\ \end{align} $ 

$ \Rightarrow\dfrac{\text{du}}{\text{dx}}\text{=(sinx}{{\text{)}}^{\text{x}}}\text{(xcotx+logsinx)} $ ….… (2)

Again, $ \text{v=si}{{\text{n}}^{\text{-1}}}\sqrt{\text{x}} $ .

Differentiating both sides of the equation with respect to  $ \text{x} $  gives  

$ \begin{align}& \dfrac{\text{dv}}{\text{dx}}\text{=}\dfrac{\text{1}}{\sqrt{\text{1-(}\sqrt{\text{x}}{{\text{)}}^{\text{2}}}}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(}\sqrt{\text{x}}\text{)} \\ & \Rightarrow \dfrac{\text{dv}}{\text{dx}}\text{=}\dfrac{\text{1}}{\sqrt{\text{1-x}}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{2}\sqrt{\text{x}}} \\ \end{align} $ 

$ \Rightarrow \dfrac{\text{dv}}{\text{dx}}\text{=}\dfrac{\text{1}}{\text{2}\sqrt{\text{x-}{{\text{x}}^{\text{2}}}}} $ 

Hence, from the equations (1), (2) and (3), gives

$\dfrac{\text{dv}}{\text{dx}}\text{=(sinx}{{\text{)}}^{\text{x}}}\text{(xcotx+logsinx)+}\dfrac{\text{1}}{\text{2}\sqrt{\text{x-}{{\text{x}}^{\text{2}}}}} $ .


9. $ \mathbf{y=}{{\mathbf{x}}^{\mathbf{sinx}}}\mathbf{+(sinx}{{\mathbf{)}}^{\mathbf{cosx}}} $ .

Ans: The given function is  $ \text{y=}{{\text{x}}^{\text{sinx}}}\text{+(sinx}{{\text{)}}^{\text{cosx}}} $ .

Then, let  $ \text{u=}{{\text{x}}^{\text{sinx}}} $  and  $ \text{v=(sinx}{{\text{)}}^{\text{cosx}}} $ .

Therefore,  $ \text{y=u+v} $ .

Differentiating both sides of the equation with respect to  $ \text{x} $  gives

 $\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{du}}{\text{dx}}\text{-}\dfrac{\text{dv}}{\text{dx}} $               …… (1)

Now,  $ \text{u=}{{\text{x}}^{\text{sinx}}} $ 

$ \begin{align}& \Rightarrow \text{logu=xlog(}{{\text{x}}^{\text{sinx}}}\text{)} \\ & \Rightarrow \text{logu=sinxlogx} \\ \end{align} $ 

Differentiating both sides of the equation with respect to  $ \text{x} $  gives
$\begin{align}&\dfrac{\text{1}}{\text{u}}\dfrac{\text{du}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{(sinx) }\!\!\times\!\!\text{ logx+sinx }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(logx)} \\ & \Rightarrow \dfrac{\text{du}}{\text{dx}}\text{=u=}\left[ \text{cosxlogx+sinx }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{x}} \right] \\ \end{align} $ 

$ \Rightarrow \dfrac{\text{du}}{\text{dx}}\text{=}{{\text{x}}^{\text{sinx}}}\left[ \text{cosxlogx+}\dfrac{\text{sinx}}{\text{x}} \right] $ …....(2)

Again,  $ \text{v=(sinx}{{\text{)}}^{\text{cosx}}} $ 

$ \begin{align} & \Rightarrow \text{logv=log(sinx}{{\text{)}}^{\text{cosx}}} \\ & \Rightarrow \text{logv=cosxlog(sinx)} \\ \end{align} $ 

Then, differentiating both sides of the equation with respect to  $ \text{x} $  gives

$ \begin{align}& \dfrac{\text{1}}{\text{v}}\dfrac{\text{dv}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{(cosx) }\!\!\times\!\!\text{ log(sinx)+cosx }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\left[ \text{log(sinx)} \right] \\ & \Rightarrow \dfrac{\text{dv}}{\text{dx}}\text{=v}\left[ \text{-sinx }\!\!\times\!\!\text{ log(sinx)+cosx }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{sinx}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(sinx)} \right] \\ & \Rightarrow\dfrac{\text{du}}{\text{dx}}\text{=(sinx}{{\text{)}}^{\text{cosx}}}\left[ \text{-sinxlogsinx+cotxcosx} \right] \\ \end{align} $ $ \Rightarrow \dfrac{\text{dv}}{\text{dx}}\text{=(sinx}{{\text{)}}^{\text{cosx}}}\text{ }\!\![\!\!\text{ cosxcotx-sinxlogsinx }\!\!]\!\!\text{ } $ …… (3)

Hence, from the equations (1), (2) and (3), gives

$ \dfrac{\text{du}}{\text{dx}}\text{=}{{\text{x}}^{\text{sinx}}}\left( \text{cosxlogx+}\dfrac{\text{sinx}}{\text{x}}\right)\text{+(sinx}{{\text{)}}^{\text{cosx}}}\text{ }\!\![\!\!\text{ cosxcotx-sinxlogsinx }\!\!]\!\!\text{ } $ .


10. $\mathbf{y=}{{\mathbf{x}}^{\mathbf{xcosx}}}\mathbf{+}\dfrac{{{\mathbf{x}}^{\mathbf{2}}}\mathbf{+1}}{{{\mathbf{x}}^{\mathbf{2}}}\mathbf{-1}} $ .

Ans:  The given function is  $\text{y=}{{\text{x}}^{\text{xcosx}}}\text{+}\dfrac{{{\text{x}}^{\text{2}}}\text{+1}}{{{\text{x}}^{\text{2}}}\text{-1}} $ .

First, let  $ \text{u=}{{\text{x}}^{\text{xcosx}}} $  and  $ \text{v=}\dfrac{{{\text{x}}^{\text{2}}}\text{+1}}{{{\text{x}}^{\text{2}}}\text{-1}} $ .

Therefore,  $ \text{y=u+v} $ .

Differentiating both sides of the equation with respect to  $ \text{x} $  gives

$\Rightarrow\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{du}}{\text{dx}}\text{-}\dfrac{\text{dv}}{\text{dx}} $ ……. (1)

Now,  $ \text{u=}{{\text{x}}^{\text{xcosx}}} $ .

Then, differentiating both sides of the equation with respect to  $ \text{x} $  gives $ \begin{aligned} &\frac{1}{u} \frac{d u}{d x}=\frac{d}{d x}(x) \times \cos x \log x+x \times \frac{d}{d x}(\cos x) \times \log x+x \cos x \times \frac{d}{d x}(\log x) \\ &\Rightarrow \frac{d u}{d x}=u\left[1 \times \cos x \times \log x+x \times(-\sin x) \log x+x \cos x \times \frac{1}{x}\right] \end{aligned} $

$\Rightarrow\dfrac{\text{du}}{\text{dx}}\text{=}{{\text{x}}^{\text{xcosx}}}\text{(cosxlogx-xsinxlogx+cosx}) $ …… (2)

Again,  $\text{v=}\dfrac{{{\text{x}}^{\text{2}}}\text{+1}}{{{\text{x}}^{\text{2}}}\text{-1}} $ 

$\Rightarrow\text{logv=log(}{{\text{x}}^{\text{2}}}\text{+1)-log(}{{\text{x}}^{\text{2}}}\text{-1)} $ 

Differentiating both sides of the equation with respect to  $ \text{x} $  gives

$ \begin{align}& \dfrac{\text{1}}{\text{v}}\text{=}\dfrac{\text{dv}}{\text{dx}}\text{=}\dfrac{\text{2x}}{{{\text{x}}^{\text{2}}}\text{+1}}\text{-}\dfrac{\text{2x}}{{{\text{x}}^{\text{2}}}\text{-1}} \\ & \Rightarrow \dfrac{\text{dv}}{\text{dx}}\text{=v}\left[ \dfrac{\text{2x(}{{\text{x}}^{\text{2}}}\text{-1)-2x(}{{\text{x}}^{\text{2}}}\text{+1)}}{\text{(}{{\text{x}}^{\text{2}}}\text{+1)(}{{\text{x}}^{\text{2}}}\text{-1)}} \right] \\ & \Rightarrow \dfrac{\text{du}}{\text{dx}}\text{=}\dfrac{{{\text{x}}^{\text{2}}}\text{+1}}{{{\text{x}}^{\text{2}}}\text{-1}}\text{ }\!\!\times\!\!\text{ }\left[ \dfrac{\text{-4x}}{\text{(}{{\text{x}}^{\text{2}}}\text{+1)(}{{\text{x}}^{\text{2}}}\text{-1)}} \right] \\ \end{align} $

$\Rightarrow\dfrac{\text{dv}}{\text{dx}}\text{=}\dfrac{\text{-4x}}{{{\text{(}{{\text{x}}^{\text{2}}}\text{-1)}}^{\text{2}}}} $ …….. (3)

Hence, from the equations (1), (2) and (3), give

$ \dfrac{\text{dv}}{\text{dx}}\text{=}{{\text{x}}^{\text{xcosx}}}\left[ \text{cosx(1+logx)-xsinxlogx}\right]\text{-}\dfrac{\text{4x}}{{{\text{(}{{\text{x}}^{\text{2}}}\text{-1)}}^{\text{2}}}} $ .


11.$\mathbf{y=(xcosx}{{\mathbf{)}}^{\mathbf{x}}}\mathbf{+(xsinx}{{\mathbf{)}}^{\dfrac{\mathbf{1}}{\mathbf{x}}}} $ .

Ans: The given function is  $ \text{y=(xcosx}{{\text{)}}^{\text{x}}}\text{+(xsinx}{{\text{)}}^{\dfrac{\text{1}}{\text{x}}}} $ .

Then, let  $ \text{u=(xcosx}{{\text{)}}^{\text{x}}} $  and  $ \text{v=(xsinx}{{\text{)}}^{\dfrac{\text{1}}{\text{x}}}} $ .

Therefore,  $ \text{y=u+v} $ .

$\Rightarrow\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{du}}{\text{dx}}\text{+}\dfrac{\text{dv}}{\text{dx}} $                                                          ……. (1)

Again,  $ \text{u=(cosx}{{\text{)}}^{\text{x}}} $ 

$ \begin{align}& \Rightarrow \text{logu=log(xcosx}{{\text{)}}^{\text{x}}} \\ & \Rightarrow \text{logu=xlog(xcosx)} \\ & \Rightarrow \text{logu=x }\!\![\!\!\text{ logx+logcosx }\!\!]\!\!\text{ } \\ & \Rightarrow \text{logu=xlogx+xlogcosx} \\ \end{align} $ 

Differentiating both sides of the equation with respect to  $ \text{x} $  gives

$ \begin{align}& \dfrac{\text{1}}{\text{u}}\dfrac{\text{du}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{(xlogx+xlogcosx)} \\ & \Rightarrow \dfrac{\text{du}}{\text{dx}}\text{=u}\left[ \left\{ \text{logx }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(x)+x }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(logx)} \right\}\text{+}\left\{ \text{logcosx }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(x)+x }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(logcosx)} \right\} \right] \\ & \Rightarrow \dfrac{\text{du}}{\text{dx}}\text{=(xcosx}{{\text{)}}^{\text{x}}}\left[ \left\{ \text{logx }\!\!\times\!\!\text{ 1+x }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{x}} \right\}\text{+}\left\{ \text{logcosx-1+x }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{cosx}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(cosx)} \right\} \right] \\ \end{align} $ 

$ \begin{align}& \Rightarrow \dfrac{\text{du}}{\text{dx}}\text{=(xcosx}{{\text{)}}^{\text{x}}}\left[ \left\{ \text{logx+1} \right\}\text{+}\left\{ \text{logcosx-1+}\dfrac{\text{x}}{\text{cosx}}\text{ }\!\!\times\!\!\text{ (-sinx)} \right\} \right] \\ & \Rightarrow \dfrac{\text{du}}{\text{dx}}\text{=(xcosx}{{\text{)}}^{\text{x}}}\left[ \text{(logx+1)+(logcosx-xtanx)} \right] \\ \end{align} $ 

$\Rightarrow\dfrac{\text{du}}{\text{dx}}\text{=(xcosx}{{\text{)}}^{\text{x}}}\left[ \text{1-xtanx+(logx+logcosx)} \right] $ 

Therefore,

$ \dfrac{\text{du}}{\text{dx}}\text{=(xcosx}{{\text{)}}^{\text{x}}}\left[ \text{1-xtanx+(logx(xcosx)} \right] $   …….. (2)

Again,  $ \text{v=(xsinx}{{\text{)}}^{\dfrac{\text{1}}{\text{x}}}} $ 

$ \begin{align}& \Rightarrow \text{logv=log(xsinx}{{\text{)}}^{\dfrac{\text{1}}{\text{x}}}} \\ & \Rightarrow \text{logv=}\dfrac{\text{1}}{\text{x}}\text{log(xsinx)} \\  & \Rightarrow \text{logv=}\dfrac{\text{1}}{\text{x}}\text{(logx+logsinx)} \\ & \Rightarrow \text{logv=}\dfrac{\text{1}}{\text{x}}\text{logx+}\dfrac{\text{1}}{\text{x}}\text{logsinx} \\ \end{align} $ 

Differentiating both sides of the equation with respect to  $ \text{x} $  gives

$ \begin{align} & \dfrac{\text{1}}{\text{v}}\dfrac{\text{dv}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\left(\dfrac{\text{1}}{\text{x}}\text{logx}\right)\text{+}\dfrac{\text{d}}{\text{dx}}\left[ \dfrac{\text{1}}{\text{x}}\text{log(sinx)} \right] \\ & \Rightarrow \dfrac{\text{1}}{\text{v}}\dfrac{\text{dv}}{\text{dx}}\text{=}\left[ \dfrac{\text{1}}{\text{x}}\text{logx }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\left( \dfrac{\text{1}}{\text{x}} \right)\text{+}\dfrac{\text{1}}{\text{x}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(logx)} \right]\text{+}\left[ \text{log(sinx) }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\left( \dfrac{\text{1}}{\text{x}} \right)\text{+}\dfrac{\text{1}}{\text{x}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\left\{ \text{(logsinx)} \right\} \right] \\ & \Rightarrow \dfrac{\text{1}}{\text{v}}\dfrac{\text{dv}}{\text{dx}}\text{=}\left[ \dfrac{\text{1}}{\text{x}}\text{logx }\!\!\times\!\!\text{ }\left( \text{-}\dfrac{\text{1}}{{{\text{x}}^{\text{2}}}} \right)\text{+}\dfrac{\text{1}}{\text{x}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{x}} \right]\text{+}\left[ \text{log(sinx) }\!\!\times\!\!\text{ }\left( \text{-}\dfrac{\text{1}}{{{\text{x}}^{\text{2}}}} \right)\text{+}\dfrac{\text{1}}{\text{x}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{sinx}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(sinx)} \right] \\ & \Rightarrow \dfrac{\text{1}}{\text{v}}\dfrac{\text{dv}}{\text{dx}}\text{=}\dfrac{\text{1}}{{{\text{x}}^{\text{2}}}}\text{(1-logx)+}\left[ \dfrac{\text{1-logx}}{{{\text{x}}^{\text{2}}}}\text{+}\dfrac{\text{1}}{\text{xsinx}}\text{ }\!\!\times\!\!\text{ cosx} \right] \\ & \Rightarrow \dfrac{\text{1}}{\text{v}}\dfrac{\text{dv}}{\text{dx}}\text{=}\dfrac{\text{1}}{{{\text{x}}^{\text{2}}}}{{\text{(xsinx)}}^{\dfrac{\text{1}}{\text{x}}}}\text{+}\left[ \dfrac{\text{1-logx}}{{{\text{x}}^{\text{2}}}}\text{+}\dfrac{\text{-log(sinx)+xcotx}}{{{\text{x}}^{\text{2}}}} \right] \\ & \Rightarrow \dfrac{\text{dv}}{\text{dx}}\text{=(xsinx}{{\text{)}}^{\dfrac{\text{1}}{\text{x}}}}\left[ \dfrac{1-\log x-\log (\sin x)+x\cot x}{{{\text{x}}^{\text{2}}}} \right] \\ \end{align} $ 

Therefore,

$\dfrac{\text{dv}}{\text{dx}}\text{=(xsinx}{{\text{)}}^{\dfrac{\text{1}}{\text{x}}}}\left[ \dfrac{\text{1-log(xsinx)+xcotx}}{{{\text{x}}^{\text{2}}}} \right] $ ……. (3)

Hence, from the equations (1), (2) and (3), gives

$\Rightarrow\dfrac{\text{dy}}{\text{dx}}\text{=(xcosx}{{\text{)}}^{\text{x}}}\left[\text{1-xtanx+log(xcosx)}\right]\text{+(xsinx}{{\text{)}}^{\dfrac{\text{1}}{\text{x}}}}\left[ \dfrac{\text{1-log(xsinx)+xcotx}}{{{\text{x}}^{\text{2}}}} \right] $ .


Find  $ \dfrac{\mathbf{dy}}{\mathbf{dx}} $  of the functions given in Exercises 12 to 15. 

12.${{\mathbf{x}}^{\mathbf{y}}}\mathbf{+}{{\mathbf{y}}^{\mathbf{x}}}\mathbf{=1} $ .

Ans: The given function is  $ {{\text{x}}^{\text{y}}}\text{+}{{\text{y}}^{\text{x}}}\text{=1} $ .

Then, let  $ {{\text{x}}^{\text{y}}}\text{=u} $  and  $ {{\text{y}}^{\text{x}}}\text{=v} $ .

Therefore,  $ \text{u+v=1} $ .

Differentiating both sides of the equation with respect to  $ \text{x} $  gives

 $ \dfrac{\text{du}}{\text{dx}}\text{+}\dfrac{\text{dv}}{\text{dy}}\text{=0} $ 

Now,  $ \text{u=}{{\text{x}}^{\text{y}}} $                             ……. (1)

 $ \begin{align} & \Rightarrow \text{logu=log(}{{\text{x}}^{\text{y}}}\text{)} \\ & \Rightarrow \text{logu=ylogx} \\ \end{align} $ 

Differentiating both sides of the equation with respect to  $ \text{x} $  gives

 $\dfrac{\text{1}}{\text{u}}\dfrac{\text{du}}{\text{dx}}\text{=logx}\dfrac{\text{dy}}{\text{dx}}\text{+y }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(logx)} $ 

 $ \Rightarrow \dfrac{\text{du}}{\text{dx}}\text{=u}\left[ \text{logx}\dfrac{\text{dy}}{\text{dx}}\text{+y }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{x}} \right] $

Therefore,  $ \dfrac{\text{du}}{\text{dx}}\text{=}{{\text{x}}^{\text{y}}}\left[ \text{logx}\dfrac{\text{dy}}{\text{dx}}\text{+}\dfrac{\text{y}}{\text{x}} \right] $ ………… (2)

Also,  $ \text{v=}{{\text{y}}^{\text{x}}} $ 

Taking logarithm both sides of the equation give

$ \begin{align} & \Rightarrow \text{logv=log(}{{\text{y}}^{\text{3}}}\text{)} \\ & \Rightarrow \text{logv=xlogy} \\ \end{align} $ 

Differentiating both sides of the equation with respect to  $ \text{x} $  gives

$ \dfrac{\text{1}}{\text{v}}\text{ }\!\!\times\!\!\text{}\dfrac{\text{dv}}{\text{dx}}\text{=logy}\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(x)+x }\!\!\times\!\!\text{}\dfrac{\text{d}}{\text{dx}}\text{(logy)} $ 

$ \Rightarrow \dfrac{\text{dv}}{\text{dx}}\text{=v}\left(\text{logy}\!\!\times\!\!\text{ 1+x}\!\!\times\!\!\text{}\dfrac{\text{1}}{\text{y}}\text{ }\!\!\times\!\!\text{}\dfrac{\text{dy}}{\text{dx}} \right) $ 

Therefore,  $ \dfrac{\text{dv}}{\text{dx}}\text{=}{{\text{y}}^{\text{x}}}\left( \text{logy+}\dfrac{\text{x}}{\text{y}}\dfrac{\text{dy}}{\text{dx}} \right) $ ……... (3)

So, from the equation (1), (2) and (3), gives

$ \begin{align}& {{\text{x}}^{\text{y}}}\left( \text{logx}\dfrac{\text{dy}}{\text{dx}}\text{+}\dfrac{\text{y}}{\text{x}} \right)\text{+}{{\text{y}}^{\text{x}}}\left(\text{logy+}\dfrac{\text{x}}{\text{y}}\dfrac{\text{dy}}{\text{dx}} \right)\text{=0} \\ & \Rightarrow \left( {{\text{x}}^{\text{2}}}\text{+logx+x}{{\text{y}}^{\text{y-1}}}\right)\dfrac{\text{dy}}{\text{dx}}\text{=-}\left(\text{y}{{\text{x}}^{\text{y-1}}}\text{+}{{\text{y}}^{\text{x}}}\text{logy} \right) \\ \end{align} $ 

Hence,  $\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{-(y}{{\text{x}}^{\text{y-1}}}\text{+}{{\text{y}}^{\text{x}}}\text{logy)}}{{{\text{x}}^{\text{y}}}\text{logx+x}{{\text{y}}^{\text{x-1}}}} $ .


13.  $ {{\text{y}}^{\text{x}}}\text{=}{{\text{x}}^{\text{y}}} $ .

Ans: The given equation is  $ {{\text{y}}^{\text{x}}}\text{=}{{\text{x}}^{\text{y}}} $ .

Then, taking logarithm both sides of the equation give

 $ \text{xlogy=ylogx} $ .

Differentiating both sides of the equation with respect to  $ \text{x} $  gives

$ \begin{align}& \text{logy }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(x)+x }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(logy)=logx }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(y)+y }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(logx)} \\ & \Rightarrow \text{logy }\!\!\times\!\!\text{ 1+x }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{y}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{dy}}{\text{dx}}\text{=logx }\!\!\times\!\!\text{ }\dfrac{\text{dy}}{\text{dx}}\text{+y }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{x}} \\ & \Rightarrow\text{logy+}\dfrac{\text{x}}{\text{y}}\dfrac{\text{dy}}{\text{dx}}\text{=logx}\dfrac{\text{dy}}{\text{dx}}\text{+}\dfrac{\text{y}}{\text{x}} \\ & \Rightarrow \left( \dfrac{\text{x}}{\text{y}}\text{-logx} \right)\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{y}}{\text{x}}\text{-logy} \\ \end{align} $ 

$ \begin{align}& \Rightarrow \left( \dfrac{\text{x-ylogx}}{\text{y}} \right)\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{y-xlogy}}{\text{x}} \\ & \Rightarrow \left( \dfrac{\text{x-ylogx}}{\text{y}} \right)\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{y-xlogy}}{\text{x}} \\ \end{align} $ 

Therefore, $ \dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{y}}{\text{x}}\left( \dfrac{\text{y-xlogy}}{\text{x-ylogx}} \right) $ .


14.  $ {{\text{(cosx)}}^{\text{y}}}\text{=(cosy}{{\text{)}}^{\text{x}}} $ .

Ans: The given equation is  $ {{\text{(cosx)}}^{\text{y}}}\text{=(cosy}{{\text{)}}^{\text{x}}} $ .

Then, taking logarithm both sides of the equation give

$ \text{ylogcosx=xlogcosy} $ .

Now, differentiating both sides of the equation with respect to  $ \text{x} $  gives

$ \begin{align}& \text{logcosx }\!\!\times\!\!\text{ }\dfrac{\text{dy}}{\text{dx}}\text{+y }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(logcosx)=logcosy }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(x)+x }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(logcosy}) \\ & \Rightarrow \text{logcosx}\dfrac{\text{dy}}{\text{dx}}\text{+}\dfrac{\text{y}}{\text{cosx}}\text{ }\!\!\times\!\!\text{ (-sinx)=logcosy+}\dfrac{\text{x}}{\text{cosy}}\text{(-siny) }\!\!\times\!\!\text{ }\dfrac{\text{dy}}{\text{dx}} \\ & \Rightarrow \text{logcosx}\dfrac{\text{dy}}{\text{dx}}\text{-ytanx=logcosy-xtany}\dfrac{\text{dy}}{\text{dx}} \\ & \Rightarrow \text{(logcosx+xtany)}\dfrac{\text{dy}}{\text{dx}}\text{=ytanx+logcosy} \\ \end{align} $ 

Therefore,$\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{ytanx+logcosy}}{\text{xtany+logcosx}} $ .


15.  $ \text{xy=}{{\text{e}}^{\text{(x-y)}}} $ .

Ans:  The given equation is  $ \text{xy=}{{\text{e}}^{\text{(x-y)}}} $ .

Then, taking logarithm both sides of the equation give

$ \begin{align}& \text{log(xy)=log(}{{\text{e}}^{\text{x-y}}}\text{)} \\ & \Rightarrow \text{logx+logy=(x-y)loge} \\ & \Rightarrow \text{logx+logy=(x-y) }\!\!\times\!\!\text{ 1} \\ & \Rightarrow \text{logx+logy=x-y} \\ \end{align} $ 

Now, differentiating both sides of the equation with respect to  $ \text{x} $  gives  $ \begin{align}& \dfrac{\text{d}}{\text{dx}}\text{(logx)+}\dfrac{\text{d}}{\text{dx}}\text{(logy)=}\dfrac{\text{d}}{\text{dx}}\text{(x)-}\dfrac{\text{dy}}{\text{dx}} \\ & \Rightarrow \dfrac{\text{1}}{\text{x}}\text{+}\dfrac{\text{1}}{\text{y}}\dfrac{\text{dy}}{\text{dx}}\text{=1-}\dfrac{\text{1}}{\text{x}} \\ & \Rightarrow \left( \text{1+}\dfrac{\text{1}}{\text{y}} \right)\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{x-1}}{\text{x}} \\ \end{align} $  

Therefore,  $\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{y(x-1)}}{\text{x(y+1)}} $ .


16. Find the derivative of the function given by  

$\text{f(x)=(1+x)(1+}{{\text{x}}^{\text{2}}}\text{)(1+}{{\text{x}}^{\text{4}}}\text{)(1+}{{\text{x}}^{\text{8}}}\text{)} $  and hence find  $ \mathbf{f'(1)} $ .

Ans: The given function is  $\text{f(x)=(1+x)(1+}{{\text{x}}^{\text{2}}}\text{)(1+}{{\text{x}}^{\text{4}}}\text{)(1+}{{\text{x}}^{\text{8}}}\text{)} $ .

By taking logarithm both sides of the equation give

$\text{logf(x)=log(1+x)+log(1+}{{\text{x}}^{\text{2}}}\text{)+log(1+}{{\text{x}}^{\text{4}}}\text{)+log(1+}{{\text{x}}^{\text{8}}}\text{)} $ 

Now, differentiating both sides of the equation with respect to  $ \text{x} $  gives  $ \begin{align}& \dfrac{\text{1}}{\text{f(x)}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{ }\!\![\!\!\text{ f(x) }\!\!]\!\!\text{ =}\dfrac{\text{d}}{\text{dx}}\text{log(1+x)+}\dfrac{\text{d}}{\text{dx}}\text{log(1+}{{\text{x}}^{\text{2}}}\text{)+}\dfrac{\text{d}}{\text{dx}}\text{log(1+}{{\text{x}}^{\text{4}}}\text{)+}\dfrac{\text{d}}{\text{dx}}\text{log(1+}{{\text{x}}^{\text{8}}}\text{)} \\ & \Rightarrow \dfrac{\text{1}}{\text{f(x)}}\text{ }\!\!\times\!\!\text{ f }\!\!'\!\!\text{ (x)=}\dfrac{\text{1}}{\text{1+x}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{dx}}\text{(1+x)+}\dfrac{\text{1}}{\text{1+}{{\text{x}}^{\text{2}}}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{log(1+}{{\text{x}}^{\text{2}}}\text{)+}\dfrac{\text{1}}{\text{1+}{{\text{x}}^{\text{4}}}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{log(1+}{{\text{x}}^{\text{4}}}\text{)} \\  & \text{+}\dfrac{\text{1}}{\text{1+}{{\text{x}}^{\text{8}}}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{log(1+}{{\text{x}}^{\text{8}}}\text{)} \\ & \Rightarrow \text{f }\!\!'\!\!\text{ (x)=f(x)}\left[ \dfrac{\text{1}}{\text{1+x}}\text{+}\dfrac{\text{1}}{\text{1+}{{\text{x}}^{\text{2}}}}\text{ }\!\!\times\!\!\text{ 2x+}\dfrac{\text{1}}{\text{1+}{{\text{x}}^{\text{4}}}}\text{ }\!\!\times\!\!\text{ 4}{{\text{x}}^{\text{3}}}\text{+}\dfrac{\text{1}}{\text{1+}{{\text{x}}^{\text{8}}}}\text{ }\!\!\times\!\!\text{ 8}{{\text{x}}^{\text{7}}} \right] \\ \end{align} $ 

Therefore,

$ \text{f }\!\!'\!\!\text{ (x)=(1+x)(1+}{{\text{x}}^{\text{2}}}\text{)(1+}{{\text{x}}^{\text{4}}}\text{)(1+}{{\text{x}}^{\text{8}}}\text{)}\left[ \dfrac{\text{1}}{\text{1+x}}\text{+}\dfrac{\text{2x}}{\text{1+}{{\text{x}}^{\text{2}}}}\text{+}\dfrac{\text{4}{{\text{x}}^{\text{3}}}}{\text{1+}{{\text{x}}^{\text{4}}}}\text{+}\dfrac{\text{8}{{\text{x}}^{\text{7}}}}{\text{1+}{{\text{x}}^{\text{8}}}} \right] $ 

So,
$ \begin{align}& \text{f }\!\!'\!\!\text{ (1)=(1+1)(1+}{{\text{1}}^{\text{2}}}\text{)(1+}{{\text{1}}^{\text{4}}}\text{)(1+}{{\text{1}}^{\text{8}}}\text{)}\left[\dfrac{\text{1}}{\text{1+1}}\text{+}\dfrac{\text{2 }\!\!\times\!\!\text{ 1}}{\text{1+}{{\text{1}}^{\text{2}}}}\text{+}\dfrac{\text{4 }\!\!\times\!\!\text{ }{{\text{1}}^{\text{3}}}}{\text{1+}{{\text{1}}^{\text{4}}}}\text{+}\dfrac{\text{8 }\!\!\times\!\!\text{ }{{\text{1}}^{\text{7}}}}{\text{1+}{{\text{1}}^{\text{8}}}} \right] \\ & \text{=2 }\!\!\times\!\!\text{ 2 }\!\!\times\!\!\text{ 2 }\!\!\times\!\!\text{ 2}\left[ \dfrac{\text{1}}{\text{2}}\text{+}\dfrac{\text{2}}{\text{2}}\text{+}\dfrac{\text{4}}{\text{2}}\text{+}\dfrac{\text{8}}{\text{2}} \right] \\ & \text{=16 }\!\!\times\!\!\text{ }\left( \dfrac{\text{1+2+4+8}}{\text{2}} \right) \\ & \text{=16 }\!\!\times\!\!\text{ }\dfrac{\text{15}}{\text{2}}\text{=120} \\ \end{align} $ 

Hence,  $ \text{{f}'}\left( \text{1} \right)\text{=120} $ .


17. Differentiate  $ \mathbf{y=(}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{-5x+8)(}{{\mathbf{x}}^{\mathbf{3}}}\mathbf{+7x+9)} $  in three ways as mentioned below. Do they all give the same answer? 

(a) by using product rule. 

Ans: The given function is  $ \text{y=}\left( {{\text{x}}^{\text{2}}}\text{-5x+8} \right)\left( {{\text{x}}^{\text{3}}}\text{+7x+9} \right) $ .

Now, let consider  $ \text{u=(}{{\text{x}}^{\text{2}}}\text{-5x+8)} $  and  $ \text{v=(}{{\text{x}}^{\text{3}}}\text{+7x+9)} $ 

Therefore,  $ \text{y=uv} $ .

$ \begin{align}& \Rightarrow \dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{du}}{\text{dv}}\text{.v+u}\text{.}\dfrac{\text{du}}{\text{dx}} \\ & \Rightarrow \dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{x}}^{\text{2}}}\text{-5x+8)}\text{.(}{{\text{x}}^{\text{3}}}\text{+7x+9)+(}{{\text{x}}^{\text{2}}}\text{-5x+8)}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{x}}^{\text{3}}}\text{+7x+9)} \\ & \Rightarrow \dfrac{\text{dy}}{\text{dx}}\text{=(2x-5)(}{{\text{x}}^{\text{3}}}\text{+7x+9)}\text{.(}{{\text{x}}^{\text{2}}}\text{-5x+8)(3}{{\text{x}}^{\text{2}}}\text{+7)} \\ & \Rightarrow\dfrac{\text{dy}}{\text{dx}}\text{=2x(}{{\text{x}}^{\text{3}}}\text{+7x+9)-5(}{{\text{x}}^{\text{2}}}\text{-5x+8)+}{{\text{x}}^{\text{2}}}\text{(3}{{\text{x}}^{\text{2}}}\text{+7)-5x(3}{{\text{x}}^{\text{2}}}\text{+7)-8(3}{{\text{x}}^{\text{2}}}\text{+7)} \\ & \Rightarrow \dfrac{\text{dy}}{\text{dx}}\text{=(2}{{\text{x}}^{\text{4}}}\text{+14}{{\text{x}}^{\text{2}}}\text{+18x)-5}{{\text{x}}^{\text{3}}}\text{-35x-45+(3}{{\text{x}}^{\text{4}}}\text{+7}{{\text{x}}^{\text{2}}}\text{)-15}{{\text{x}}^{\text{3}}}\text{-35x+24}{{\text{x}}^{\text{2}}}\text{+56} \\ \end{align} $ 

Hence,  $ \dfrac{\text{dy}}{\text{dx}}\text{=5}{{\text{x}}^{\text{4}}}\text{-20}{{\text{x}}^{\text{3}}}\text{+45}{{\text{x}}^{\text{2}}}\text{+52x+11} $ .


(b) by expanding the product to obtain a single polynomial.

Ans: The given function is

$\text{y=(}{{\text{x}}^{\text{2}}}\text{-5x+8)(}{{\text{x}}^{\text{3}}}\text{+7x+9)} $ .

Then, calculating the product, gives

$\begin{align}&\text{y=}{{\text{x}}^{\text{2}}}\text{(}{{\text{x}}^{\text{3}}}\text{+7x+9)-5}{{\text{x}}^{\text{4}}}\text{(}{{\text{x}}^{\text{3}}}\text{+7x+9)+8(}{{\text{x}}^{\text{3}}}\text{+7x+9)} \\ & \Rightarrow \text{y=}{{\text{x}}^{\text{5}}}\text{+7}{{\text{x}}^{\text{3}}}\text{+9}{{\text{x}}^{\text{2}}}\text{-5}{{\text{x}}^{\text{3}}}\text{-26}{{\text{x}}^{\text{2}}}\text{+11x+72} \\ \end{align} $ 

Now, differentiating both sides of the equation with respect to  $ \text{x} $  gives

$\begin{align}&\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{x}}^{\text{5}}}\text{+7}{{\text{x}}^{\text{3}}}\text{+9}{{\text{x}}^{\text{2}}}\text{-5}{{\text{x}}^{\text{3}}}\text{-26}{{\text{x}}^{\text{2}}}\text{+11x+72)} \\ & =\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{x}}^{\text{5}}}\text{)-5}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{x}}^{\text{4}}}\text{)+15}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{x}}^{\text{3}}}\text{)-26}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{x}}^{\text{3}}}\text{)+11}\dfrac{\text{d}}{\text{dx}}\text{(x)+}\dfrac{\text{d}}{\text{dx}}\text{(72)} \\ & \text{=5}{{\text{x}}^{\text{4}}}\text{-5 }\!\!\times\!\!\text{ 4}{{\text{x}}^{\text{3}}}\text{+15 }\!\!\times\!\!\text{ 3}{{\text{x}}^{\text{2}}}\text{-26 }\!\!\times\!\!\text{ 2x+11 }\!\!\times\!\!\text{ 1+0}  \end{align} $ 

Hence,  $\dfrac{\text{dy}}{\text{dx}}\text{=5}{{\text{x}}^{\text{4}}}\text{-20}{{\text{x}}^{\text{3}}}\text{+45}{{\text{x}}^{\text{2}}}\text{-52x+11} $ .


(c) by logarithmic differentiation.

Ans:     The given function is

$\text{y=(}{{\text{x}}^{\text{2}}}\text{-5x+8)(}{{\text{x}}^{\text{3}}}\text{+7x+9}$ .

Now, taking logarithm both sides of the function give

$\text{logy=log(}{{\text{x}}^{\text{2}}}\text{-5x+8)+log(}{{\text{x}}^{\text{3}}}\text{+7x+9)} $ 

Differentiating both sides of the equation with respect to  $ \text{x} $  gives  $ \begin{align} & \dfrac{\text{1}}{\text{y}}\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{log(}{{\text{x}}^{\text{2}}}\text{-5x+8)+}\dfrac{\text{d}}{\text{dx}}\text{log(}{{\text{x}}^{\text{3}}}\text{+7x+9)} \\  & \Rightarrow \dfrac{\text{1}}{\text{y}}\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{1}}{{{\text{x}}^{\text{2}}}\text{-5x+8}}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{x}}^{\text{2}}}\text{-5x+8)+}\dfrac{\text{1}}{{{\text{x}}^{\text{3}}}\text{+7x+9}}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{x}}^{\text{3}}}\text{+7x+9)} \\ & \Rightarrow \dfrac{\text{dy}}{\text{dx}}\text{=y}\left[ \dfrac{\text{1}}{{{\text{x}}^{\text{2}}}\text{-5x+8}}\text{ }\!\!\times\!\!\text{ (2x-5)+}\dfrac{\text{1}}{{{\text{x}}^{\text{3}}}\text{+7x+9}}\text{ }\!\!\times\!\!\text{ (3}{{\text{x}}^{\text{2}}}\text{+7)} \right] \\ & \Rightarrow \dfrac{\text{dy}}{\text{dx}}\text{=(}{{\text{x}}^{\text{2}}}\text{-5x+8)(}{{\text{x}}^{\text{3}}}\text{+7x+9)}\left[ \dfrac{\text{2x-5}}{{{\text{x}}^{\text{3}}}\text{-5x+8}}\text{+}\dfrac{\text{3}{{\text{x}}^{\text{2}}}\text{+7}}{{{\text{x}}^{\text{3}}}\text{+7x+9}} \right] \\  & \Rightarrow\dfrac{\text{dy}}{\text{dx}}\text{=(}{{\text{x}}^{\text{2}}}\text{-5x+8)(}{{\text{x}}^{\text{3}}}\text{+7x+9)}\left[ \dfrac{\text{(2x-5)(}{{\text{x}}^{\text{3}}}\text{+7x+9)+(3}{{\text{x}}^{\text{2}}}\text{+7)(}{{\text{x}}^{\text{2}}}\text{-5x+8)}}{\text{(}{{\text{x}}^{\text{3}}}\text{-5x+8)+(}{{\text{x}}^{\text{3}}}\text{+7x+9)}} \right] \\ & \Rightarrow \dfrac{\text{dy}}{\text{dx}}\text{=2x(}{{\text{x}}^{\text{3}}}\text{+7x+9}{{\text{x}}^{\text{2}}}\text{)-5(}{{\text{x}}^{\text{3}}}\text{+7x+9)+3}{{\text{x}}^{\text{2}}}\text{(}{{\text{x}}^{\text{2}}}\text{-5x+8)+7(}{{\text{x}}^{\text{3}}}\text{+7x+9)} \\ & \Rightarrow \dfrac{\text{dy}}{\text{dx}}\text{=(2}{{\text{x}}^{\text{4}}}\text{+14}{{\text{x}}^{\text{2}}}\text{+18x)+(5}{{\text{x}}^{\text{3}}}\text{-35x+45)+(3}{{\text{x}}^{\text{4}}}\text{-15}{{\text{x}}^{\text{3}}}\text{+24}{{\text{x}}^{\text{2}}}\text{)+(7}{{\text{x}}^{\text{2}}}\text{+35x+56)} \\ \end{align} $ 

Therefore,  $\dfrac{\text{dy}}{\text{dx}}\text{=5}{{\text{x}}^{\text{2}}}\text{-20}{{\text{x}}^{\text{3}}}\text{+45}{{\text{x}}^{\text{2}}}\text{-52x+11} $ .

Hence, comparing the above three results, it is concluded that the derivative  $ \dfrac{\text{dy}}{\text{dx}} $  are the same for all methods.


18. If  $ \mathbf{u} $ ,  $ \mathbf{v} $ , and  $ \mathbf{w} $  are functions of  $ \mathbf{x} $  , then show that  $ \dfrac{\text{d}}{\text{dx}}\text{(u}\text{.v}\text{.w)=}\dfrac{\text{du}}{\text{dx}}\text{v}\text{.w+u}\dfrac{\text{du}}{\text{dx}}\text{.w+u}\text{.v}\dfrac{\text{dw}}{\text{d}\mathbf{x}} $  in two ways - first by using repeated application of product rule, second by logarithmic differentiation. 

Ans: Let the function  $ \text{y=u}\text{.v}\text{.w=u}\text{.(v}\text{.w)} $ .

Then applying the product rule of derivatives, give

 $ \dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{du}}{\text{dx}}\text{.(v}\text{.w)+u}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(v}\text{.w)} $           

$\Rightarrow\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{du}}{\text{dx}}\text{v}\text{.w+u}\left[\dfrac{\text{dv}}{\text{dx}}\text{.w+v}\text{.}\dfrac{\text{dv}}{\text{dx}} \right] $         (Using the product rule again)

Thus,

$\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{du}}{\text{dx}}\text{v}\text{.w+u}\text{.}\dfrac{\text{dv}}{\text{dx}}\text{.w+u}\text{.v}\dfrac{\text{dw}}{\text{dx}} $ .

Now, take logarithm both sides of the function  $ \text{y=u}\text{.v}\text{.w} $.

Then, we have  $ \text{logy=logu+logv+logw} $ .

Differentiating both sides of the equation with respect to  $ \text{x} $  gives  $ \begin{align}&\dfrac{\text{1}}{\text{y}}\text{.}\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{(logu)+}\dfrac{\text{d}}{\text{dx}}\text{(logv)+}\dfrac{\text{d}}{\text{dx}}\text{(logw)} \\ & \Rightarrow \dfrac{\text{1}}{\text{y}}\text{.}\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{1}}{\text{u}}\dfrac{\text{du}}{\text{dx}}\text{+}\dfrac{\text{1}}{\text{v}}\dfrac{\text{dv}}{\text{dx}}\text{+}\dfrac{\text{1}}{\text{w}}\dfrac{\text{dw}}{\text{dx}} \\ & \Rightarrow \dfrac{\text{dy}}{\text{dx}}\text{=y}\left( \dfrac{\text{1}}{\text{u}}\dfrac{\text{du}}{\text{dx}}\text{+}\dfrac{\text{1}}{\text{v}}\dfrac{\text{dv}}{\text{dx}}\text{+}\dfrac{\text{1}}{\text{w}}\dfrac{\text{dw}}{\text{dx}} \right) \\ & \Rightarrow \dfrac{\text{dy}}{\text{dx}}\text{=u}\text{.v}\text{.w}\left( \dfrac{\text{1}}{\text{u}}\dfrac{\text{du}}{\text{dx}}\text{+}\dfrac{\text{1}}{\text{v}}\dfrac{\text{dv}}{\text{dx}}\text{+}\dfrac{\text{1}}{\text{w}}\dfrac{\text{dw}}{\text{dx}} \right) \\ \end{align} $ 

Hence, $\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{du}}{\text{dx}}\text{v}\text{.w+u}\dfrac{\text{dv}}{\text{dx}}\text{.w+u}\text{.v}\dfrac{\text{dw}}{\text{dx}} $.

Exercise 5.6 

If  $ \mathbf{x,}\,\mathbf{y} $  are connected parametrically by the equation given below, without eliminating the parameter, Find  $ \dfrac{\mathbf{dy}}{\mathbf{dx}} $ .

1.$\text{x=2a}{{\text{t}}^{\text{2}}}\text{,}\,\,\text{y=a}{{\text{t}}^{\text{4}}} $ .

Ans: The given equations are 

 $ \text{x=2a}{{\text{t}}^{\text{2}}} $                                                                             …… (1)

and  $ \text{y=a}{{\text{t}}^{\text{4}}} $                                                                       …… (2)

Then, differentiating both sides of the equation (1) with respect to  $ \text{t} $  gives

$\dfrac{\text{dx}}{\text{dt}}\text{=}\dfrac{\text{d}}{\text{dt}}\text{(2a}{{\text{t}}^{\text{2}}}\text{)=2a }\!\!\times\!\!\text{}\dfrac{\text{d}}{\text{dt}}\text{(}{{\text{t}}^{\text{2}}}\text{)=2a }\!\!\times\!\!\text{ 2t=4at} $ .                        …… (3)

Also, differentiating both sides of the equation (2) with respect to  $ \text{t} $  gives

$\dfrac{\text{dy}}{\text{dt}}\text{=}\dfrac{\text{d}}{\text{dt}}\text{(a}{{\text{t}}^{\text{4}}}\text{)=a }\!\!\times\!\!\text{}\dfrac{\text{d}}{\text{dt}}\text{(}{{\text{t}}^{\text{4}}}\text{)=a }\!\!\times\!\!\text{ 4 }\!\!\times\!\!\text{}{{\text{t}}^{\text{3}}}\text{=4a}{{\text{t}}^{\text{3}}} $                           …… (4)

Now, dividing the equations (4) by (3) gives

$ \dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\left( \dfrac{\text{dy}}{\text{dt}} \right)}{\left(\dfrac{\text{dx}}{\text{dt}}\right)}\text{=}\dfrac{\text{4a}{{\text{t}}^{\text{3}}}}{\text{4at}}\text{=}{{\text{t}}^{\text{2}}} $ .

Hence,  $ \dfrac{\text{dy}}{\text{dx}}\text{=}{{\text{t}}^{\text{2}}} $ .


 2.  $ \mathbf{x=acos\theta ,}\,\,\mathbf{y=bcos\theta } $ .

Ans:  The given equations are 

$ \text{x=acos }\!\;\theta\;\!\text{ } $                                                                             …… (1)

and  $ \text{y=bcos }\!\;\theta\;\!\text{ } $                                                                      …… (2)

Then, differentiating both sides of the equation (1) with respect to  $ \text{ }\!\!\theta\!\!\text{ } $  gives

$ \dfrac{\text{dx}}{\text{d }\!\!\theta\!\!\text{ }}\text{=}\dfrac{\text{d}}{\text{d }\!\!\theta\!\!\text{ }}\text{(acos }\!\!\theta\!\!\text{ )=a(-sin }\!\!\theta\!\!\text{ )=-asin }\!\!\theta\!\!\text{ } $ .                                      …… (3)

Also, differentiating both sides of the equation (1) with respect to  $ \text{ }\!\!\theta\!\!\text{ } $  gives

$ \dfrac{\text{dy}}{\text{d }\!\!\theta\!\!\text{}}\text{=}\dfrac{\text{d}}{\text{d }\!\!\theta\!\!\text{ }}\text{(bcos }\!\!\theta\!\!\text{ )=b(-sin }\!\!\theta\!\!\text{ )=-bsin }\!\!\theta\!\!\text{ } $                                        …… (4)

Therefore, dividing the equation (4) by (3) gives

$ \dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\left( \dfrac{\text{dy}}{\text{d }\!\!\theta\!\!\text{ }}\right)}{\left(\dfrac{\text{dx}}{\text{d}\!\!\theta\!\!\text{ }} \right)}\text{=}\dfrac{\text{-bsin }\!\!\theta\!\!\text{}}{\text{-asin}\!\!\theta\!\!\text{}}\text{=}\dfrac{\text{b}}{\text{a}} $ .

Hence,   $ \dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{b}}{\text{a}} $ .


3.  $ \mathbf{x=sint,}\,\,\,\mathbf{y=cos2t} $ .

Ans:  

The given equations are

$ \text{x=sint} $                                                                         …… (1)

and  $ \text{y=cos2t} $                                                                …… (2)

Then, differentiating both sides of the equation (1) with respect to  $ \text{t} $  gives

$ \dfrac{\text{dx}}{\text{dt}}\text{=}\dfrac{\text{d}}{\text{dt}}\text{(sint)=cost} $ .                                                    …… (3)

Also, differentiating both sides of the equation (2) with respect to  $ \text{t} $  gives

$\dfrac{\text{dy}}{\text{dt}}\text{=}\dfrac{\text{d}}{\text{dt}}\text{(cos2t)=sin2t }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dt}}\text{(2t)=-2sin2t} $                           …… (4)

Therefore, by dividing the equation (4) by (3) gives

$ \dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\left( \dfrac{\text{dy}}{\text{dt}} \right)}{\left( \dfrac{\text{dx}}{\text{dt}} \right)}\text{=}\dfrac{\text{-2sin2t}}{\text{cost}}\text{=}\dfrac{\text{-2 }\!\!\times\!\!\text{ 2sintcost}}{\text{cost}}\text{=-4sint} $ 

Hence,  $ \dfrac{\text{dy}}{\text{dx}}\text{=-4sint} $ .


4.  $ \mathbf{x=4t, }\,\mathbf{y=}\dfrac{\mathbf{4}}{\mathbf{t}} $ .

Ans: The given equations are

$ \text{x=4t} $                                                                    …… (1)

 and  $ \text{y=}\dfrac{\text{4}}{\text{t}} $                                                             …… (2)

Now, differentiating both sides of the equation (1) with respect to  $ \text{t} $  gives

 $ \dfrac{\text{dx}}{\text{dt}}\text{=}\dfrac{\text{d}}{\text{dt}}\text{(4t)=4} $ .                                                    …… (3)

Also, differentiating both sides of the equation (2) with respect to  $ \text{t} $  gives

$ \dfrac{\text{dy}}{\text{dt}}\text{=}\dfrac{\text{d}}{\text{dt}}\left( \dfrac{\text{4}}{\text{t}} \right)\text{=4 }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dt}}\left( \dfrac{\text{1}}{\text{t}} \right)\text{=4 }\!\!\times\!\!\text{ }\left( \dfrac{\text{-1}}{{{\text{t}}^{\text{2}}}} \right)\text{=}\dfrac{\text{-4}}{{{\text{t}}^{\text{2}}}} $                    …… (4)

Therefore, dividing the equation (4) by (3) gives

 $ \dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\left( \dfrac{\text{dy}}{\text{dt}} \right)}{\left( \dfrac{\text{dx}}{\text{dt}} \right)}\text{=}\dfrac{\left( \dfrac{\text{-4}}{{{\text{t}}^{\text{2}}}} \right)}{\text{4}}\text{=}\dfrac{\text{-1}}{{{\text{t}}^{\text{2}}}} $ .

Hence,  $ \dfrac{\text{dy}}{\text{dx}}\text{=-}\dfrac{\text{1}}{{{\text{t}}^{\text{2}}}} $ .


5.  $ \mathbf{x=cos\theta -cos2\theta , y=sin\theta -sin2\theta } $ .

Ans: The given equations are

$ \text{x}\,\text{=}\,\text{cos }\!\!\theta\!\!\text{ }\,\text{-}\,\text{2cos }\!\!\theta\!\!\text{ } $                                                                                       …… (1)

 and  $ \text{y=sin }\!\!\theta\!\!\text{ -sin2 }\!\!\theta\!\!\text{ } $                                                                                    …… (2)

Then, differentiating both sides of the equation (1) with respect to  $ \text{ }\!\!\theta\!\!\text{ } $  gives

 $ \dfrac{\text{dx}}{\text{d }\!\!\theta\!\!\text{ }}\text{=}\dfrac{\text{d}}{\text{d }\!\!\theta\!\!\text{ }}\text{(cos }\!\!\theta\!\!\text{ -cos2 }\!\!\theta\!\!\text{ )=}\dfrac{\text{d}}{\text{d }\!\!\theta\!\!\text{ }}\text{(cos }\!\!\theta\!\!\text{ )-}\dfrac{\text{d}}{\text{d }\!\!\theta\!\!\text{ }}\text{(cos2 }\!\!\theta\!\!\text{ )=-sin }\!\!\theta\!\!\text{ (-2sin2 }\!\!\theta\!\!\text{ )=2sin2 }\!\!\theta\!\!\text{ -sin }\!\!\theta\!\!\text{ } $  … (3)

Also, differentiating both sides of the equation (2) with respect to  $ \text{ }\!\!\theta\!\!\text{ } $  gives

 $ \dfrac{\text{dy}}{\text{d }\!\!\theta\!\!\text{ }}\text{=}\dfrac{\text{d}}{\text{d }\!\!\theta\!\!\text{ }}\text{(sin }\!\!\theta\!\!\text{ -sin2 }\!\!\theta\!\!\text{ )=}\dfrac{\text{d}}{\text{d }\!\!\theta\!\!\text{ }}\text{(sin }\!\!\theta\!\!\text{ )-}\dfrac{\text{d}}{\text{d }\!\!\theta\!\!\text{ }}\text{(sin2 }\!\!\theta\!\!\text{ )=cos }\!\!\theta\!\!\text{ -2cos2 }\!\!\theta\!\!\text{ } $                            ….. (4)

Therefore, dividing the equation (4) by (3) gives

 $ \dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\left( \dfrac{\text{dy}}{\text{d }\!\!\theta\!\!\text{ }} \right)}{\left( \dfrac{\text{dx}}{\text{d }\!\!\theta\!\!\text{ }} \right)}\text{=}\dfrac{\text{cos }\!\!\theta\!\!\text{ -2cos2 }\!\!\theta\!\!\text{ }}{\text{2sin2 }\!\!\theta\!\!\text{ -sin }\!\!\theta\!\!\text{ }} $ .

Hence,  $ \dfrac{\text{dy}}{\text{dx}}=\dfrac{\text{cos}\theta\text{ -2cos}2\theta\text{}}{\text{2sin2}\theta\text{-sin}\theta\text{}} $ . 


6.  $ \mathbf{x=a(\theta -sin\theta ), y=a(1+cos\theta )} $ .

Ans:  The given equations are

  $ \text{x=a( }\!\!\theta\!\!\text{ -sin }\!\!\theta\!\!\text{ )} $                                                                       …… (1)

and  $ \text{y=a(1+cos }\!\!\theta\!\!\text{ )} $                                                                …… (2)

Then, differentiating on both sides of the equation (1) with respect to  $ \text{ }\!\!\theta\!\!\text{ } $  gives

$ \dfrac{\text{dx}}{\text{d }\!\!\theta\!\!\text{ }}\text{=a}\left[ \dfrac{\text{d}}{\text{d }\!\!\theta\!\!\text{ }}\text{( }\!\!\theta\!\!\text{ )-}\dfrac{\text{d}}{\text{d }\!\!\theta\!\!\text{ }}\text{(sin }\!\!\theta\!\!\text{ )} \right]\text{=a(1-cos }\!\!\theta\!\!\text{ )} $                                   …… (3)

Also, differentiating both sides of the equation (2) with respect to  $ \text{ }\!\!\theta\!\!\text{ } $  gives

$ \dfrac{\text{dy}}{\text{d }\!\!\theta\!\!\text{ }}\text{=a}\left[ \dfrac{\text{d}}{\text{d }\!\!\theta\!\!\text{ }}\text{(1)+}\dfrac{\text{d}}{\text{d }\!\!\theta\!\!\text{ }}\text{(cos }\!\!\theta\!\!\text{ )} \right]\text{=a}\left[ \text{0+(-sin }\!\!\theta\!\!\text{ )} \right]\text{=-asin }\!\!\theta\!\!\text{ } $                  …… (4)

Therefore, by dividing the equation (4) by (3) gives

$ \dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\left( \dfrac{\text{dy}}{\text{d }\!\!\theta\!\!\text{ }} \right)}{\left( \dfrac{\text{dx}}{\text{d }\!\!\theta\!\!\text{ }} \right)}\text{=}\dfrac{\text{-asin }\!\!\theta\!\!\text{ }}{\text{a(1-cos }\!\!\theta\!\!\text{ )}}\text{=}\dfrac{\text{-2sin}\dfrac{\text{ }\!\!\theta\!\!\text{ }}{\text{2}}\text{cos}\dfrac{\text{ }\!\!\theta\!\!\text{ }}{\text{2}}}{\text{2si}{{\text{n}}^{\text{2}}}\dfrac{\text{ }\!\!\theta\!\!\text{ }}{\text{2}}}\text{=}\dfrac{\text{-cos}\dfrac{\text{ }\!\!\theta\!\!\text{ }}{\text{2}}}{\text{sin}\dfrac{\text{ }\!\!\theta\!\!\text{ }}{\text{2}}}\text{=-cot}\dfrac{\text{ }\!\!\theta\!\!\text{ }}{\text{2}} $ .

Hence,  $ \dfrac{\text{dy}}{\text{dx}}\text{=-cot}\dfrac{\text{ }\!\!\theta\!\!\text{ }}{\text{2}} $ .


7.$\mathbf{x=-}\dfrac{\mathbf{si}{{\mathbf{n}}^{\mathbf{3}}}\mathbf{t}}{\sqrt{\mathbf{cos2t}}}\mathbf{,y=}\dfrac{\mathbf{co}{{\mathbf{s}}^{\mathbf{3}}}\mathbf{t}}{\sqrt{\mathbf{cos2t}}} $ .

Ans: The given equations are,

$ \text{x=-}\dfrac{\text{si}{{\text{n}}^{\text{3}}}\text{t}}{\sqrt{\text{cos2t}}} $                                                                                …… (1)

 and  $ \text{y=}\dfrac{\text{co}{{\text{s}}^{\text{3}}}\text{t}}{\sqrt{\text{cos2t}}} $                                                                         …… (2)

Then, differentiating both sides of the equation (1) with respect to  $ \text{t} $  gives

$ \dfrac{\text{dx}}{\text{dt}}\text{=}\dfrac{\text{d}}{\text{dt}}\left[ \dfrac{\text{si}{{\text{n}}^{\text{3}}}\text{t}}{\sqrt{\text{cos2t}}} \right] $ 

$\begin{align}&=\dfrac{\sqrt{\text{cos2t}}\dfrac{\text{d}}{\text{dt}}\text{(si}{{\text{n}}^{\text{3}}}\text{t)-si}{{\text{n}}^{\text{3}}}\text{t }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dt}}\sqrt{\text{cos2t}}}{\text{cos2t}} \\ & =\dfrac{\sqrt{\text{cos2t}}\text{ }\!\!\times\!\!\text{ 3si}{{\text{n}}^{\text{2}}}\text{t }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dt}}\text{(sint)-si}{{\text{n}}^{\text{3}}}\text{t }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{2}\sqrt{\text{cos2t}}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dt}}\text{(cos2t)}}{\text{cos2t}} \\ & =\dfrac{\text{3cos2t }\!\!\times\!\!\text{ si}{{\text{n}}^{\text{2}}}\text{tcost-}\dfrac{\text{si}{{\text{n}}^{\text{3}}}\text{t}}{2}\text{ }\!\!\times\!\!\text{ (-2sin2t)}}{\text{cos2t}\sqrt{\text{cos2t}}} \\ \end{align} $ 

Also, differentiating both sides of the equation (2) with respect to  $ \text{t} $  gives

$\dfrac{\text{dx}}{\text{dt}}=\dfrac{\text{3cos2tsi}{{\text{n}}^{\text{2}}}\text{tcost+si}{{\text{n}}^{3}}\text{tsin2t}}{\text{cos2t}\sqrt{\text{cos2t}}} $ .                                                       …… (3)

$ \begin{align}& \dfrac{\text{dy}}{\text{dt}}=\dfrac{\text{d}}{\text{dt}}\left[ \dfrac{\text{co}{{\text{s}}^{\text{3}}}\text{t}}{\sqrt{\text{cos2t}}} \right] \\ & =\dfrac{\sqrt{\text{cos2t}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dt}}\text{(co}{{\text{s}}^{\text{3}}}\text{t)-co}{{\text{s}}^{\text{3}}}\text{t }\!\!\times\!\!\text{}\dfrac{\text{d}}{\text{dt}}\text{(}\sqrt{\text{cos2t}}\text{)}}{\text{cos2t}} \\ & =\dfrac{\text{3}\sqrt{\text{cos2t}}\text{co}{{\text{s}}^{\text{2}}}\text{t(-sint)-co}{{\text{s}}^{\text{3}}}\text{t }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{2(}\sqrt{\text{cos2t}}\text{)}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dt}}\text{(cos2t)}}{\text{cos2t}} \\ \end{align} $ 

$ \dfrac{\text{dy}}{\text{dt}}\text{=}\dfrac{\text{-3cos2t }\!\!\times\!\!\text{ co}{{\text{s}}^{\text{2}}}\text{t }\!\!\times\!\!\text{sint+co}{{\text{s}}^{\text{3}}}\text{tsin2t}}{\text{cos2t }\!\!\times\!\!\text{ }\sqrt{\text{cos2t}}} $                                                   …… (4)

Thus, dividing the equation (4) by the equation (3) gives

$ \dfrac{\text{dy}}{\text{dx}}=\dfrac{\left( \dfrac{\text{dx}}{\text{dt}} \right)}{\left( \dfrac{\text{dx}}{\text{dt}} \right)}\text{=}\dfrac{\text{-3cos2t }\!\!\times\!\!\text{ co}{{\text{s}}^{\text{2}}}\text{t }\!\!\times\!\!\text{ sint+co}{{\text{s}}^{\text{3}}}\text{tsin2t}}{\text{3cos2tcostsi}{{\text{n}}^{\text{2}}}\text{t+si}{{\text{n}}^{\text{3}}}\text{tsin2t}} $ 

$ =\dfrac{\text{sintcost}\left[ \text{-3cos2t }\!\!\times\!\!\text{ cost+2co}{{\text{s}}^{\text{3}}}\text{t} \right]}{\text{sintcost}\left[ \text{3cos2tsint+2si}{{\text{n}}^{\text{3}}}\text{t} \right]} $ 

$=\dfrac{\left[\text{-3(2co}{{\text{s}}^{\text{2}}}\text{t-1)cost+2co}{{\text{s}}^{\text{3}}}\text{t}\right]}{\left[\text{3(1-2si}{{\text{n}}^{\text{3}}}\text{t)sint+2si}{{\text{n}}^{\text{3}}}\text{t} \right]} $  

$ \left[ \begin{align}& \text{cos2t=(2co}{{\text{s}}^{\text{2}}}\text{t-1)} \\ & \text{cos2t=(1-2si}{{\text{n}}^{\text{2}}}\text{t)} \\ \end{align} \right] $ 

$ \begin{align}& =\dfrac{\text{-4co}{{\text{s}}^{\text{3}}}\text{t+3cost}}{\text{3sint-4si}{{\text{n}}^{\text{3}}}\text{t}} \\ & =\dfrac{\text{-cos3t}}{\text{sin3t}} \\ \end{align} $
$ \left[ \begin{align}& \text{cos3t=4co}{{\text{s}}^{\text{3}}}\text{t-3cost} \\ & \text{sin3t=3sint-4si}{{\text{n}}^{\text{2}}}\text{t} \\ \end{align} \right] $ 

Hence,  $ \dfrac{\text{dy}}{\text{dx}}\text{=-cot3t} $ .


8.  $ \mathbf{x=a}\left( \mathbf{cost+logtan}\dfrac{\mathbf{t}}{\mathbf{2}} \right)\mathbf{, y=asint} $ .

Ans: The given equations are

$ \text{x=a}\left( \text{cost+logtan}\dfrac{\text{t}}{\text{2}} \right) $                                                              …… (1)

and  $ \text{y=asint} $                                                                            …… (2)

Then, differentiating both sides of the equation (1) with respect to  $ \text{t} $  gives

$ \dfrac{\text{dx}}{\text{dt}}\text{=a }\!\!\times\!\!\text{ }\left[ \dfrac{\text{d}}{\text{d }\!\!\theta\!\!\text{}}\text{(cost)+}\dfrac{\text{d}}{\text{d }\!\!\theta\!\!\text{ }}\text{(logtan}\dfrac{\text{t}}{\text{2}}\text{)} \right] $ 

$ \begin{align}& \text{=a}\left[ \text{-sint+}\dfrac{\text{1}}{\text{tan}\dfrac{\text{t}}{\text{2}}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dt}}\left( \text{tan}\dfrac{\text{t}}{\text{2}} \right) \right] \\ & \text{=a}\left[ \text{-sint+cot}\dfrac{\text{t}}{\text{2}}\text{ }\!\!\times\!\!\text{ se}{{\text{c}}^{\text{2}}}\dfrac{\text{t}}{\text{2}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dt}}\left( \dfrac{\text{t}}{\text{2}} \right) \right] \\ \end{align} $ 

$\text{=a}\left(\text{-sint+}\dfrac{\text{cos}\dfrac{\text{t}}{\text{2}}}{\text{sin}\dfrac{\text{t}}{\text{2}}}\text{ }\!\!\times\!\!\text{}\dfrac{\text{1}}{\text{co}{{\text{s}}^{\text{2}}}\dfrac{\text{t}}{\text{2}}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{2}} \right) $ 

$\begin{align}&\text{=a}\left(\text{-sint+}\dfrac{\text{1}}{\text{2sin}\dfrac{\text{t}}{\text{2}}\text{cos}\dfrac{\text{t}}{\text{2}}} \right) \\ & \text{=a}\left( \text{-sint+}\dfrac{\text{1}}{\text{sint}} \right) \\ & \text{=a}\left( \dfrac{\text{-si}{{\text{n}}^{\text{2}}}\text{t+1}}{\text{sint}} \right) \\ \end{align} $ 

Therefore,  $\dfrac{\text{dx}}{\text{dt}}\text{=a}\dfrac{\text{co}{{\text{s}}^{\text{2}}}\text{t}}{\text{sint}} $                                                     …… (3)

Also, differentiating both sides of the equation (2) with respect to  $ \text{t} $  gives

$\dfrac{\text{dy}}{\text{dt}}\text{=a}\dfrac{\text{d}}{\text{dt}}\text{(sint)=acost} $                                                         …… (4)

Thus, dividing the equation (4) by the equation (3) gives

$ \dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\left( \dfrac{\text{dy}}{\text{dt}} \right)}{\left(\dfrac{\text{dx}}{\text{dt}}\right)}\text{=}\dfrac{\text{acost}}{\left( \text{a}\dfrac{\text{co}{{\text{s}}^{\text{2}}}\text{t}}{\text{sint}} \right)}\text{=}\dfrac{\text{sint}}{\text{cost}}\text{=tant} $ .

Hence,  $ \dfrac{\text{dy}}{\text{dx}}\text{=tant} $ .


9.  $ \mathbf{x=asec\theta , }\,\mathbf{y=btan\theta } $ . 

Ans: The given equations are 

$ \text{x=asec} $                                                                                        …… (1)

 and  $ \text{y=btan }\!\!\theta\!\!\text{ } $                                                                              …… (2)

Then, differentiating both sides of the equation (1) with respect to  $ \text{ }\!\!\theta\!\!\text{ } $  gives

$ \dfrac{\text{dx}}{\text{d }\!\!\theta\!\!\text{ }}\text{=a }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{d }\!\!\theta\!\!\text{ }}\text{(sec }\!\!\theta\!\!\text{ )=asec }\!\!\theta\!\!\text{ tan }\!\!\theta\!\!\text{ } $                                                      …… (3)

Also, differentiating both sides of the equation (2) with respect to  $ \text{ }\!\!\theta\!\!\text{ } $  gives

$ \dfrac{\text{dy}}{\text{d }\!\!\theta\!\!\text{ }}\text{=b }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{d }\!\!\theta\!\!\text{ }}\text{(tan }\!\!\theta\!\!\text{ )=bse}{{\text{c}}^{\text{2}}}\text{ }\!\!\theta\!\!\text{ } $                                                           …… (4)

Thus, dividing the equation (4) by the equation (3) gives

$ \dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\left( \dfrac{\text{dy}}{\text{d }\!\!\theta\!\!\text{ }} \right)}{\left( \dfrac{\text{dx}}{\text{d }\!\!\theta\!\!\text{ }} \right)}\text{=}\dfrac{\text{bse}{{\text{c}}^{\text{2}}}\text{ }\!\!\theta\!\!\text{ }}{\text{asec }\!\!\theta\!\!\text{ tan }\!\!\theta\!\!\text{ }}\text{=}\dfrac{\text{b}}{\text{a}}\text{sec }\!\!\theta\!\!\text{ tan }\!\!\theta\!\!\text{ =-}\dfrac{\text{bcos }\!\!\theta\!\!\text{ }}{\text{acos }\!\!\theta\!\!\text{ sin }\!\!\theta\!\!\text{ }}\text{=}\dfrac{\text{b}}{\text{a}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{sin }\!\!\theta\!\!\text{ }}\text{=}\dfrac{\text{b}}{\text{a}}\text{cosec }\!\!\theta\!\!\text{ } $ 

Hence,  $ \dfrac{\text{dy}}{\text{dx}}=\dfrac{\text{b}}{\text{a}}\text{cosec }\!\!\theta\!\!\text{ } $ .


10.  $ \text{x=a(cos }\!\!\theta\!\!\text{ + }\!\!\theta\!\!\text{ sin }\!\!\theta\!\!\text{ ), y=a(sin }\!\!\theta\!\!\text{ - }\!\!\theta\!\!\text{ cos }\!\!\theta\!\!\text{ )} $ . 

Ans:  The given equations are

$ \text{x}\,\text{=}\,\text{a(cos }\!\!\theta\!\!\text{ + }\!\!\theta\!\!\text{ sin }\!\!\theta\!\!\text{ )} $                                                                          …… (1)

and  $ \text{y}\,\text{=}\,\text{a(sin }\!\!\theta\!\!\text{ - }\!\!\theta\!\!\text{ cos }\!\!\theta\!\!\text{ )} $                                                                   …… (2)

Then, differentiating both sides of the equation (1) with respect to  $ \text{ }\!\!\theta\!\!\text{ } $  gives

$ \begin{align}& \dfrac{\text{dx}}{\text{d }\!\!\theta\!\!\text{ }}\,\text{=}\,\text{a}\left[ \dfrac{\text{d}}{\text{d }\!\!\theta\!\!\text{ }}\text{cos }\!\!\theta\!\!\text{ +}\dfrac{\text{d}}{\text{d }\!\!\theta\!\!\text{ }}\text{( }\!\!\theta\!\!\text{ sin }\!\!\theta\!\!\text{ )} \right]\, \\ & \text{=}\,\text{a}\left[ \text{-sin }\!\!\theta\!\!\text{ + }\!\!\theta\!\!\text{ }\dfrac{\text{d}}{\text{d }\!\!\theta\!\!\text{ }}\text{(sin }\!\!\theta\!\!\text{ )+sin }\!\!\theta\!\!\text{ }\dfrac{\text{d}}{\text{d }\!\!\theta\!\!\text{ }}\text{( }\!\!\theta\!\!\text{ )} \right] \\ \end{align} $ 

$ \text{=a}\left[ \text{-sin }\!\!\theta\!\!\text{ + }\!\!\theta\!\!\text{ cos }\!\!\theta\!\!\text{ +sin }\!\!\theta\!\!\text{ } \right] $ .

Therefore,  $ \dfrac{\text{dx}}{\text{d }\!\!\theta\!\!\text{ }}\text{=a }\!\!\theta\!\!\text{ cos }\!\!\theta\!\!\text{ } $                                                                 …… (3)

Also, differentiating both sides of the equation (2) with respect to  $ \text{ }\!\!\theta\!\!\text{ } $  gives

$ \begin{align}& \dfrac{\text{dy}}{\text{d }\!\!\theta\!\!\text{ }}\,\text{=}\,\text{a}\left[ \dfrac{\text{d}}{\text{d }\!\!\theta\!\!\text{ }}\text{(sin }\!\!\theta\!\!\text{ )-}\dfrac{\text{d}}{\text{d }\!\!\theta\!\!\text{ }}\text{( }\!\!\theta\!\!\text{ cos }\!\!\theta\!\!\text{ )} \right]\, \\ & \text{=}\,\text{a}\left[ \text{cos }\!\!\theta\!\!\text{ -}\left\{ \text{ }\!\!\theta\!\!\text{ }\dfrac{\text{d}}{\text{d }\!\!\theta\!\!\text{ }}\text{(cos }\!\!\theta\!\!\text{ )+cos }\!\!\theta\!\!\text{  }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{d }\!\!\theta\!\!\text{ }}\text{( }\!\!\theta\!\!\text{ )} \right\} \right] \\ \end{align} $ 

$ \Rightarrow \dfrac{\text{dy}}{\text{d }\!\!\theta\!\!\text{ }}\,\text{=}\,\text{a}\left[ \text{cos }\!\!\theta\!\!\text{ + }\!\!\theta\!\!\text{ sin }\!\!\theta\!\!\text{ -cos }\!\!\theta\!\!\text{ } \right] $ 

Therefore,  $ \dfrac{\text{dy}}{\text{d }\!\!\theta\!\!\text{ }}\,\text{=}\,\text{a }\!\!\theta\!\!\text{ sin }\!\!\theta\!\!\text{ } $                                                                   …… (4)                                     

Thus, dividing the equation (4) by the equation (3) gives

$ \begin{align}& \dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\left( \dfrac{\text{dy}}{\text{d }\!\!\theta\!\!\text{ }} \right)}{\left( \dfrac{\text{dx}}{\text{d }\!\!\theta\!\!\text{ }} \right)} \\ & \text{=}\dfrac{\text{a }\!\!\theta\!\!\text{ sin }\!\!\theta\!\!\text{ }}{\text{a }\!\!\theta\!\!\text{ sin }\!\!\theta\!\!\text{ }} \\ & \text{=tan }\!\!\theta\!\!\text{ } \\ \end{align} $ .

Hence,  $ \dfrac{\text{dy}}{\text{dx}}\text{=tan }\!\!\theta\!\!\text{ } $ .


11.If$\mathbf{x=}\sqrt{{{\mathbf{a}}^{\mathbf{si}{{\mathbf{n}}^{\mathbf{-1}}}\mathbf{t}}}}\mathbf{, y=}\sqrt{{{\mathbf{a}}^{\mathbf{co}{{\mathbf{s}}^{\mathbf{-1}}}\mathbf{t}}}} $ ,  show that,  $ \dfrac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=-}\dfrac{\mathbf{y}}{\mathbf{x}} $ .

Ans: The given parametric equations are  $ \text{x=}\sqrt{{{\text{a}}^{\text{si}{{\text{n}}^{\text{-1}}}\text{t}}}} $  and  $ \text{y=}\sqrt{{{\text{a}}^{\text{co}{{\text{s}}^{\text{-1}}}\text{t}}}} $ .

Now,  $ \text{x=}\sqrt{{{\text{a}}^{\text{si}{{\text{n}}^{\text{-1}}}\text{t}}}} $  and  $ \text{y=}\sqrt{{{\text{a}}^{\text{co}{{\text{s}}^{\text{-1}}}\text{t}}}} $ 

$ \Rightarrow \text{x=}\left( {{\text{a}}^{\text{si}{{\text{n}}^{\text{-1}}}\text{t}}} \right) $  and  $ \text{y=}{{\left( {{\text{a}}^{\text{co}{{\text{s}}^{\text{-1}}}\text{t}}} \right)}^{\dfrac{\text{1}}{\text{2}}}} $ 

$\Rightarrow\text{x=}{{\text{a}}^{\dfrac{\text{1}}{\text{2}}\text{si}{{\text{n}}^{\text{-1t}}}}} $  and  

$\text{y=}{{\text{a}}^{\dfrac{\text{1}}{\text{2}}\text{co}{{\text{s}}^{\text{-1}}}\text{t}}} $ 

Therefore, first consider  $ \text{x=}{{\text{a}}^{\dfrac{\text{1}}{\text{2}}\text{si}{{\text{n}}^{\text{-1}}}\text{t}}} $ .

Take logarithms on both sides of the equation.

Then, we have

$\text{logx=}\dfrac{\text{1}}{\text{2}}\text{si}{{\text{n}}^{\text{-1}}}\text{tloga} $ .

Then, differentiating both sides of the equation with respect to  $ \text{t} $  gives

$ \begin{align}& \dfrac{\text{1}}{\text{x}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{dx}}{\text{dt}}\text{=}\dfrac{\text{1}}{\text{2}}\text{loga }\!\!\times\!\!\text{}\dfrac{\text{d}}{\text{dt}}\text{(si}{{\text{n}}^{\text{-1}}}\text{t)} \\ & \Rightarrow \dfrac{\text{dx}}{\text{dt}}\text{=}\dfrac{\text{x}}{\text{2}}\text{loga }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\sqrt{\text{1-}{{\text{t}}^{\text{2}}}}} \\ \end{align} $ 

Therefore,  $ \dfrac{\text{dx}}{\text{dt}}\text{=}\dfrac{\text{xloga}}{\text{2}\sqrt{\text{1-}{{\text{t}}^{\text{2}}}}} $ .                                               …… (1)

Again, consider the equation $\text{y=}{{\text{a}}^{\dfrac{\text{1}}{\text{2}}\text{co}{{\text{s}}^{\text{-1}}}\text{t}}} $ .

Take logarithm both sides of the equation.

Then, we have

$\text{logy=}\dfrac{\text{1}}{\text{2}}\text{co}{{\text{s}}^{\text{-1}}}\text{tloga} $ 

Differentiating both sides of the equation with respect to  $ \text{t} $  gives

 $ \begin{align}& \dfrac{\text{1}}{\text{y}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{dx}}{\text{dt}}\text{=}\dfrac{\text{1}}{\text{2}}\text{loga }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dt}}\text{(co}{{\text{s}}^{\text{-1}}}\text{t)} \\ & \Rightarrow \dfrac{\text{dx}}{\text{dt}}\text{=}\dfrac{\text{yloga}}{\text{2}}\text{ }\!\!\times\!\!\text{ }\left( \dfrac{\text{1}}{\sqrt{\text{1-}{{\text{t}}^{\text{2}}}}} \right) \\ \end{align} $ 

Therefore,  $ \dfrac{\text{dx}}{\text{dt}}\text{=}\dfrac{\text{-yloga}}{\text{2}\sqrt{\text{1-}{{\text{t}}^{\text{2}}}}} $ .                                             …… (2)

Thus, dividing the equation (2) by the equation (1) gives 

 $ \begin{align}& \dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\left( \dfrac{\text{dy}}{\text{dt}} \right)}{\left( \dfrac{\text{dx}}{\text{dt}} \right)} \\ &\text{=}\dfrac{\left(\dfrac{\text{-yloga}}{\text{2}\sqrt{\text{1-}{{\text{t}}^{\text{2}}}}}\right)}{\left(\dfrac{\text{xloga}}{\text{2}\sqrt{\text{1-}{{\text{t}}^{\text{2}}}}} \right)} \\ & \text{=}\dfrac{\text{y}}{\text{x}} \\ \end{align} $ .

Hence,$ \dfrac{\text{dy}}{\text{dx}}\text{=}-\dfrac{\text{y}}{\text{x}} $ .


Exercise 5.7 

Find the second order derivatives of the functions given in Exercises 1 to 10. 

1.  $ \text{y=}{{\text{x}}^{\text{2}}}\text{+3x+2} $ .

Ans:  The given function is  $ \text{y=}{{\text{x}}^{\text{2}}}\text{+3x+2} $ .

Then, differentiating both sides with respect to  $ \text{x} $  gives

$\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{x}}^{\text{2}}}\text{)+}\dfrac{\text{d}}{\text{dx}}\text{(3x)+}\dfrac{\text{d}}{\text{dx}}\text{(2)=2x+3+0=2x+3} $ 

That is,

$ \dfrac{\text{dy}}{\text{dx}}=2\text{x+3} $ .

Again, differentiating both sides with respect to  $ \text{x} $  gives

$\dfrac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{(2x+3)=}\dfrac{\text{d}}{\text{dx}}\text{(2x)+}\dfrac{\text{d}}{\text{dx}}\text{(3)=2+0=2} $ 

Hence,  $\dfrac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=2} $.

2.  $ \mathbf{y=}{{\mathbf{x}}^{\mathbf{20}}} $ .

Ans: The given function is  $ \text{y=}{{\text{x}}^{\text{20}}} $ .

Then, differentiating both sides with respect to  $ \text{x} $  gives $\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{x}}^{\text{20}}}\text{)=20}{{\text{x}}^{\text{19}}} $ 

Again, differentiating both sides with respect to  $ \text{x} $  gives

$\dfrac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{(20}{{\text{x}}^{\text{19}}}\text{)=20}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{x}}^{\text{19}}}\text{)=20}\left(19\right){{\text{x}}^{\text{18}}}\text{=380}{{\text{x}}^{\text{18}}} $ .

Hence, $\dfrac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=380}{{\text{x}}^{\text{18}}} $ .


3.  $ \mathbf{y=x}\cdot \mathbf{cosx} $ .

Ans: The given function is  $ \text{y=x}\text{.cosx} $ .

Then, differentiating both sides with respect to  $ \text{x} $  gives

$\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{(x}\text{.cosx)=cosx}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(x)+x}\dfrac{\text{d}}{\text{dx}}\text{(cosx)=cosx}\text{.1+x(-sinx)=cosx-xsinx} $ 

That is,  $ \dfrac{\text{dy}}{\text{dx}}\text{=cosx-xsinx} $ .

Again, differentiating both sides with respect to  $ \text{x} $  gives

$\begin{align}&\dfrac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{(cosx-xsinx)=}\dfrac{\text{d}}{\text{dx}}\text{(cosx)-}\dfrac{\text{d}}{\text{dx}}\text{(xsinx)} \\ & \text{=-sinx- }\!\![\!\!\text{ sinx}\cdot \dfrac{\text{d}}{\text{dx}}\text{(x)+x}\cdot \dfrac{\text{d}}{\text{dx}}\text{(sinx) }\!\!]\!\!\text{ } \\ & \text{=-sinx-(sinx+xcosx)} \\ \end{align} $ 

Hence,  $ \dfrac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=-(xcosx+2sinx)} $ .


4.  $ \mathbf{y=logx} $ .

Ans: The given function is  $ \text{y=logx} $ .

Then, differentiating both sides with respect to  $ \text{x} $  gives

$\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{(logx)=}\dfrac{\text{1}}{\text{x}} $ 

Again, differentiating both sides with respect to  $ \text{x} $  gives

$\dfrac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}\dfrac{\text{d}}{\text{dx}}\left(\dfrac{\text{1}}{\text{x}}\right)\text{=}\dfrac{\text{-1}}{{{\text{x}}^{\text{2}}}} $ 

Hence,  $\dfrac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=-}\dfrac{\text{1}}{{{\text{x}}^{\text{2}}}} $ .


5.  $ \text{y=}{{\text{x}}^{\text{3}}}\text{logx} $ .

Ans: The given function is  $ \text{y=}{{\text{x}}^{\text{3}}}\text{logx} $ .

Then, differentiating both sides with respect to  $ \text{x} $  gives

$ \dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\left[ {{\text{x}}^{\text{3}}}\text{logx}\right]\text{=logx}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{x}}^{\text{3}}}\text{)+}{{\text{x}}^{\text{3}}}\dfrac{\text{d}}{\text{dx}}\text{(logx)=logx}\text{.3}{{\text{x}}^{\text{2}}}\text{+}{{\text{x}}^{\text{3}}}\text{.}\dfrac{\text{1}}{\text{x}}\text{=logx}\text{.3}{{\text{x}}^{\text{2}}}\text{+}{{\text{x}}^{\text{2}}} $ 

That is,  $\dfrac{\text{dy}}{\text{dx}}\text{=}{{\text{x}}^{\text{2}}}\text{(1+3logx)} $ . 

Again, differentiating both sides with respect to  $ \text{x} $  gives
$ \begin{align}& \dfrac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{x}}^{\text{2}}}\text{(1+3logx))} \\ & \text{=(1+3logx)}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{x}}^{\text{2}}}\text{)+}{{\text{x}}^{\text{2}}}\dfrac{\text{d}}{\text{dx}}\text{(1+3logx)} \\ & \text{=(1+3logx)}\text{.2x+}{{\text{x}}^{\text{3}}}\text{.}\dfrac{\text{3}}{\text{x}} \\ & \text{=2x+6logx+3x} \\ & \text{=5x+6xlogx}  \end{align} $ 

Hence,  $\dfrac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=x(5+6logx)} $ .


6.  $ \mathbf{y=}{{\mathbf{e}}^{\mathbf{x}}}\mathbf{sin5x} $ 

Ans: The given function is  $ \text{y=}{{\text{e}}^{\text{x}}}\text{sin5x} $ .

Then, differentiating both sides with respect to  $ \text{x} $  gives

 $ \begin{align}& \dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\left[ {{\text{e}}^{\text{x}}}\text{sin5x} \right]\text{=sinx}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{e}}^{\text{x}}}\text{)+}{{\text{e}}^{\text{x}}}\dfrac{\text{d}}{\text{dx}}\text{(sin5x)} \\ & \Rightarrow\dfrac{\text{dy}}{\text{dx}}\text{=sin5x}\text{.}{{\text{e}}^{\text{x}}}\text{+}{{\text{e}}^{\text{x}}}\text{.cos5x}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(5x)} \\ \end{align} $ 

That is,  $\dfrac{\text{dy}}{\text{dx}}\text{=}{{\text{e}}^{\text{x}}}\text{(sin5x+5cos5x)} $ .

Again, differentiating both sides with respect to  $ \text{x} $  gives

$ \begin{align}& \dfrac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}\dfrac{\text{d}}{\text{dx}}\left[ {{\text{e}}^{\text{x}}}\text{(sin5x+5cos5x)} \right] \\&\text{=(sin5x+5cos5x)}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{e}}^{\text{x}}}\text{)+}{{\text{e}}^{\text{x}}}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(sin5x+5cos5x)} \\ & \text{=(sin5x+5cos5x)(}{{\text{e}}^{\text{x}}}\text{)+}{{\text{e}}^{\text{x}}}\left[ \text{cos5x}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(5x)+5(-sin5x)}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(5x)} \right] \\ & \text{=}{{\text{e}}^{\text{x}}}\text{(sin5x+5cos5x)+}{{\text{e}}^{\text{x}}}\text{(5cos5x-25sin5x)} \\ \end{align} $ 

$ \text{=}{{\text{e}}^{\text{x}}}\text{(10cos5x-24sin5x)} $ .

Hence,  $\dfrac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=2}{{\text{e}}^{\text{x}}}\text{(5cox5x-12sin5x)} $ .


7.  $ \text{y=}{{\text{e}}^{\text{6x}}}\text{cos3x} $ .

Ans:  The given function is  $ \text{y=}{{\text{e}}^{\text{6x}}}\text{cos3x} $ .

Then, differentiating both sides with respect to  $ \text{x} $  gives

$ \begin{align}& \dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{e}}^{\text{6x}}}\text{cos3x)=cos3x }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{e}}^{\text{6x}}}\text{)+}{{\text{e}}^{\text{6x}}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(cos3x)} \\ & \Rightarrow \dfrac{\text{dy}}{\text{dx}}=\text{cos3x }\!\!\times\!\!\text{ }{{\text{e}}^{\text{6x}}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(6x)+}{{\text{e}}^{\text{6x}}}\text{ }\!\!\times\!\!\text{ (-sin3x) }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(3x)} \\ \end{align} $ 

Therefore,

$\dfrac{\text{dy}}{\text{dx}}\text{=6}{{\text{e}}^{\text{6x}}}\text{cos3x-3}{{\text{e}}^{\text{6x}}}\text{sin3x} $ …… (1)

Again, differentiating both sides with respect to  $ \text{x} $  gives

$\dfrac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{(6}{{\text{e}}^{\text{6x}}}\text{cos3x-3}{{\text{e}}^{\text{6x}}}\text{sin3x)=6 }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{e}}^{\text{6x}}}\text{cos3x)-3 }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{e}}^{\text{6x}}}\text{sin3x)} $ 

$\text{=6}\!\!\times\!\!\text{}\left[\text{6}{{\text{e}}^{\text{6x}}}\text{cos3x-3}{{\text{e}}^{\text{6x}}}\text{sin3x} \right]\text{-3 }\!\!\times\!\!\text{ }\left[ \text{sin3x }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{e}}^{\text{6x}}}\text{)+}{{\text{e}}^{\text{6x}}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(sin3x)} \right] $ [using (1)]

$\begin{align}&\text{=36}{{\text{e}}^{\text{6x}}}\text{cos3x-18}{{\text{e}}^{\text{6x}}}\text{sin3x-3}\left[ \text{sin3x }\!\!\times\!\!\text{ }{{\text{e}}^{\text{6x}}}\text{ }\!\!\times\!\!\text{ 6+}{{\text{e}}^{\text{6x}}}\text{ }\!\!\times\!\!\text{ cos3x-3} \right] \\ & \text{=36}{{\text{e}}^{\text{6x}}}\text{cos3x-18}{{\text{e}}^{\text{6x}}}\text{sin3x-18}{{\text{e}}^{\text{6x}}}\text{sin3x-9}{{\text{e}}^{\text{6x}}}\text{cos3x} \\ \end{align} $ 

Hence,  $\dfrac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=9}{{\text{e}}^{\text{6x}}}\text{(3cos3x-4sin3x)} $ .


8.  $ \mathbf{y=ta}{{\mathbf{n}}^{\mathbf{-1}}}\mathbf{x} $ .

Ans: The given function is  $ \text{y=ta}{{\text{n}}^{\text{-1}}}\text{x} $ .

Then, differentiating both sides with respect to  $ \text{x} $  gives

$\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{ta}{{\text{n}}^{\text{-1}}}\text{x=}\dfrac{\text{1}}{\text{1-}{{\text{x}}^{\text{2}}}} $ 

Again, differentiating both sides with respect to  $ \text{x} $  gives

$ \begin{align}& \dfrac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}\dfrac{\text{d}}{\text{dx}}\left( \dfrac{\text{1}}{\text{1+}{{\text{x}}^{\text{2}}}} \right)\text{=}\dfrac{\text{d}}{\text{dx}}{{\text{(1+}{{\text{x}}^{\text{2}}}\text{)}}^{\text{-1}}}\text{=(-1) }\!\!\times\!\!\text{ (1+}{{\text{x}}^{\text{2}}}{{\text{)}}^{\mathbf{-2}}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(1+}{{\text{x}}^{\text{2}}}\text{)} \\  & \text{=-}\dfrac{\text{1}}{{{\text{(1+}{{\text{x}}^{\text{2}}}\text{)}}^{2}}}\text{ }\!\!\times\!\!\text{ 2x} \\ \end{align} $ 

Hence,  

$\dfrac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}\dfrac{\text{-2x}}{{{\text{(1+}{{\text{x}}^{\text{2}}}\text{)}}^{2}}} $ .


9.  $ \text{y=log(logx)} $ .

Ans: The given function is  $ \text{y=log(logx)} $ .

Now, differentiating both sides with respect to  $ \text{x} $  gives

 $ \begin{align}& \dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{ }\!\![\!\!\text{ log(logx) }\!\!]\!\!\text{ } \\ &=\dfrac{\text{1}}{\text{logx}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(logx)} \\ & \Rightarrow \dfrac{\text{dy}}{\text{dx}}\text{=(xlogx}{{\text{)}}^{\text{-1}}} \\ \end{align} $ 

Again, differentiating both sides with respect to  $ \text{x} $  gives

$ \begin{align}& \dfrac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}\dfrac{\text{d}}{\text{dx}}\left[ {{\text{(xlogx)}}^{\text{-1}}} \right]\text{=(-1) }\!\!\times\!\!\text{ (xlogx}{{\text{)}}^{\text{-2}}}\dfrac{\text{d}}{\text{dx}}\text{(xlogx)} \\ & =\dfrac{\text{-1}}{{{\text{(xlogx)}}^{\text{2}}}}\text{ }\!\!\times\!\!\text{ }\left[ \text{logx }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(x)+x }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(logx)} \right] \\ & =\dfrac{\text{-1}}{{{\text{(xlogx)}}^{\text{2}}}}\text{ }\!\!\times\!\!\text{ }\left[ \text{logx }\!\!\times\!\!\text{ 1+x }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{x}} \right] \\ \end{align} $ 

Hence,$\dfrac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}\dfrac{\text{-(1+logx)}}{{{\text{(xlogx)}}^{\text{2}}}} $ .


10.  $ \text{y=sin(logx)} $ .

Ans: The given function is  $ \text{y=sin(logx)} $ .

Now, differentiating both sides with respect to  $ \text{x} $  gives

 $ \dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{ }\!\![\!\!\text{ sin(logx) }\!\!]\!\!\text{ =cos(logx) }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(logx)}=\dfrac{\text{cos(logx)}}{\text{x}} $ 

Again, differentiating both sides with respect to  $ \text{x} $  gives

$\begin{align}&\dfrac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}\dfrac{\text{d}}{\text{dx}}\left[ \dfrac{\text{cos(logx)}}{\text{x}} \right] \\ & =\dfrac{\text{x}\left[ \text{cos(logx)} \right]\text{-cos(logx) }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(x)}}{{{\text{x}}^{\text{2}}}} \\ & =\dfrac{\text{x}\left[ \text{-sin(logx) }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(logx)} \right]\text{-cos(logx) }\!\!\times\!\!\text{ 1}}{{{\text{x}}^{\text{2}}}} \\ & =\dfrac{\text{-xsin(logx) }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{x}}\text{-cos(logx)}}{{{\text{x}}^{\text{2}}}} \\ \end{align} $ 

Hence,  $\dfrac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}\dfrac{\left[ \text{-sin(logx)-cos(logx)} \right]}{{{\text{x}}^{\text{2}}}} $ .


11. If  $ \text{y=5cosx-3sinx} $ , prove that  $ \dfrac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{+y=0} $ . 

Ans: The given equation is  $ \text{y=5cosx-3sinx} $ .

Then, differentiating both sides with respect to  $ \text{x} $  gives

$\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{(5cosx)-}\dfrac{\text{d}}{\text{dx}}\text{(3sinx)=5}\dfrac{\text{d}}{\text{dx}}\text{(cosx)-3}\dfrac{\text{d}}{\text{dx}}\text{(sinx)=5(-sinx)-3cosx} $ 

Therefore,  $ \dfrac{\text{dy}}{\text{dx}}\text{=-(5sinx+3cosx)} $ .

Again, differentiating both sides with respect to  $ \text{x} $  gives

$\dfrac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{ }\!\![\!\!\text{ -(5sinx+3cosx) }\!\!]\!\!\text{ } $ 

$ \begin{align}& \text{=-}\left[ \text{5 }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(sinx)+3 }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(cosx)} \right] \\ & \text{= }\!\![\!\!\text{ 5cosx+3(-sinx) }\!\!]\!\!\text{ } \\ & \text{=-y} \\ \end{align} $ 

That is, $\dfrac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=-y} $ .

Hence,  $\dfrac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{+y=0} $ .


12. If  $ \mathbf{y=co}{{\mathbf{s}}^{\mathbf{-1}}}\mathbf{x} $ , Find  $ \dfrac{{{\mathbf{d}}^{\mathbf{2}}}\mathbf{y}}{\mathbf{d}{{\mathbf{x}}^{\mathbf{2}}}} $  in the terms of  $ \mathbf{y} $  alone.

Ans: The given function is  $ \text{y=co}{{\text{s}}^{\text{-1}}}\text{x} $ .

Now, differentiating both sides with respect to  $ \text{x} $  gives

$\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{(co}{{\text{s}}^{\text{-1}}}\text{x)=}\dfrac{\text{-1}}{\sqrt{\text{1-}{{\text{x}}^{\text{2}}}}}\text{=-(1-}{{\text{x}}^{\text{2}}}{{\text{)}}^{\dfrac{\text{-1}}{\text{2}}}} $ 

Again, differentiating both sides with respect to  $ \text{x} $  gives

$ \begin{align}& \dfrac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}\dfrac{\text{d}}{\text{dx}}\left[ \text{-(1-}{{\text{x}}^{\text{2}}}{{\text{)}}^{\dfrac{\text{-1}}{\text{2}}}} \right] \\ & =\left( \dfrac{\text{-1}}{\text{2}} \right)\text{ }\!\!\times\!\!\text{ (1-}{{\text{x}}^{\text{2}}}{{\text{)}}^{\dfrac{\text{-3}}{\text{2}}}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(1-}{{\text{x}}^{\text{2}}}\text{)} \\ & =\dfrac{\text{1}}{\sqrt[\text{2}]{{{\text{(1-}{{\text{x}}^{\text{2}}}\text{)}}^{\text{3}}}}}\text{ }\!\!\times\!\!\text{ (-2x)} \\ \end{align} $ 

$\Rightarrow\dfrac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}\dfrac{\text{-x}}{\sqrt{{{\text{(1-}{{\text{x}}^{\text{2}}}\text{)}}^{\text{3}}}}} $                                                                        …… (1)                                                                                                          

Now,  $ \text{y=co}{{\text{s}}^{\text{-1}}}\text{x}\Rightarrow \text{x=cosy} $ .

Therefore, substituting  $ \text{x=cosy} $  into equation (1), gives

$\begin{align}&\dfrac{{{\text{d}}^{\text{2}}}\text{x}}{\text{d}{{\text{y}}^{\text{2}}}}\text{=}\dfrac{\text{-cosy}}{\sqrt{{{\text{(1-co}{{\text{s}}^{\text{2}}}\text{y)}}^{\text{3}}}}} \\ &=\dfrac{\text{-cosy}}{\text{si}{{\text{n}}^{\text{3}}}\text{y}} \\ & =\dfrac{\text{-cosy}}{\text{siny}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{si}{{\text{n}}^{\text{2}}}\text{y}} \\ \end{align} $ 

Hence,$\dfrac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{= -coty }\!\!\times\!\!\text{ cose}{{\text{c}}^{\text{2}}}\text{y} $ .


13. If  $ \mathbf{y=3cos(logx)+4sin(logx)} $ , show that  $ {{\mathbf{x}}^{\mathbf{2}}}{{\mathbf{y}}_{\mathbf{2}}}\mathbf{+x}{{\mathbf{y}}_{\mathbf{1}}}\mathbf{+y=0} $ .

Ans: The given equations are  $ \text{y=3cos(logx)+4sin(logx)} $                                …… (1)

and  $ {{\text{x}}^{\text{2}}}{{\text{y}}_{\text{2}}}\text{+x}{{\text{y}}_{\text{1}}}\text{+y=0} $                                                                                 …… (2)

Then, differentiating both sides of the equation (1) with respect to  $ \text{x} $  gives

 $ \begin{align}& {{\text{y}}_{\text{1}}}\text{=3 }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{ }\!\![\!\!\text{ cos(logx) }\!\!]\!\!\text{ +4 }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{ }\!\![\!\!\text{ sin(logx) }\!\!]\!\!\text{ } \\ & \text{=3 }\!\!\times\!\!\text{ }\left[ \text{-sin(logx) }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(logx)} \right]\text{+4 }\!\!\times\!\!\text{ }\left[ \text{cos(logx) }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(logx)} \right] \\ \end{align} $ 

${{\text{y}}_{\text{1}}}\text{=}\dfrac{\text{-3sin(logx)}}{\text{x}}\text{+}\dfrac{\text{4cos(logx)}}{\text{x}}\text{=}\dfrac{\text{4cos(logx)-3sin(logx)}}{\text{x}} $ 

Again, differentiating both sides with respect to  $ \text{x} $  gives

$ {{\text{y}}_{\text{2}}}\text{=}\dfrac{\text{d}}{\text{dx}}\left( \dfrac{\text{4cos(logx)-3sin(logx)}}{\text{x}} \right) $ 

$ \begin{align}& \text{=}\dfrac{\text{x}\cdot \dfrac{d}{dx}\left[ \text{4 }\!\!\{\!\!\text{ cos(logx) }\!\!\}\!\!\text{ - }\!\!\{\!\!\text{ 3sin(logx) }\!\!\}\!\!\text{ } \right]\text{- }\!\!\{\!\!\text{ 4cos(logx)-3sin(logx) }\!\!\}\!\!\text{ }\times \text{1}}{{{\text{x}}^{\text{2}}}} \\ & \text{=}\dfrac{\text{x}\left[\text{-4sin(logx)}\dfrac{d}{dx}\text{(logx)-3cos(logx)}\dfrac{d}{dx}\text{(logx)}\right]\text{-4cos(logx)+3sin(logx)}}{{{\text{x}}^{\text{2}}}} \\ \end{align} $ 

$ =\dfrac{\text{x}\left[ \text{-4sin(logx)}\cdot \dfrac{\text{1}}{\text{x}}\text{-3cos(logx)}\cdot \dfrac{\text{1}}{\text{x}} \right]\text{-4cos(logx)+3sin(logx)}}{{{\text{x}}^{\text{2}}}} $ 

$=\dfrac{\text{-4sin(logx)-3cos(logx)-4cos(logx)+3sin(logx)}}{{{\text{x}}^{\text{2}}}} $ 

Therefore,  

${{\text{y}}_{\text{2}}}\text{=}\dfrac{\text{-sin(logx)-7cos(logx)}}{{{\text{x}}^{\text{2}}}} $ .

Now, substituting the derivatives  $ {{\text{y}}_{1}} $  , $ {{\text{y}}_{2}} $  and  $ \text{y} $  into the LHS of the equation (2) gives

$ \begin{align}& {{\text{x}}^{\text{2}}}{{\text{y}}_{\text{2}}}\text{+x}{{\text{y}}_{\text{1}}}\text{+y} \\ & \text{=}{{\text{x}}^{\text{2}}}\left( \dfrac{\text{-sin(logx)-7cos(logx)}}{{{\text{x}}^{\text{2}}}} \right)\text{+x}\left( \dfrac{\text{4cos(logx)-3sin(logx)}}{{{\text{x}}^{\text{2}}}} \right)\text{+3cos(logx)+4sin(logx)} \\ & \text{=-sin(logx)-7cos(logx)+4cos(logx)-3sin(logx)+4sin(logx)} \\ & \text{=0} \\ \end{align} $ 

Hence, it has been proved that  $ {{\text{x}}^{\text{2}}}{{\text{y}}_{\text{2}}}\text{+x}{{\text{y}}_{\text{1}}}\text{+y}=0 $ .


14. If  $\mathbf{y=A}{{\mathbf{e}}^{\mathbf{mx}}}\mathbf{+B}{{\mathbf{e}}^{\mathbf{nx}}} $ , show that  $\dfrac{{{\mathbf{d}}^{\mathbf{2}}}\mathbf{y}}{\mathbf{d}{{\mathbf{x}}^{\mathbf{2}}}}\mathbf{-(m+n)}\dfrac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+mny=0} $ .

Ans: The given equations are  $ \text{y=A}{{\text{e}}^{\text{mx}}}\text{+B}{{\text{e}}^{\text{nx}}} $                                      …… (1)

and  $\dfrac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{-(m+n)}\dfrac{\text{dy}}{\text{dx}}\text{+mny=0} $                                                       ……. (2)

Then, differentiating both sides of the equation (1) with respect to  $ \text{x} $  gives

$\dfrac{\text{dy}}{\text{dx}}\text{=A}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{e}}^{\text{mx}}}\text{)+B}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{e}}^{\text{mx}}}\text{)=A}\text{.}{{\text{e}}^{\text{mx}}}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(mx)+B}\text{.}{{\text{e}}^{\text{nx}}}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(nx)=Am}{{\text{e}}^{\text{mx}}}\text{+Bn}{{\text{e}}^{\text{nx}}} $ 

Again, differentiating both sides with respect to  $ \text{x} $  gives

$ \begin{align}& \dfrac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{(Am}{{\text{e}}^{\text{mx}}}\text{+Bn}{{\text{e}}^{\text{nx}}}\text{)=Am}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{e}}^{\text{mx}}}\text{)+Bn}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{e}}^{\text{nx}}}\text{)} \\ & \text{=Am}\text{.}{{\text{e}}^{\text{mx}}}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(mx)+Bn}\text{.}{{\text{e}}^{\text{nx}}}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(nx)} \\ \end{align} $ 

Therefore,  $\dfrac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=A}{{\text{m}}^{\text{2}}}{{\text{e}}^{\text{mx}}}\text{+B}{{\text{n}}^{\text{2}}}{{\text{e}}^{\text{nx}}} $ .

Thus, substituting the derivatives  $ {{\text{y}}_{1}} $  , $ {{\text{y}}_{2}} $  and  $ \text{y} $  into the LHS of the equation (2) gives

$\begin{align}&\dfrac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{-(m+n)}\dfrac{\text{dy}}{\text{dx}}\text{+mny} \\ & \text{=A}{{\text{m}}^{\text{2}}}\text{e}{{\text{x}}^{\text{mx}}}\text{+B}{{\text{n}}^{\text{2}}}{{\text{e}}^{\text{nx}}}\text{-(m+n)}\text{.(Am}{{\text{e}}^{\text{mx}}}\text{+Bn}{{\text{e}}^{\text{nx}}}\text{)+mn(A}{{\text{e}}^{\text{mx}}}\text{+B}{{\text{e}}^{\text{nx}}}\text{)} \\ & \text{=A}{{\text{m}}^{\text{2}}}\text{e}{{\text{x}}^{\text{mx}}}\text{+B}{{\text{n}}^{\text{2}}}{{\text{e}}^{\text{nx}}}\text{-Ame}{{\text{x}}^{\text{mx}}}\text{+Bmn}{{\text{e}}^{\text{nx}}}\text{+Amn}{{\text{e}}^{\text{mx}}}\text{+B}{{\text{n}}^{\text{2}}}{{\text{e}}^{\text{nx}}}\text{+Amn}{{\text{e}}^{\text{mx}}}\text{+Bmn}{{\text{e}}^{\text{nx}}} \\ & \text{=0} \\ \end{align} $ 

Thus, it has been proved that  $ \dfrac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{-(m+n)}\dfrac{\text{dy}}{\text{dx}}\text{+mny=0} $ .


15. If  $\mathbf{y=500}{{\mathbf{e}}^{\mathbf{7x}}}\mathbf{+600}{{\mathbf{e}}^{\mathbf{-7x}}} $ , show that  $\dfrac{{{\mathbf{d}}^{\mathbf{2}}}\mathbf{y}}{\mathbf{d}{{\mathbf{x}}^{\mathbf{2}}}}\mathbf{=49y} $ .

Ans: The given equation is  $\text{y=500}{{\text{e}}^{\text{7x}}}\text{+600}{{\text{e}}^{\text{-7x}}} $ .                                     …… (1)

Then, differentiating both sides with respect to  $ \text{x} $  gives

 $ \begin{align}& \dfrac{\text{dy}}{\text{dx}}\text{=500 }\!\!\times\!\!\text{ (}{{\text{e}}^{\text{7x}}}\text{)+600 }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(-7x)} \\ & \text{=500 }\!\!\times\!\!\text{ }{{\text{e}}^{\text{7x}}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(7x)+600 }\!\!\times\!\!\text{ }{{\text{e}}^{\text{-7x}}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(-7x)} \\ & \text{=3500}{{\text{e}}^{\text{7x}}}\text{-4200}{{\text{e}}^{\text{-7x}}} \\ \end{align} $ 

Again, differentiating both sides with respect to  $ \text{x} $  gives

$ \begin{align}& \dfrac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=3500 }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{e}}^{\text{7x}}}\text{)-4200 }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{e}}^{\text{-7x}}}\text{)} \\ & \text{=3500 }\!\!\times\!\!\text{ }{{\text{e}}^{\text{7x}}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(7x)-4200 }\!\!\times\!\!\text{ }{{\text{e}}^{\text{-7x}}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(-7x)} \\ & \text{=7 }\!\!\times\!\!\text{ 3500 }\!\!\times\!\!\text{ }{{\text{e}}^{\text{7x}}}\text{+7 }\!\!\times\!\!\text{ 4200 }\!\!\times\!\!\text{ }{{\text{e}}^{\text{-7x}}} \\ & \text{=49 }\!\!\times\!\!\text{ 500}{{\text{e}}^{\text{7x}}}\text{+49 }\!\!\times\!\!\text{ 600}{{\text{e}}^{\text{-7x}}} \\ & \text{=49(500}{{\text{e}}^{\text{7x}}}\text{+600}{{\text{e}}^{\text{-7x}}}\text{)} \\ \end{align} $ 

$ \text{=49y} $ , using the equation (1).

Thus, it has been proved that  $ \dfrac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=49y} $ .


16. If  $ {{\mathbf{e}}^{\mathbf{y}}}\mathbf{(x+1)=1} $ , show that  $ \dfrac{{{\mathbf{d}}^{\mathbf{2}}}\mathbf{y}}{\mathbf{d}{{\mathbf{x}}^{\mathbf{2}}}}\mathbf{=}{{\left( \dfrac{\mathbf{dy}}{\mathbf{dx}} \right)}^{\mathbf{2}}} $ .

Ans: The given equation is  $ {{\text{e}}^{\text{y}}}\text{(x+1)=1} $ .

Now,  $ {{\text{e}}^{\text{y}}}\text{(x+1)=1}\Rightarrow {{\text{e}}^{\text{y}}}\text{=}\dfrac{\text{1}}{\text{x+1}} $ .

So, taking logarithm bth sides of the equation gives

 $ \text{y=log}\dfrac{\text{1}}{\text{(x+1)}} $ 

Therefore, differentiating both sides with respect to  $ \text{x} $  gives

$ \dfrac{\text{dy}}{\text{dx}}\text{=(x+1)}\dfrac{\text{d}}{\text{dx}}\left( \dfrac{\text{1}}{\text{x+1}} \right)\text{=(x+1) }\!\!\times\!\!\text{ }\dfrac{\text{-1}}{{{\text{(x+1)}}^{\text{2}}}}\text{=}\dfrac{\text{-1}}{\text{x+1}} $ 

That is,

$ \dfrac{\text{dy}}{\text{dx}}=\dfrac{\text{-1}}{\text{x+1}} $                                                      …… (1)

Again, differentiating both sides with respect to  $ \text{x} $  gives

$ \begin{align}& \dfrac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{=}\left( \dfrac{\text{1}}{\text{x+1}} \right)\text{=-}\left( \dfrac{\text{-1}}{{{\text{(x+1)}}^{\text{2}}}} \right)\text{=}\dfrac{\text{1}}{{{\text{(x+1)}}^{\text{2}}}} \\ & \Rightarrow \dfrac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}{{\left( \dfrac{\text{-1}}{\text{x+1}} \right)}^{\text{2}}} \\ \end{align} $ 

$\Rightarrow\dfrac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}{{\left( \dfrac{\text{dy}}{\text{dx}} \right)}^{\text{2}}} $ , using the equation (1).

Thus, it is proved that  $\dfrac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}{{\left( \dfrac{\text{dy}}{\text{dx}} \right)}^{\text{2}}} $ .


17.If$\mathbf{y=(ta}{{\mathbf{n}}^{\mathbf{-1}}}\mathbf{x}{{\mathbf{)}}^{\mathbf{2}}} $ , show that  $ {{\mathbf{(}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{+1)}}^{\mathbf{2}}}{{\mathbf{y}}_{\mathbf{2}}}\mathbf{+2x(}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{+1)}{{\mathbf{y}}_{\mathbf{1}}}\mathbf{=2} $ .

Ans:  The given equations are  $ \text{y=(ta}{{\text{n}}^{\text{-1}}}\text{x}{{\text{)}}^{\text{2}}} $ .

Then, differentiating both sides with respect to  $ \text{x} $  gives

$ \begin{align}& {{\text{y}}_{\text{1}}}\text{=2ta}{{\text{n}}^{\text{-1}}}\text{x}\dfrac{\text{d}}{\text{dx}}\text{(ta}{{\text{n}}^{\text{-1}}}\text{x)} \\ & \Rightarrow {{\text{y}}_{\text{1}}}\text{=2ta}{{\text{n}}^{\text{-1}}}\text{x }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{1+}{{\text{x}}^{\text{2}}}} \\ & \Rightarrow\text{(1+}{{\text{x}}^{\text{2}}}\text{)}{{\text{y}}_{\text{1}}}\text{=2ta}{{\text{n}}^{\text{-1}}}\text{x} \\ \end{align} $ 

Again, differentiating both sides with respect to  $ \text{x} $  gives

$\begin{align}&\text{(1+}{{\text{x}}^{\text{2}}}\text{)}{{\text{y}}_{\text{2}}}\text{+2x}{{\text{y}}_{\text{1}}}\text{=2}\left(\dfrac{\text{1}}{\text{1+}{{\text{x}}^{\text{2}}}}\right)\\&\Rightarrow\text{(1+}{{\text{x}}^{\text{2}}}\text{)}{{\text{y}}_{\text{2}}}\text{+2x(1+}{{\text{x}}^{\text{2}}}\text{)}{{\text{y}}_{\text{1}}}\text{=2} \\ \end{align} $ 

Thus, it has been proved that  $ \text{(1+}{{\text{x}}^{\text{2}}}\text{)}{{\text{y}}_{\text{2}}}\text{+2x(1+}{{\text{x}}^{\text{2}}}\text{)}{{\text{y}}_{\text{1}}}\text{=2} $ .


Exercise 5.8 

1. Verify Rolle’s theorem for the function   $ \mathbf{f(x)=}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{+2x-8, x\in [-4,2]} $ .

Ans: The given function is  $ \text{f(x)=}{{\text{x}}^{\text{2}}}\text{+2x-8} $ .

Note that, the  $ \text{f}\left( \text{x} \right) $  is a polynomial function and so it is continuous in the closed interval  $ \text{ }\!\![\!\!\text{ -4,2 }\!\!]\!\!\text{ } $  and differentiable in the open interval  $ \text{(-4,2)} $  .

Now,  $ \text{f(-4)=(-4}{{\text{)}}^{\text{2}}}\text{+2x(-4)-8=16-8-8=0} $  and

 $ \text{f(2)=(2}{{\text{)}}^{\text{2}}}\text{+2 }\!\!\times\!\!\text{ 2-8=4+4-8=0} $ .

Therefore,  $ \text{f(-4)=f(2)=0} $ .

Thus, we observed that the value of at  $ \text{x=-4,2} $  are the same. 

Now, according to the Rolle’s Theorem, if a function  $ \text{f} $  is continuous on  $ \left[ \text{a,b} \right] $  and differentiable on  $ \left( \text{a,b} \right) $  such that  $ \text{f}\left( \text{a} \right)\text{=f}\left( \text{b} \right) $ , then for any  $ \text{c}\in \left( \text{a,b} \right) $ ,  $ \text{{f}'}\left( \text{c} \right)\text{=0} $ .

So, differentiating the function  $ \text{f}\left( \text{x} \right) $  both sides give

 $ \text{f }\!\!'\!\!\text{ (x)=2x+2} $ .

Substituting  $ \text{x=c} $  into the above equation gives

 $ \begin{align}& \text{f }\!\!'\!\!\text{ (c)=0} \\  & \Rightarrow \text{2c+2=0} \\&\Rightarrow\text{c=-1}\\&\Rightarrow\text{c=-1}\in(\text{-4,2)}\\\end{align} $ 

Thus, the function  $ \text{f(x)=}{{\text{x}}^{\text{2}}}\text{+2x-8} $  satisfies the Rolle’s Theorem.


2. Examine if Rolle's Theorem is applicable to any of the following functions. Can you say something about the converse of Rolle’s theorem from these examples?

(i)  $ \text{f(x)= }\!\![\!\!\text{ x }\!\!]\!\!\text{  for x}\in [\text{5,9 }\!\!]\!\!\text{ } $ 

Ans: According to the Rolle’s Theorem, for  $ \text{f. }\!\![\!\!\text{ a,b }\!\!]\!\!\text{ }\to \text{R} $ , if 

  1.  $ \text{f} $  is continuous on  $ \text{ }\!\![\!\!\text{ a,b }\!\!]\!\!\text{ } $ 

  2.  $ \text{f} $  is continuous on  $ \text{(a,b)} $ 

  3.   $ \text{f(a)=f(b)} $ ,

then, there exists any  $ \text{c}\in (\text{a,b)} $  such that  $ \text{f }\!\!'\!\!\text{ (c)=0} $ . 

Therefore, the functions that do not satisfy any of the three conditions given in the Rolle’s Theorem, do not satisfy the Rolle’s Theorem.

The given function is  $ \text{f(x)= }\!\![\!\!\text{ x }\!\!]\!\!\text{  for x}\in [5,9] $ .

Note that,  $ \text{f(x)} $  is not continuous at every integral point. 

In fact,  $ \text{f(x)} $  is not continuous at the points  $ \text{x=5} $  and   $ \text{x=9} $ .

Thus,  $ \text{f(x)} $  is not continuous in the closed interval  $ [5,9] $ . 

Also,  $ \text{f(5)= }\!\![\!\!\text{ 5 }\!\!]\!\!\text{ =5} $  and  $ \text{f(9)= }\!\![\!\!\text{ 9 }\!\!]\!\!\text{ =9} $ .

So,  $ \text{f}\left( 5 \right)\ne \text{f}\left( \text{9} \right) $ . 

For the differentiability of  $ \text{f} $  in  $ (5,9) $ , let  $ \text{n} $  be an integer in the open interval  $ (5,9) $ .

Then, the left-hand-derivative of  $ \text{f} $  at  $ \text{x=n} $  is

$\underset{h\to{{0}^{-}}}{\mathop{\text{lim}}}\,\dfrac{\text{f(n+h)-f(n)}}{\text{h}}\text{=}\underset{h\to{{0}^{-}}}{\mathop{\text{lim}}}\,\dfrac{\text{}\!\![\!\!\text{n+h}\!\!]\!\!\text{-}\!\![\!\!\text{n}\!\!]\!\!\text{}}{\text{h}}\text{=}\underset{h\to{{0}^{-}}}{\mathop{\text{lim}}}\,\dfrac{\text{n-1-n}}{\text{h}}\text{=}\underset{h\to{{0}^{-}}}{\mathop{\text{lim}}}\,\dfrac{\text{-1}}{\text{h}}\text{=}\infty  $  and

the right-hand-derivative of  $ \text{f} $  at  $ \text{x=n} $  is

$\underset{\text{h}\to{{\text{0}}^{+}}}{\mathop{\text{lim}}}\,\dfrac{\text{f(n+h)-f(n)}}{\text{h}}\text{=}\underset{\text{h}\to{{\text{0}}^{+}}}{\mathop{\text{lim}}}\,\dfrac{\text{}\!\![\!\!\text{n+h}\!\!]\!\!\text{-}\!\![\!\!\text{n}\!\!]\!\!\text{}}{\text{h}}\text{=}\underset{\text{h}\to{{\text{0}}^{+}}}{\mathop{\text{lim}}}\,\dfrac{\text{n-n}}{\text{h}}\text{=}\underset{\text{h}\to{{\text{0}}^{+}}}{\mathop{\text{lim}}}\,\dfrac{0}{\text{h}}\text{=0} $ .

Thus, it has been noticed that  $ \underset{h\to {{0}^{-}}}{\mathop{\text{lim}}}\,\dfrac{\text{f(n+h)-f(n)}}{\text{h}}\ne \underset{h\to {{0}^{+}}}{\mathop{\text{lim}}}\,\dfrac{\text{f(n+h)-f(n)}}{\text{h}} $ .

So, the function  $ \text{f} $  is not differentiable at  $ \text{x=n} $ .

Therefore, the function  $ \text{f} $  is not differentiable in the open interval  $ (5,9) $ . 

Hence, from the above discuss it can be concluded that since the function does not satisfy the conditions of Rolle’s Theorem, so the function  $ \text{f(x)= }\!\![\!\!\text{ x }\!\!]\!\!\text{  for x}\in [5,9] $  does not satisfy Rolle's Theorem.


(ii)  $ \text{f(x)= }\!\![\!\!\text{ x }\!\!]\!\!\text{  for x}\in [\text{-2,2 }\!\!]\!\!\text{ } $ 

Ans: According to the Rolle’s Theorem, for  $ \text{f. }\!\![\!\!\text{ a,b }\!\!]\!\!\text{ }\to \text{R} $ , if 

  1.  $ \text{f} $  is continuous on  $ \text{ }\!\![\!\!\text{ a,b }\!\!]\!\!\text{ } $ 

  2.  $ \text{f} $  is continuous on  $ \text{(a,b)} $ 

  3.   $ \text{f(a)=f(b)} $ ,

then, there exists any  $ \text{c}\in (\text{a,b)} $  such that  $ \text{f }\!\!'\!\!\text{ (c)=0} $ . 

Therefore, the functions that do not satisfy any of the three conditions given in the Rolle’s Theorem, do not satisfy the Rolle’s Theorem.

 $ \text{f(x)= }\!\![\!\!\text{ x }\!\!]\!\!\text{  for x}\in [-2,2] $ 

Note that,  $ \text{f(x)} $  is not continuous at all integral points. 

In fact, the function  $ \text{f(x)} $  is not continuous at the end points  $ \text{x=-2} $  and  $ \text{x=2} $ .

Therefore, the function  $ \text{f=(x)} $  is not continuous in the closed interval  $ \text{ }\!\![\!\!\text{ -2,2 }\!\!]\!\!\text{ } $ .

Also, the function values at the endpoints, that is  $ \text{f(-2)= }\!\![\!\!\text{ 2 }\!\!]\!\!\text{ =-2} $  and  $ \text{f(2)= }\!\![\!\!\text{ 2 }\!\!]\!\!\text{ =2} $ .

So,  $ \text{f}\left( \text{-2} \right)\ne \text{f}\left( \text{2} \right) $ .

Now, for the differentiability of in  $ \text{(-2,2)} $ , let  $ \text{n} $  be an integer in the open interval  $ (\text{-2,2)} $ . 

So, the left-hand-derivative of the function  $ \text{f} $  at  $ \text{x=n} $  is

$\underset{\text{h}\to{{\text{0}}^{-}}}{\mathop{\text{lim}}}\,\dfrac{\text{f(n+h)-f(n)}}{\text{h}}\text{=}\underset{\text{h}\to{{\text{0}}^{-}}}{\mathop{\text{lim}}}\,\dfrac{\text{}\!\![\!\!\text{n+h}\!\!]\!\!\text{-}\!\![\!\!\text{n}\!\!]\!\!\text{}}{\text{h}}\text{=}\underset{\text{h}\to{{\text{0}}^{-}}}{\mathop{\text{lim}}}\,\dfrac{\text{n-1-n}}{\text{h}}\text{=}\underset{\text{h}\to{{\text{0}}^{-}}}{\mathop{\text{lim}}}\,\dfrac{\text{-1}}{\text{h}}\text{=}\infty$

andtheright-hand-derivative of the function  $ \text{f} $  at  $ \text{x=n} $  is

$\underset{\text{h}\to{{\text{0}}^{+}}}{\mathop{\text{lim}}}\,\dfrac{\text{f(n+h)-f(n)}}{\text{h}}\text{=}\underset{\text{h}\to{{\text{0}}^{+}}}{\mathop{\text{lim}}}\,\dfrac{\text{}\!\![\!\!\text{n+h}\!\!]\!\!\text{-}\!\![\!\!\text{n}\!\!]\!\!\text{}}{\text{h}}\text{=}\underset{\text{h}\to{{\text{0}}^{+}}}{\mathop{\text{lim}}}\,\dfrac{\text{n-n}}{\text{h}}\text{=}\underset{\text{h}\to{{\text{0}}^{+}}}{\mathop{\text{lim}}}\,\text{0=0} $ .

Thus, it has been noticed that

$\underset{h\to{{0}^{-}}}{\mathop{\text{lim}}}\,\dfrac{\text{f(n+h)-f(n)}}{\text{h}}\ne\underset{h\to{{0}^{+}}}{\mathop{\text{lim}}}\,\dfrac{\text{f(n+h)-f(n)}}{\text{h}} $ .

So, the function  $ \text{f} $  is not differentiable at the point  $ \text{x=n} $ . 

Therefore, the function  $ \text{f} $  is not continuous in the open interval  $ \text{(-2,2)} $ . 

Hence, from the above discuss it can be concluded that since the function does not satisfy the conditions of Rolle’s Theorem, so the function  $ \text{f(x)= }\!\![\!\!\text{ x }\!\!]\!\!\text{  for x}\in \text{ }\!\![\!\!\text{ -2,2 }\!\!]\!\!\text{ } $  does not satisfy Rolle's Theorem.


(iii)  $ \text{f(x)=}{{\text{x}}^{\text{2}}}\text{-1 for }\mathbf{x}\in \mathbf{[1,2]} $ 

Ans: According to the Rolle’s Theorem, for  $ \text{f. }\!\![\!\!\text{ a,b }\!\!]\!\!\text{ }\to \text{R} $ , if 

  1.  $ \text{f} $  is continuous on  $ \text{ }\!\![\!\!\text{ a,b }\!\!]\!\!\text{ } $ 

  2.  $ \text{f} $  is continuous on  $ \text{(a,b)} $ 

  3.   $ \text{f(a)=f(b)} $ ,

then, there exists any  $ \text{c}\in (\text{a,b)} $  such that  $ \text{f }\!\!'\!\!\text{ (c)=0} $ . 

Therefore, the functions that do not satisfy any of the three conditions given in the Rolle’s Theorem, do not satisfy the Rolle’s Theorem.

The given function is  $ \text{f(x)=}{{\text{x}}^{\text{2}}}\text{-1 for x}\in[1,2]$.

Note that,  $ \text{f} $  is a polynomial function, and so it is continuous in closed interval  $ [1,2] $  and differentiable on the open interval  $ (1,2) $ .

Also,  $ \text{f(1)=(1}{{\text{)}}^{\text{2}}}\text{-1=0} $  and

 $ \text{f(2)=(2}{{\text{)}}^{\text{2}}}\text{-1=3} $ .

Therefore,  $ \text{f(1)}\ne \text{f(2)} $ .

Hence, from the above discussion it can be concluded that since the function does not satisfy one of the conditions of Rolle’s Theorem, so the function  $ \text{f(x)=}{{\text{x}}^{\text{2}}}\text{-1 for x}\in [1,2] $  does not satisfy the Rolle’s Theorem.


3. If  $ \text{f. }\!\![\!\!\text{ -5,5 }\!\!]\!\!\text{ }\to \mathbf{R} $  is a differentiable function and if  $ \mathbf{f^{\prime}(x)} $  does not vanish anywhere, then prove that  $ \text{f(-5)}\ne \text{f(5)} $ .

Ans: The given function  $ \text{f. }\!\![\!\!\text{ -5,5 }\!\!]\!\!\text{ }\to \text{R} $  is differentiable.

Now, remember that every differentiable function is a continuous function.

Therefore,  $ \text{f} $  is continuous on  $ \text{ }\!\![\!\!\text{ -5,5 }\!\!]\!\!\text{ } $  as well as on  $ \text{(-5,5)} $ . 

So, according to the Mean Value Theorem, there exists any  $ \text{c}\in \text{(-5,5)} $  for which

 $ \text{{f}'}\left( \text{c} \right)=\dfrac{\text{f}\left( 5 \right)-\text{f}\left( -5 \right)}{5-\left( -5 \right)} $ . 

Now, it is provided that the function  $ \text{f(x)}\ne \text{0} $  for any  $ \text{x} $ .

Therefore,  $ \text{{f}'(c)}\ne \text{0} $ .

$\begin{align}&\Rightarrow\text{10{f}'(c)}\ne\text{0}\\&\Rightarrow\text{f(5)-f(-5)}\ne \text{0} \\ & \Rightarrow \text{f(5)}\ne \text{f(-5)} \\ \end{align} $ 

Thus, it has been proved that  $ \text{f(5)}\ne \text{f(-5)} $ .


 4. Verify Mean Value Theorem, if

$\mathbf{f(x)=}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{-4x-3} $  in the interval  $ \text{ }\!\![\!\!\text{ a,b }\!\!]\!\!\text{ } $ , where  $ \mathbf{a=1} $ , and  $ \mathbf{b=4} $ .

Ans: The given function is  $ \text{f(x)=}{{\text{x}}^{\text{2}}}\text{-4x-3} $ .

Note that, the function  $ \text{f} $  is a polynomial function and so, it is continuous in the closed interval  $ \text{ }\!\![\!\!\text{ 1,4 }\!\!]\!\!\text{ } $  and differentiable in the open interval  $ \text{(1,4)} $  .


Differentiating the function with respect to  $ \text{x} $  gives

 $ \text{{f}'}\left( \text{x} \right)=2\text{x-4} $ .

Also,  $ \text{f(1)=(1}{{\text{)}}^{\text{2}}}\text{-4 }\!\!\times\!\!\text{ 1-3=6,} $ 

 $ \text{f(4)=(4}{{\text{)}}^{\text{2}}}\text{-4 }\!\!\times\!\!\text{ 4-3=-3} $  and

$\dfrac{\text{f(b)-f(a)}}{\text{b-a}}\text{=}\dfrac{\text{f(4)-f(1)}}{\text{4-1}}\text{=}\dfrac{\text{-3-(-6)}}{\text{3}}\text{=}\dfrac{\text{3}}{\text{3}}\text{=1} $ .

Therefore, according to the Mean Value Theorem there exists a point  $ \text{c}\in \text{(1,4)} $  for which  $ \text{f }\!\!'\!\!\text{ (c)=1} $ .

So,  $ \text{f }\!\!'\!\!\text{ (c)=1} $ 

  $ \Rightarrow \text{2c-4=1} $ 

 $ \Rightarrow \text{c=}\dfrac{\text{5}}{\text{2}} $   $ \in \text{(1,4)} $ .

Hence, the function  $ \text{f(x)=}{{\text{x}}^{\text{2}}}\text{-4x-3} $  satisfies the Mean Value Theorem in the interval  $ \left[ \text{1,4} \right] $ .


5. Verify Mean Value Theorem, if

$\mathbf{f(x)=}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{-5}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{-3x} $  in the interval  $ \text{ }\!\![\!\!\text{ a,b }\!\!]\!\!\text{ } $ , where   $ \mathbf{a=1} $  and  $ \mathbf{b=3} $ . Find all  $ \mathbf{c}\in \mathbf{(1,3)} $  for which  $ \text{f }\!\!'\!\!\text{ (c)=0} $ .

Ans: The given function is

$\text{f(x)=}{{\text{x}}^{\text{2}}}\text{-5}{{\text{x}}^{\text{2}}}\text{-3x} $ .

Note that, the function  $ \text{f} $  is a polynomial function, and so it is continuous in the closed interval  $ \text{ }\!\![\!\!\text{ 1,3 }\!\!]\!\!\text{ } $  and differentiable in the open interval  $ \text{(1,3)} $ .

Now, differentiating the function  $ \text{f}\left( \text{x} \right) $  with respect to  $ \text{x} $  gives

 $ \text{{f}'}\left( \text{x} \right)\text{=3}{{\text{x}}^{\text{2}}}\text{-10x-3} $ .

Also, the function values at the endpoints are

 $ \begin{align}& \text{f(1)=}{{\text{1}}^{\text{2}}}\text{-5 }\!\!\times\!\!\text{ }{{\text{1}}^{\text{2}}}\text{-3}\!\!\times\!\!\text{1=-7,}\\&\text{f(3)=}{{\text{3}}^{\text{2}}}\text{-5 }\!\!\times\!\!\text{ }{{\text{3}}^{\text{2}}}\text{-3 }\!\!\times\!\!\text{ 3=27} \\ \end{align} $ 

Therefore,

$\dfrac{\text{f(b)-f(a)}}{\text{b-a}}\text{=}\dfrac{\text{f(3)-f(1)}}{\text{3-1}}\text{=}\dfrac{\text{-27-(-7)}}{\text{3-1}}\text{=-10} $ .

So, according to the Mean Value Theorem there is at least a point  $ \text{c}\in \text{(1,3)} $  for which  $ \text{f }\!\!'\!\!\text{ (c)=-10} $ .

Then,

$\begin{align}&\text{f}\!\!'\!\!\text{(c)=-10}\\&\Rightarrow\text{3}{{\text{c}}^{\text{2}}}\text{-10c-3=10}\\&\Rightarrow\text{3}{{\text{c}}^{\text{2}}}\text{-10c+7=0} \\  & \Rightarrow \text{3}{{\text{c}}^{\text{2}}}\text{-3c-7c+7=0} \\  & \Rightarrow \text{3c(c-1)-7(c-1)=0} \\  & \Rightarrow \text{(c-1)(3c-7)=0} \\  & \Rightarrow \text{c=1,}\dfrac{7}{\text{3}} \\ \end{align} $ 

Therefore,  $ \text{c=}\dfrac{\text{7}}{\text{3}}\in \text{(1,3)} $ .

Hence, the function

$\text{f(x)=}{{\text{x}}^{\text{2}}}\text{-5}{{\text{x}}^{\text{2}}}\text{-3x} $  satisfies the Mean Value Theorem in the interval  $ \text{ }\!\![\!\!\text{ 1,3 }\!\!]\!\!\text{ } $  and  $ \text{c=}\dfrac{\text{7}}{\text{3}}\in \text{(1,3)} $  is the only point such that  $ \text{f }\!\!'\!\!\text{ (c)=0} $ .


6. Examine the applicability of Mean Value Theorem for all three functions given in the above exercise 2. (Not present in Current syllabus)

(i)   $ \mathbf{f(x)=[x] }\,\mathbf{for}\,\mathbf{ x}\in \mathbf{[5,9]} $  

Ans: According to the Rolle’s Theorem, for  $ \text{f. }\!\![\!\!\text{ a,b }\!\!]\!\!\text{ }\to \text{R} $ , if 

(i) $ \text{f} $  is continuous on  $ \text{ }\!\![\!\!\text{ a,b }\!\!]\!\!\text{ } $ 

(ii)  $ \text{f} $  is continuous on  $ \text{(a,b)} $ 

(iii)   $ \text{f(a)=f(b)} $ ,

then, there exists any  $ \text{c}\in (\text{a,b)} $  such that  $ \text{f }\!\!'\!\!\text{ (c)=0} $ . 

Therefore, the functions that do not satisfy any of the three conditions given in the Rolle’s Theorem, do not satisfy the Rolle’s Theorem. 

Now, note that here  $ \text{f(x)} $  is not continuous at all the integral points. 

In fact, the function  $ \text{f(x)} $  is not continuous at the endpoints  $ \text{x=5} $  and  $ \text{x=9} $ .

Therefore, the function  $ \text{f(x)} $  is not continuous in the closed interval  $ \text{ }\!\![\!\!\text{ 5,9 }\!\!]\!\!\text{ } $ . 

For the differentiability of  $ \text{f} $  in  $ (5,9) $ , let  $ \text{n} $  be an integer in the open interval  $ (5,9) $ .

Then, the left-hand-derivative of  $ \text{f} $  at  $ \text{x=n} $  is

$\underset{h\to{{0}^{-}}}{\mathop{\text{lim}}}\,\dfrac{\text{f(n+h)-f(n)}}{\text{h}}\text{=}\underset{h\to{{0}^{-}}}{\mathop{\text{lim}}}\,\dfrac{\text{}\!\![\!\!\text{n+h}\!\!]\!\!\text{-}\!\![\!\!\text{n}\!\!]\!\!\text{}}{\text{h}}\text{=}\underset{h\to{{0}^{-}}}{\mathop{\text{lim}}}\,\dfrac{\text{n-1-n}}{\text{h}}\text{=}\underset{h\to{{0}^{-}}}{\mathop{\text{lim}}}\,\dfrac{\text{-1}}{\text{h}}\text{=}\infty  $  and

the right-hand-derivative of  $ \text{f} $  at  $ \text{x=n} $  is

$\underset{\text{h}\to{{\text{0}}^{+}}}{\mathop{\text{lim}}}\,\dfrac{\text{f(n+h)-f(n)}}{\text{h}}\text{=}\underset{\text{h}\to{{\text{0}}^{+}}}{\mathop{\text{lim}}}\,\dfrac{\text{}\!\![\!\!\text{n+h}\!\!]\!\!\text{-}\!\![\!\!\text{n}\!\!]\!\!\text{}}{\text{h}}\text{=}\underset{\text{h}\to{{\text{0}}^{+}}}{\mathop{\text{lim}}}\,\dfrac{\text{n-n}}{\text{h}}\text{=}\underset{\text{h}\to{{\text{0}}^{+}}}{\mathop{\text{lim}}}\,\dfrac{0}{\text{h}}\text{=0} $ .

Thus, it has been noticed that

$\underset{h\to{{0}^{-}}}{\mathop{\text{lim}}}\,\dfrac{\text{f(n+h)-f(n)}}{\text{h}}\ne\underset{h\to{{0}^{+}}}{\mathop{\text{lim}}}\,\dfrac{\text{f(n+h)-f(n)}}{\text{h}} $ .

So, the function  $ \text{f} $  is not differentiable at  $ \text{x=n} $ .

From the above discussions we can conclude that  $ \text{f} $  does not satisfy all the conditions of the Mean Value Theorem.

Hence, the function  $ \text{f(x)= }\!\![\!\!\text{ x }\!\!]\!\!\text{ } $  does not satisfy the Mean Value Theorem  $ \text{for x}\in \text{ }\!\![\!\!\text{ 5,9 }\!\!]\!\!\text{ } $ .


(ii)  $ \mathbf{f(x)=[x] }\,\mathbf{for}\,\mathbf{ x}\in \mathbf{[-2,2]} $ 

Ans: According to the Rolle’s Theorem, for  $ \text{f. }\!\![\!\!\text{ a,b }\!\!]\!\!\text{ }\to \text{R} $ , if 

(i) $ \text{f} $  is continuous on  $ \text{ }\!\![\!\!\text{ a,b }\!\!]\!\!\text{ } $ 

(ii)  $ \text{f} $  is continuous on  $ \text{(a,b)} $ 

(iii)   $ \text{f(a)=f(b)} $ ,

then, there exists any  $ \text{c}\in (\text{a,b)} $  such that  $ \text{f }\!\!'\!\!\text{ (c)=0} $ . 

Therefore, the functions that do not satisfy any of the three conditions given in the Rolle’s Theorem, do not satisfy the Rolle’s Theorem.

Note that,  $ \text{f(x)} $  is not continuous at all integral points. 

In fact, the function  $ \text{f(x)} $  is not continuous at the end points  $ \text{x=-2} $  and  $ \text{x=2} $ .

Therefore, the function  $ \text{f=(x)} $  is not continuous in the closed interval  $ \text{ }\!\![\!\!\text{ -2,2 }\!\!]\!\!\text{ } $ .

Now, for the differentiability of in  $ \text{(-2,2)} $ , let  $ \text{n} $  be an integer in the open interval  $ (\text{-2,2)} $ . 

So, the left-hand-derivative of the function  $ \text{f} $  at  $ \text{x=n} $  is

$\underset{\text{h}\to{{\text{0}}^{-}}}{\mathop{\text{lim}}}\,\dfrac{\text{f(n+h)-f(n)}}{\text{h}}\text{=}\underset{\text{h}\to{{\text{0}}^{-}}}{\mathop{\text{lim}}}\,\dfrac{\text{}\!\![\!\!\text{n+h}\!\!]\!\!\text{-}\!\![\!\!\text{n}\!\!]\!\!\text{}}{\text{h}}\text{=}\underset{\text{h}\to{{\text{0}}^{-}}}{\mathop{\text{lim}}}\,\dfrac{\text{n-1-n}}{\text{h}}\text{=}\underset{\text{h}\to{{\text{0}}^{-}}}{\mathop{\text{lim}}}\,\dfrac{\text{-1}}{\text{h}}\text{=}\infty  $  and the right-hand-derivative of the function  $ \text{f} $  at  $ \text{x=n} $  is

$\underset{\text{h}\to{{\text{0}}^{+}}}{\mathop{\text{lim}}}\,\dfrac{\text{f(n+h)-f(n)}}{\text{h}}\text{=}\underset{\text{h}\to{{\text{0}}^{+}}}{\mathop{\text{lim}}}\,\dfrac{\text{}\!\![\!\!\text{n+h}\!\!]\!\!\text{-}\!\![\!\!\text{n}\!\!]\!\!\text{}}{\text{h}}\text{=}\underset{\text{h}\to{{\text{0}}^{+}}}{\mathop{\text{lim}}}\,\dfrac{\text{n-n}}{\text{h}}\text{=}\underset{\text{h}\to{{\text{0}}^{+}}}{\mathop{\text{lim}}}\,\text{0=0} $ .

Thus, it has been noticed that

$\underset{h\to{{0}^{-}}}{\mathop{\text{lim}}}\,\dfrac{\text{f(n+h)-f(n)}}{\text{h}}\ne\underset{h\to{{0}^{+}}}{\mathop{\text{lim}}}\,\dfrac{\text{f(n+h)-f(n)}}{\text{h}} $ .

So, the function  $ \text{f} $  is not differentiable at the point  $ \text{x=n} $ . 

From the above discussions we can conclude that  $ \text{f} $  does not satisfy all the conditions of the Mean Value Theorem. 

Thus, the function  $ \text{f(x)= }\!\![\!\!\text{ x }\!\!]\!\!\text{ } $  does not satisfy the Mean Value Theorem  $ \text{for x}\in \text{ }\!\![\!\!\text{ -2,2 }\!\!]\!\!\text{ } $ .


(iii) $\mathbf{f(x)=}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{-1}\,\,\mathbf{ for }\,\,\mathbf{x}\in \mathbf{[1,2]} $ 

According to the Rolle’s Theorem, for  $ \text{f. }\!\![\!\!\text{ a,b }\!\!]\!\!\text{ }\to \text{R} $ , if 

(i) $ \text{f} $  is continuous on  $ \text{ }\!\![\!\!\text{ a,b }\!\!]\!\!\text{ } $ 

(ii)  $ \text{f} $  is continuous on  $ \text{(a,b)} $ 

(iii)   $ \text{f(a)=f(b)} $ ,

then, there exists any  $ \text{c}\in (\text{a,b)} $  such that  $ \text{f }\!\!'\!\!\text{ (c)=0} $ . 

Therefore, the functions that do not satisfy any of the three conditions given in the Rolle’s Theorem, do not satisfy the Rolle’s Theorem.

The given function is  $ \text{f(x)=}{{\text{x}}^{\text{2}}}\text{-1 for x}\in[1,2]$.

Note that,  $ \text{f} $  is a polynomial function, and so it is continuous in closed interval  $ [1,2] $  and differentiable on the open interval  $ (1,2) $ .

Thus, the function  $ \text{f} $  satisfies all the conditions of the Mean Value Theorem.

Hence, Mean Value Theorem holds for the function

$\text{f(x)=}{{\text{x}}^{\text{2}}}\text{-1 for x}\in \text{ }\!\![\!\!\text{ 1,2 }\!\!]\!\!\text{ } $ . 

Now follow the procedure for proving it.

Here,  $ \text{f(1)=(1}{{\text{)}}^{\text{2}}}\text{-1=0,} $  and

 $ \text{f(2)=(2}{{\text{)}}^{\text{2}}}\text{-1=3} $ .

Therefore,$\dfrac{\text{f(b)-f(a)}}{\text{b-a}}\text{=}\dfrac{\text{f(2)-f(1)}}{\text{2-1}}\text{=}\dfrac{\text{3-0}}{\text{1}}\text{=3} $ .

Then, differentiating the function  $ \text{f}\left( \text{x} \right) $  with respect to  $ \text{x} $  gives

 $ \text{f }\!\!'\!\!\text{ (x)=2x} $ 

Thus, at  $ \text{x=c} $ ,

$\begin{align}&\text{f}\!\!'\!\!\text{(c)=3}\\&\Rightarrow\text{2c=3}\\&\Rightarrow\text{c=}\dfrac{\text{3}}{\text{2}}\text{=1}\text{.5}\in\text{}\!\![\!\!\text{ 1,2 }\!\!]\!\!\text{ } \\ \end{align} $ 

Hence, it is proved that the Mean Value Theorem holds for the function  $ \text{f(x)=}{{\text{x}}^{\text{2}}}\text{-1 for x}\in \text{ }\!\![\!\!\text{ 1,2 }\!\!]\!\!\text{ } $ .


Miscellaneous Exercise

Differentiate w.r.t. to  $ \mathbf{x} $ , the following function. 

1.$\mathbf{y=(3}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{-9x+5}{{\mathbf{)}}^{\mathbf{9}}} $ .

Ans: The given function is

$\text{y=(3}{{\text{x}}^{\text{2}}}\text{-9x+5}{{\text{)}}^{\text{9}}} $ .

Differentiating both sides with respect to  $ \text{x} $  gives

$\begin{align}&\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}{{\text{(3}{{\text{x}}^{\text{2}}}\text{-9x+5)}}^{\text{9}}}\\&\text{=9(3}{{\text{x}}^{\text{2}}}\text{-9x+5}{{\text{)}}^{\text{8}}}\text{}\!\!\times\!\!\text{}\dfrac{\text{d}}{\text{dx}}\text{(3}{{\text{x}}^{\text{2}}}\text{-9x+5)}\\&\text{=9(3}{{\text{x}}^{\text{2}}}\text{-9x+5}{{\text{)}}^{\text{8}}}\text{}\!\!\times\!\!\text{(6x-9x)} \\  & \text{=9(3}{{\text{x}}^{\text{2}}}\text{-9x+5}{{\text{)}}^{\text{8}}}\text{ }\!\!\times\!\!\text{3(2x-3)}\\&\text{=27(3}{{\text{x}}^{\text{2}}}\text{-9x+5}{{\text{)}}^{\text{8}}}\text{(2x-3)} \\ \end{align} $ 


2.$\mathbf{y=si}{{\mathbf{n}}^{\mathbf{3}}}\mathbf{x+co}{{\mathbf{s}}^{\mathbf{6}}}\mathbf{x} $ .

Ans: The given function is

$\text{y=si}{{\text{n}}^{\text{3}}}\text{x+co}{{\text{s}}^{\text{6}}}\text{x} $ .

Differentiating both sides with respect to  $ \text{x} $  gives

$\begin{align}&\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{=(si}{{\text{n}}^{\text{3}}}\text{x)+}\dfrac{\text{d}}{\text{dx}}\text{(co}{{\text{s}}^{\text{6}}}\text{x)}\\&\text{=3si}{{\text{n}}^{\text{2}}}\text{x}\!\!\times\!\!\text{}\dfrac{\text{d}}{\text{dx}}\text{(sinx)+6co}{{\text{s}}^{\text{5}}}\text{x}\dfrac{\text{d}}{\text{dx}}\text{(cosx)}\\&\text{=3si}{{\text{n}}^{\text{2}}}\text{x}\!\!\times\!\!\text{cosx+6co}{{\text{s}}^{\text{5}}}\text{x(-sinx)}\\&\text{=3sin}^2\text{xcosx(sinx-2co}{{\text{s}}^{\text{4}}}\text{x)} \\ \end{align} $


3.  $ \mathbf{y=(5x}{{\mathbf{)}}^{\mathbf{3cos2x}}} $ .

Ans: The given function is  $ \text{y=(5x}{{\text{)}}^{\text{3cos2x}}} $ .

First, take the logarithm of both sides of the function.

 $ \log y=3\cos 2x\log 5x $ .

Then, differentiating both sides with respect to  $ \text{x} $  gives

$\begin{align}&\dfrac{\text{1}}{\text{y}}\dfrac{\text{dy}}{\text{dx}}\text{=3}\left[\text{log5}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(cos2x)+cos2x}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(log5x)}\right]\\&\Rightarrow\dfrac{\text{dy}}{\text{dx}}\text{=3y}\left[\text{log5x(-sin2x)}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(2x)+cos2x}\text{.}\dfrac{\text{1}}{\text{5x}}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(5x)} \right] \\  & \Rightarrow \dfrac{\text{dy}}{\text{dx}}\text{=3y}\left[ \text{-2sin2xlog5x+}\dfrac{\text{cos2x}}{\text{x}} \right] \\  & \Rightarrow \dfrac{\text{dy}}{\text{dx}}\text{=3y}\left[\dfrac{\text{3cos2x}}{\text{x}}\text{-6sin2xlog5x} \right] \\ \end{align} $ 

Hence,  $ \dfrac{\text{dy}}{\text{dx}}\text{=(5x}{{\text{)}}^{\text{3cos2x}}}\left[ \dfrac{\text{3cos2x}}{\text{x}}\text{-6sin2xlog5x} \right] $ .

 

4.  $ \text{y=si}{{\text{n}}^{\text{-1}}}\left( \text{x}\sqrt{\text{x}} \right)\text{,}\,\,\,\,\text{0}\le \text{x}\le 1 $ .

Ans: The given function is  $ \text{y=si}{{\text{n}}^{\text{-1}}}\left( \text{x}\sqrt{\text{x}} \right) $ .

Then, differentiating both sides with respect to  $ \text{x} $  by using the chain rule gives

$\begin{align}&\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{si}{{\text{n}}^{\text{-1}}}\left(\text{x}\sqrt{\text{x}}\right)\\&=\dfrac{\text{1}}{\sqrt{\text{1-}{{\left(\text{x}\sqrt{\text{x}}\right)}^{2}}}}\text{}\!\!\times\!\!\text{}\dfrac{\text{d}}{\text{dx}}\left(\text{x}\sqrt{\text{x}}\right)\\&=\dfrac{\text{1}}{\sqrt{\text{1-}{{\text{x}}^{\text{3}}}}}\text{.}\dfrac{\text{d}}{\text{dx}}\left({{\text{x}}^{\dfrac{3}{2}}}\right)\\&=\dfrac{\text{1}}{\sqrt{\text{1-}{{\text{x}}^{\text{3}}}}}\text{}\!\!\times\!\!\text{}\dfrac{\text{3}}{\text{2}}\text{.}{{\text{x}}^{\dfrac{\text{1}}{\text{2}}}}\\&=\dfrac{\text{3}\sqrt{\text{x}}}{\text{2}\sqrt{\text{1-}{{\text{x}}^{\text{3}}}}} \\ \end{align} $ 

Hence,

$\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{3}}{\text{2}}\sqrt{\dfrac{\text{x}}{\text{1-}{{\text{x}}^{\text{3}}}}} $ .


5.$\text{y=}\dfrac{\text{co}{{\text{s}}^{\text{-1}}}\dfrac{\text{x}}{\text{2}}}{\sqrt{\text{2x+7}}}\text{,}\,\,\,\,\,\text{-2}$<$x$<$2$ .

Ans: The given function is 

$\text{y=}\dfrac{\text{co}{{\text{s}}^{\text{-1}}}\dfrac{\text{x}}{\text{2}}}{\sqrt{\text{2x+7}}} $ .

Then, differentiating both sides with respect to  $ \text{x} $  using the quotient rule gives

$\begin{align}&\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\sqrt{\text{2x+7}}\dfrac{\text{d}}{\text{dx}}\left(\text{co}{{\text{s}}^{\text{-1}}}\dfrac{\text{x}}{\text{2}} \right)\text{-}\left( \text{co}{{\text{s}}^{\text{-1}}}\dfrac{\text{x}}{\text{2}} \right)\dfrac{\text{d}}{\text{dx}}\left(\sqrt{\text{2x+7}}\right)}{{{\left(\sqrt{\text{2x+7}}\right)}^{\text{2}}}}\\&=\dfrac{\sqrt{\text{2x+7}}\left[\dfrac{\text{-1}}{\sqrt{\text{1-}{{\left(\dfrac{\text{x}}{\text{2}}\right)}^{\text{2}}}}}\text{.}\dfrac{\text{d}}{\text{dx}}\left( \dfrac{\text{x}}{\text{2}} \right) \right]\text{-}\left( \text{co}{{\text{s}}^{\text{-1}}}\dfrac{\text{x}}{\text{2}}\right)\dfrac{\text{1}}{\text{2}\sqrt{\text{2x+7}}}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(2x+7)}}{\text{2x+7}}\\&=\dfrac{\sqrt{\text{2x+7}}\dfrac{\text{-1}}{\sqrt{\text{4-}{{\text{x}}^{\text{2}}}}}\text{-}\left( \text{co}{{\text{s}}^{\text{-1}}}\dfrac{\text{x}}{\text{2}} \right)\dfrac{\text{2}}{\text{2}\sqrt{\text{2x+7}}}}{\text{2x+7}}\\&=\dfrac{\text{-}\sqrt{\text{2x+7}}}{\sqrt{\text{4-}{{\text{x}}^{\text{2}}}}\text{}\!\!\times\!\!\text{(2x+7)}}\text{-}\dfrac{\text{co}{{\text{s}}^{\text{-1}}}\dfrac{\text{x}}{\text{2}}}{\left( \sqrt{\text{2x+7}} \right)\left( \text{2x+7} \right)} \\ \end{align} $ 

Hence,  $ \dfrac{\text{dy}}{\text{dx}}\text{=-}\left[ \dfrac{\text{1}}{\sqrt{4-x^2}\sqrt{2x+7}}\text{+}\dfrac{\text{co}{{\text{s}}^{\text{-1}}}\dfrac{\text{x}}{\text{2}}}{{{\left( \text{2x+7} \right)}^{\dfrac{\text{3}}{\text{2}}}}} \right] $ .


6.  $ \mathbf{y=co}{{\mathbf{t}}^{\mathbf{-1}}}\left[ \dfrac{\sqrt{\mathbf{1+sinx}}\mathbf{+}\sqrt{\mathbf{1-sinx}}}{\sqrt{\mathbf{1+sinx}}\mathbf{-}\sqrt{\mathbf{1-sinx}}} \right]\mathbf{,}\text{  }\mathbf{0}<\mathbf{x}<\mathbf{2} $ .

Ans: The given function is  $ \text{y=co}{{\text{t}}^{\text{-1}}}\left[ \dfrac{\sqrt{\text{1+sinx}}\text{+}\sqrt{\text{1-sinx}}}{\sqrt{\text{1+sinx}}\text{-}\sqrt{\text{1-sinx}}} \right] $                 ……. (1)

Now,

$\dfrac{\sqrt{\text{1+sinx}}\text{+}\sqrt{\text{1-sinx}}}{\sqrt{\text{1+sinx}}\text{-}\sqrt{\text{1-sinx}}} $

$\begin{align}&=\dfrac{\left(\sqrt{\text{1+sinx}}\text{+}\sqrt{\text{1-sinx}}\right)}{\left(\sqrt{\text{1+sinx}}\text{-}\sqrt{\text{1-sinx}}\right)\sqrt{\text{1+sinx}}\text{+}\sqrt{\text{1-sinx}}}\\&=\dfrac{\text{(1+sinx)+(1-sinx)+2}\sqrt{\text{(1+sinx)-(1-sinx)}}}{\text{(1+sinx)-(1-sinx)}}\\&=\dfrac{\text{2+2}\sqrt{\text{1-si}{{\text{n}}^{\text{2}}}\text{x}}}{\text{2sinx}}\\&=\dfrac{\text{1+cosx}}{\text{sinx}}\\&=\dfrac{\text{2co}{{\text{s}}^{\text{2}}}\dfrac{\text{x}}{\text{2}}}{\text{2sinx}\dfrac{\text{x}}{\text{2}}\text{cos}\dfrac{\text{x}}{\text{2}}}\\\end{align}$ 

Therefore,

$\dfrac{\sqrt{\text{1+sinx}}\text{+}\sqrt{\text{1-sinx}}}{\sqrt{\text{1+sinx}}\text{-}\sqrt{\text{1-sinx}}}\text{=cot}\dfrac{\text{x}}{\text{2}} $ .                                                 …… (2)

So, from the equations (1) and (2) we obtain,

$\begin{align}&\text{y=co}{{\text{t}}^{\text{-1}}}\left(\text{co}{{\text{t}}^{\dfrac{\text{x}}{\text{2}}}}\right)\\&\Rightarrow\text{y=}\dfrac{\text{x}}{\text{2}}\\\end{align} $ 

Now, differentiating both sides with respect to $ \text{x} $ gives $\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{1}}{\text{2}}\dfrac{\text{d}}{\text{dx}}\text{(x)} $ Hence, $ \dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{1}}{\text{2}} $ .


7.  $ \text{y=(logx}{{\text{)}}^{\text{logx}}}\text{,}\,\,\,\text{x}>1 $ .

Ans: The given function is  $ \text{y=(logx}{{\text{)}}^{\text{logx}}} $ .

First take logarithm both sides of the function.

 $ \text{logy=logx }\!\!\times\!\!\text{ log(logx)} $ .

Now, differentiating both sides with respect to  $ \text{x} $  gives

$\begin{align}&\dfrac{\text{1}}{\text{y}}\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\left[\text{logx}\!\!\times\!\!\text{log(logx)}\right]\\&\Rightarrow \dfrac{\text{1}}{\text{y}}\dfrac{\text{dy}}{\text{dx}}\text{=log(logx) }\!\!\times\!\!\text{}\dfrac{\text{d}}{\text{dx}}\text{(logx)+}\dfrac{\text{d}}{\text{dx}}\left[\text{log(logx)}\right]\\&\Rightarrow\dfrac{\text{dy}}{\text{dx}}\text{=y}\left[\text{log(logx)}\!\!\times\!\!\text{}\dfrac{\text{1}}{\text{x}}\text{+logx }\!\!\times\!\!\text{ }\dfrac{\text{1}}{\text{logx}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(logx)}\right]\\&\Rightarrow\dfrac{\text{dy}}{\text{dx}}\text{=y}\left[\dfrac{\text{1}}{\text{x}}\text{log(logx)+}\dfrac{\text{1}}{\text{x}} \right] \\ \end{align} $ 

Hence,  $ \dfrac{\text{dy}}{\text{dx}}\text{=(logx}{{\text{)}}^{\text{logx}}}\left[ \dfrac{\text{1}}{\text{x}}\text{+}\dfrac{\text{log(logx)}}{\text{x}} \right] $ .


8.  $ \mathbf{y=cos(acosx+bsinx)} $ , for some constants  $ \text{a} $  and  $ \mathbf{b} $ .

Ans: The given function is  $ \text{y=cos(acosx+bsinx)} $ .

Now, differentiating both sides with respect to  $ \text{x} $  by using the chain rule of derivatives gives

$\begin{align}&\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{cos(acosx+bsinx)}\\&\Rightarrow\dfrac{\text{dy}}{\text{dx}}\text{=-sin(acosx+bsinx)}\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{(acosx+bsinx)} \\ & =\text{-sin(acosx+bsinx)}\!\!\times\!\!\text{}\left[\text{a(-sinx)+bcosx}\right] \\ \end{align} $ 

Hence,  $ \dfrac{\text{dy}}{\text{dx}}=\text{(asinx-bcosx) }\!\!\times\!\!\text{ sin(acosx+bsinx)} $ .


9.$\mathbf{y=(sinx-cosx}{{\mathbf{)}}^{\mathbf{(sinx-cosx)}}}\mathbf{,}\,\,\,\dfrac{\mathbf{\pi }}{\mathbf{4}}<\mathbf{x}<\dfrac{\mathbf{3\pi }}{\mathbf{4}} $ .

Ans: The given function is  $ \text{y=(sinx-cosx}{{\text{)}}^{\text{(sinx-cosx)}}}$.

First take logarithm both sides of the function.

 $ \begin{align}& \text{logy=log}\left[ {{\text{(sinx-cosx)}}^{\text{(sinx-cosx)}}} \right] \\ &\Rightarrow\text{logy=(sinx-cosx)}\!\!\times\!\!\text{log(sinx-cosx)} \\ \end{align} $ 

Now, differentiating both sides with respect to  $ \text{x} $  gives

$\begin{align}&\dfrac{\text{1}}{\text{y}}\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\left[ \text{(sinx-cosx) }\!\!\times\!\!\text{ log(sinx-cosx)} \right]\\&\Rightarrow\dfrac{\text{1}}{\text{y}}\dfrac{\text{dy}}{\text{dx}}\text{=log(sinx-cosx)}\!\!\times\!\!\text{}\dfrac{\text{d}}{\text{dx}}\text{(sinx-cosx)+(sinx-cosx) }\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\text{log(sinx-cosx)} \\&\Rightarrow\dfrac{\text{1}}{\text{y}}\dfrac{\text{dy}}{\text{dx}}\text{=log(sinx-cosx)}\!\!\times\!\!\text{(cosx+sinx)+(sinx-cosx)}\!\!\times\!\!\text{}\dfrac{\text{1}}{\text{(sinx-cosx)}}\text{}\!\!\times\!\!\text{}\dfrac{\text{d}}{\text{dx}}\text{(sinx-cosx)}\\&\Rightarrow\dfrac{\text{dy}}{\text{dx}}\text{=(sinx-cosx}{{\text{)}}^{\text{(sinx-cosx)}}}\left[\left(\text{cosx+sinx}\right)\text{}\!\!\times\!\!\text{ log(sinx-cosx)+(cosx+sinx)} \right] \\ \end{align} $  

Hence, the required derivative is 

$\dfrac{\text{dy}}{\text{dx}}\text{=(sinx-cosx}{{\text{)}}^{\text{(sinx-cosx)}}}\text{(cosx+sinx)}\left[ \text{1+log(sinx-cosx)} \right] $ .


10.$\mathbf{y=}{{\mathbf{x}}^{\mathbf{x}}}\mathbf{+}{{\mathbf{x}}^{\mathbf{a}}}\mathbf{+}{{\mathbf{a}}^{\mathbf{x}}}\mathbf{+}{{\mathbf{a}}^{\mathbf{a}}} $ , for some fixed  $ \text{a}>0 $  and  $ \text{x}>0 $ .

Ans: The given function is

$\text{y=}{{\text{x}}^{\text{x}}}\text{+}{{\text{x}}^{\text{a}}}\text{+}{{\text{a}}^{\text{x}}}\text{+}{{\text{a}}^{\text{a}}} $ .

Now, assume that  $ {{\text{x}}^{\text{x}}}\text{=u} $ ,

${{\text{x}}^{\text{a}}}\text{=v} $ ,  $ {{\text{a}}^{\text{x}}}\text{=w} $  and  $ {{\text{a}}^{\text{a}}}\text{=s} $ 

Therefore, we have  $ \text{y=u+v+w+s} $ .

So, differentiating both sides with respect to  $ \text{x} $  gives

$\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{du}}{\text{dx}}\text{+}\dfrac{\text{dv}}{\text{dx}}\text{+}\dfrac{\text{dw}}{\text{dx}}\text{+}\dfrac{\text{ds}}{\text{dx}} $                                            …… (1)

Also,  $ \text{u=}{{\text{x}}^{\text{x}}} $ 

 $ \begin{align}  & \Rightarrow \text{logu=log}{{\text{x}}^{\text{x}}} \\  & \Rightarrow \text{logu=xlogx} \\ \end{align} $ 

Then, differentiating both sides with respect to  $ \text{x} $  gives 

$\begin{align}&\dfrac{\text{1}}{\text{u}}\dfrac{\text{du}}{\text{dx}}\text{=logx}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(x)+x}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(logx)}\\&\Rightarrow\dfrac{\text{du}}{\text{dx}}\text{=u}\left[\text{logx}\text{.1+x}\text{.}\dfrac{\text{1}}{\text{x}} \right] \\ \end{align} $ 

Thus,  $ \dfrac{\text{du}}{\text{dx}}\text{=}{{\text{x}}^{\text{x}}}\text{ }\!\![\!\!\text{ logx+1 }\!\!]\!\!\text{ =}{{\text{x}}^{\text{x}}}\text{(1+logx)} $                           ……. (2)

Again,  $ \text{v=}{{\text{x}}^{\text{a}}} $ 

Then, differentiating both sides with respect to  $ \text{x} $  gives

$\dfrac{\text{du}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{x}}^{\text{a}}}\text{)} $ 

 $ \Rightarrow \dfrac{\text{dv}}{\text{dx}}\text{=a}{{\text{x}}^{\text{a-1}}} $                                                                  …… (3)

Also,  $ \text{w=}{{\text{a}}^{\text{x}}} $

$\begin{align}&\Rightarrow\text{logw=log}{{\text{a}}^{\text{x}}}\\&\Rightarrow \text{logw=xloga} \\ \end{align} $ 

So, differentiating both sides with respect to  $ \text{x} $  gives

$\begin{align}&\dfrac{\text{1}}{\text{w}}\text{.}\dfrac{\text{dw}}{\text{dx}}\text{=loga}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(x)}\\&\Rightarrow\dfrac{\text{dw}}{\text{dx}}\text{=wloga} \\ \end{align} $ 

$\Rightarrow\dfrac{\text{dw}}{\text{dx}}\text{=}{{\text{a}}^{\text{x}}}\text{loga} $                                                          ……… (4)

and

 $ \text{s=}{{\text{a}}^{\text{a}}} $ 

Then differentiating both sides with respect to  $ \text{x} $  gives

 $ \dfrac{\text{ds}}{\text{dx}}\text{=0} $ ,                                                                         ……(5)

as  $ \text{a} $  is constant, and so  $ {{\text{a}}^{\text{a}}} $  is also a constant.

Now, from the equations (1), (2), (3), (4), and (5) we have

$\dfrac{\text{dy}}{\text{dx}}\text{=}{{\text{x}}^{\text{x}}}\text{(1+logx)+a}{{\text{x}}^{\text{a-1}}}\text{+}{{\text{a}}^{\text{x}}}\text{loga+0} $ 

Hence,$\dfrac{\text{dy}}{\text{dx}}\text{=}{{\text{x}}^{\text{x}}}\text{(1+logx)+a}{{\text{x}}^{\text{a-1}}}\text{+}{{\text{a}}^{\text{x}}}\text{loga} $ .


11.$\mathbf{y=}{{\mathbf{x}}^{{{\mathbf{x}}^{\mathbf{2}}}\mathbf{-3}}}\mathbf{+(x-3}{{\mathbf{)}}^{{{\mathbf{x}}^{\mathbf{2}}}}} $ , for  $ \text{x}>3 $ .

Ans: The given function is

$\text{y=}{{\text{x}}^{{{\text{x}}^{\text{2}}}\text{-3}}}\text{+(x-3}{{\text{)}}^{{{\text{x}}^{\text{2}}}}} $ . 

Now suppose that  $ \text{u=}{{\text{x}}^{{{\text{x}}^{\text{2}}}\text{-3}}} $  and  $ \text{v=(x-3}{{\text{)}}^{{{\text{x}}^{\text{2}}}}} $ 

Therefore,  $ \text{y=u+v} $ .

Now, differentiating both sides with respect to  $ \text{x} $  gives

$\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{du}}{\text{dx}}\text{+}\dfrac{\text{dv}}{\text{dx}} $                                                ……. (1)

Also,  $ \text{u=}{{\text{x}}^{{{\text{x}}^{\text{2}}}\text{-3}}} $ .

Take logarithm both sides of the equation.

$\begin{align}&\Rightarrow\text{logu=log(}{{\text{x}}^{{{\text{x}}^{\text{2}}}\text{-3}}}\text{)} \\&\Rightarrow\text{logu=(}{{\text{x}}^{\text{2}}}\text{-3)logx} \\ \end{align} $ 

Differentiating both sides with respect to  $ \text{x} $  gives

$\begin{align}&\dfrac{\text{1}}{\text{u}}\dfrac{\text{du}}{\text{dx}}\text{=logx}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{x}}^{\text{2}}}\text{-3)+(}{{\text{x}}^{\text{2}}}\text{-3)}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(logx)}\\&\Rightarrow\dfrac{\text{1}}{\text{u}}\dfrac{\text{du}}{\text{dx}}\text{=logx}\text{.2x+(}{{\text{x}}^{\text{2}}}\text{-3)}\text{.}\dfrac{\text{1}}{\text{x}} \\ \end{align} $ 

Hence,  $\dfrac{\text{du}}{\text{dx}}\text{=}{{\text{x}}^{{{\text{x}}^{\text{2}}}\text{-3}}}\text{.}\left[ \dfrac{{{\text{x}}^{\text{2}}}\text{-3}}{\text{x}}\text{+2 }\!\!\times\!\!\text{ logx} \right] $ .                            …… (2)

Again,  $ \text{v=(x-3}{{\text{)}}^{{{\text{x}}^{\text{2}}}}} $ .

Take logarithm both sides of the equation.

$\begin{align}&\Rightarrow\text{logv=log(x-3}{{\text{)}}^{{{\text{x}}^{\text{2}}}}}\\&\Rightarrow \text{logv=}{{\text{x}}^{\text{2}}}\text{log(x-3)} \\ \end{align} $ 

Now, differentiating both sides with respect to  $ \text{x} $  gives

$\begin{align}&\dfrac{\text{1}}{\text{u}}\text{.}\dfrac{\text{dv}}{\text{dx}}\text{=log(x-3)}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{x}}^{\text{2}}}\text{)+}{{\text{x}}^{\text{2}}}\text{.}\dfrac{\text{d}}{\text{dx}}\text{}\!\![\!\!\text{log(x-3)}\!\!]\!\!\text{}\\&\Rightarrow\dfrac{\text{1}}{\text{u}}\text{.}\dfrac{\text{dv}}{\text{dx}}\text{=log(x-3)}\text{.2x+}{{\text{x}}^{\text{2}}}\text{.}\dfrac{\text{1}}{\text{x-3}}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(x-3)}\\&\Rightarrow\dfrac{\text{dv}}{\text{dx}}\text{=v}\text{.}\left[\text{2xlog(x-3)+}\dfrac{{{\text{x}}^{\text{2}}}}{\text{x-3}}\text{.1} \right] \\ \end{align} $ 

Hence,$\dfrac{\text{dv}}{\text{dx}}\text{=(x-3}{{\text{)}}^{{{\text{x}}^{\text{2}}}}}\left[ \dfrac{{{\text{x}}^{\text{2}}}}{\text{x-3}}\text{+2xlog(x-3)} \right] $                                    …… (3)

Thus, from the equations (1), (2) and (3) we obtain

$\dfrac{\text{dy}}{\text{dx}}\text{=}{{\text{x}}^{{{\text{x}}^{\text{2}}}\text{-3}}}\left[ \dfrac{{{\text{x}}^{\text{2}}}\text{-3}}{\text{x}}\text{+2xlogx} \right]\text{+(x-3}{{\text{)}}^{{{\text{x}}^{\text{2}}}}}\left[ \dfrac{{{\text{x}}^{\text{2}}}}{\text{x-3}}\text{+2xlog(x-3)} \right] $ .


12. Find  $ \dfrac{\mathbf{dy}}{\mathbf{dx}} $  if  $\mathbf{y=12(1-cost),x=10(t-sint),}\,\,-\dfrac{\mathbf{\pi }}{\mathbf{2}}<\mathbf{t}<\dfrac{\mathbf{\pi }}{\mathbf{2}} $ .

Ans: The given equations are  $ \text{y=12(1-cost),} $                                 …… (1)

and  $ \text{x=10(t-sint)} $                                                                    …… (2)

Then differentiating the equations (1) and (2) with respect to  $ \text{x} $  gives

$\begin{align}&\dfrac{\text{dx}}{\text{dt}}\text{=}\dfrac{\text{d}}{\text{dt}}\left[\text{10(t-sint)}\right]\text{=10}\!\!\times\!\!\text{}\dfrac{\text{d}}{\text{dt}}\text{(t-sint)=10(1-cost)}\\&\dfrac{\text{dy}}{\text{dt}}\text{=}\dfrac{\text{d}}{\text{dt}}\left[\text{12(1-cost)}\right]\text{=12}\!\!\times\!\!\text{}\dfrac{\text{d}}{\text{dt}}\text{(1-cost)=12 }\!\!\times\!\!\text{ }\left[ \text{0-(-sint)} \right]\text{=12sint} \\ \end{align} $ 

Therefore, by dividing  $ \dfrac{\text{dy}}{\text{dt}} $  by  $ \dfrac{\text{dx}}{\text{dt}} $  we have,

$\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\left(\dfrac{\text{dy}}{\text{dt}}\right)}{\left(\dfrac{\text{dx}}{\text{dt}}\right)}\text{=}\dfrac{\text{12sint}}{\text{10(1-cost)}}\text{=}\dfrac{\text{12 }\!\!\times\!\!\text{ 2sin}\dfrac{\text{t}}{\text{2}}\text{ }\!\!\times\!\!\text{ cos}\dfrac{\text{t}}{\text{2}}}{\text{10}\!\!\times\!\!\text{2si}{{\text{n}}^{\text{2}}}\dfrac{\text{t}}{\text{2}}} $

Hence,  $\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{6}}{\text{5}}\text{cot}\dfrac{\text{t}}{\text{2}} $ .


13. Find  $ \dfrac{\text{dy}}{\text{dx}} $ , if  $ \text{y=si}{{\text{n}}^{\text{-1}}}\text{x+si}{{\text{n}}^{\text{-1}}}\sqrt{\text{1-}{{\text{x}}^{\text{2}}}}\text{,}\,\,\,\text{0}<\text{x}<\text{1} $ .

Ans: The given equation is  $\text{y=si}{{\text{n}}^{\text{-1}}}\text{x+si}{{\text{n}}^{\text{-1}}}\sqrt{\text{1-}{{\text{x}}^{\text{2}}}} $ .

Differentiating both sides of the equation with respect to  $ \text{x} $  gives

$\begin{align}&\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\left[\text{si}{{\text{n}}^{\text{-1}}}\text{x+si}{{\text{n}}^{\text{-1}}}\sqrt{\text{1-}{{\text{x}}^{\text{2}}}}\right]\\&\Rightarrow\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{(si}{{\text{n}}^{\text{-1}}}\text{x)+}\dfrac{\text{d}}{\text{dx}}\left(\text{si}{{\text{n}}^{\text{-1}}}\sqrt{\text{1-}{{\text{x}}^{\text{2}}}} \right) \\ \end{align} $ 

$\begin{align}&\Rightarrow\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{1}}{\sqrt{\text{1-}{{\text{x}}^{\text{2}}}}}\text{+}\dfrac{\text{1}}{\sqrt{\text{1-}{{\left( \sqrt{\text{1-}{{\text{x}}^{\text{2}}}} \right)}^{2}}}}\text{}\!\!\times\!\!\text{ }\dfrac{\text{d}}{\text{dx}}\left( \sqrt{\text{1-}{{\text{x}}^{\text{2}}}} \right) \\  &\Rightarrow\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{1}}{\sqrt{\text{1-}{{\text{x}}^{\text{2}}}}}\text{-}\dfrac{1}{\sqrt{1-1+{{\text{x}}^{2}}}}\dfrac{\text{1}}{\text{2 }\!\!\times\!\!\text{ }\sqrt{\text{1-}{{\text{x}}^{\text{2}}}}}\times 2\text{x}\\&\Rightarrow\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{1}}{\sqrt{\text{1-}{{\text{x}}^{\text{2}}}}}\text{-}\dfrac{\text{1}}{\sqrt{\text{1-}{{\text{x}}^{\text{2}}}}} \\ \end{align} $ 

Hence,  $ \dfrac{\text{dy}}{\text{dx}}\text{=0} $ .


14. If  $\mathbf{x}\sqrt{\mathbf{1+y}}\mathbf{+y}\sqrt{\mathbf{1+x}}\mathbf{=0} $, for  $-1$<$x$<$1$ ,  prove that  

$\dfrac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=-}\dfrac{\mathbf{1}}{{{\mathbf{(1+x)}}^{\mathbf{2}}}} $ .

Ans: The given equation is

 $ \begin{align}& \text{x}\sqrt{\text{1+y}}\text{+y}\sqrt{\text{1+x}}\text{=0} \\  & \Rightarrow \text{x}\sqrt{\text{1+y}}\text{=y}\sqrt{\text{1+x}} \\ \end{align} $ 

Now, squaring both sides of the equation, gives

 $ \begin{align}& {{\text{x}}^{\text{2}}}\text{(1+y)=}{{\text{y}}^{\text{2}}}\text{(1+x)} \\  & \Rightarrow {{\text{x}}^{\text{2}}}\text{+}{{\text{x}}^{\text{2}}}\text{y=}{{\text{y}}^{\text{2}}}\text{+x}{{\text{y}}^{\text{2}}} \\  & \Rightarrow {{\text{x}}^{\text{2}}}\text{-}{{\text{y}}^{\text{2}}}\text{=x}{{\text{y}}^{\text{2}}}\text{-}{{\text{x}}^{\text{2}}}\text{y} \\  & \Rightarrow {{\text{x}}^{\text{2}}}\text{-}{{\text{y}}^{\text{2}}}\text{=xy(y-x)} \\  & \Rightarrow \text{(x+y)(x-y)=xy(y-x)} \\  & \therefore \text{x+y=-xy} \\  & \Rightarrow \text{(1+x)y=-x} \\  & \Rightarrow \text{y=}\dfrac{\text{-x}}{\text{(1+x)}} \\ \end{align} $ 

Now, differentiating both sides of the equation with respect to  $ \text{x} $  gives

$\dfrac{\text{dy}}{\text{dx}}\text{=-}\dfrac{\text{(1+x)}\dfrac{\text{d}}{\text{dx}}\text{(x)-x}\dfrac{\text{d}}{\text{dx}}\text{(1+x)}}{{{\text{(1+x)}}^{\text{2}}}}\text{=-}\dfrac{\text{(1+x)-x}}{{{\text{(1+x)}}^{\text{2}}}} $ 

Hence,  $ \dfrac{\text{dy}}{\text{dx}}\text{=-}\dfrac{\text{1}}{{{\text{(1+x)}}^{\text{2}}}} $.


15. If  $ {{\mathbf{(x-a)}}^{\mathbf{2}}}\mathbf{+(y-b}{{\mathbf{)}}^{\mathbf{2}}}\mathbf{=}{{\mathbf{c}}^{\mathbf{2}}} $ , for some constant  $ \text{c}>0 $ , prove that  $ \dfrac{{{\left[ \mathbf{1+}{{\left( \dfrac{\mathbf{dy}}{\mathbf{dx}} \right)}^{\mathbf{2}}} \right]}^{\dfrac{\mathbf{3}}{\mathbf{2}}}}}{\dfrac{{{\mathbf{d}}^{\mathbf{2}}}\mathbf{y}}{\mathbf{d}{{\mathbf{x}}^{\mathbf{2}}}}} $  is a constant independent of  $ \text{a} $  and  $ \text{b} $ .

Ans: The given equation is  ${{\text{(x-a)}}^{\text{2}}}\text{+(y-b}{{\text{)}}^{\text{2}}}\text{=}{{\text{c}}^{\text{2}}} $  .

Differentiating both sides of the equation with respect to  $ \text{x} $  gives  $ \begin{align}& \dfrac{\text{d}}{\text{dx}}\text{= }\!\![\!\!\text{ (x-a}{{\text{)}}^{\text{2}}}\text{ }\!\!]\!\!\text{ +}\dfrac{\text{d}}{\text{dx}}\text{ }\!\![\!\!\text{ (y-b}{{\text{)}}^{\text{2}}}\text{ }\!\!]\!\!\text{ =}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{c}}^{\text{2}}}\text{)} \\  & \Rightarrow \text{2(x-a)}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(x-a)+2(y-b)}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(y-b)=0} \\ & \Rightarrow \text{2(x-a)}\text{.1+2(y-b)}\text{.}\dfrac{\text{dy}}{\text{dx}}\text{=0} \\ \end{align} $ 

Hence,  $ \dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{-(x-a)}}{\text{y-b}} $                                                  ..…... (1)

Again, differentiating both sides of the equation with respect to  $ \text{x} $  gives

 $ \begin{align}& \dfrac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}\dfrac{\text{d}}{\text{dx}}\left[ \dfrac{\text{-(x-a)}}{\text{y-b}} \right] \\ & \text{=-}\dfrac{\left[ \text{(y-b)}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(x-a)-(x-a)}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(y-b)} \right]}{{{\text{(y-b)}}^{\text{2}}}} \\ & \text{=-}\left[ \dfrac{\text{(y-b)-(x-a)}\text{.}\dfrac{\text{dy}}{\text{dx}}}{{{\text{(y-b)}}^{\text{2}}}} \right] \\  & \text{=-}\left[ \dfrac{\text{(y-b)-(x-a)}\text{.}\left\{ \dfrac{\text{-(x-a)}}{\text{y-b}} \right\}}{{{\text{(y-b)}}^{\text{2}}}} \right] \\  & \text{=-}\left[ \dfrac{{{\text{(y-b)}}^{\text{2}}}\text{+(x+a}{{\text{)}}^{\text{2}}}}{{{\text{(y-b)}}^{\text{2}}}} \right] \\ \end{align} $ 

Therefore,

 $ {{\left[ \dfrac{\text{1+}{{\left( \dfrac{\text{dy}}{\text{dx}} \right)}^{\text{2}}}}{\dfrac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}} \right]}^{\dfrac{\text{3}}{\text{2}}}}\text{=}\dfrac{{{\left[ \left( \text{1+}\dfrac{{{\text{(x-a)}}^{\text{2}}}}{{{\text{(y-b)}}^{\text{2}}}} \right) \right]}^{\dfrac{\text{3}}{\text{2}}}}}{\text{-}\left[ \dfrac{{{\text{(y-a)}}^{\text{2}}}\text{+(x-a}{{\text{)}}^{\text{2}}}}{{{\text{(y-a)}}^{\text{3}}}} \right]}\text{=}\dfrac{{{\left[ \dfrac{{{\text{(y-b)}}^{\text{2}}}\text{+(x-a}{{\text{)}}^{\text{2}}}}{{{\text{(y-b)}}^{\text{2}}}} \right]}^{\dfrac{\text{3}}{\text{2}}}}}{\text{-}\left[ \dfrac{{{\text{(y-a)}}^{\text{2}}}\text{+(x-a}{{\text{)}}^{\text{2}}}}{{{\text{(y-a)}}^{\text{3}}}} \right]}=\dfrac{{{\left[ \dfrac{{{\text{c}}^{\text{2}}}}{{{\text{(y-b)}}^{\text{2}}}} \right]}^{\dfrac{\text{3}}{\text{2}}}}}{\text{-}\dfrac{{{\text{c}}^{\text{2}}}}{{{\text{(y-b)}}^{\text{3}}}}} $ 

$\Rightarrow{{\left[\dfrac{\text{1+}{{\left(\dfrac{\text{dy}}{\text{dx}}\right)}^{\text{2}}}}{\dfrac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}}\right]}^{\dfrac{\text{3}}{\text{2}}}}\text{=}\dfrac{\dfrac{{{\text{c}}^{\text{2}}}}{{{\text{(y-b)}}^{\text{3}}}}}{\dfrac{{{\text{c}}^{\text{2}}}}{{{\text{(y-b)}}^{\text{3}}}}}=\text{-c} $ , is a constant, and is independent of  $ \text{a} $  and  $ \text{b} $ .


16. If  $ \text{cosy=xcos(a+y)} $ , with  $\text{cosa}\ne\text{}\!\!\pm\!\!\text{ 1} $ , prove that  $\dfrac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\dfrac{\mathbf{co}{{\mathbf{s}}^{\mathbf{2}}}\mathbf{(a+y)}}{\mathbf{sina}} $ .

Ans: The given equation is  $ \text{cosy=xcos(a+y)} $ .

Then, differentiating both sides of the equation with respect to  $ \text{x} $  gives

$\begin{align}&\dfrac{\text{d}}{\text{dx}}\text{}\!\![\!\!\text{cosy}\!\!]\!\!\text{=}\dfrac{\text{d}}{\text{dx}}\text{}\!\![\!\!\text{xcos(a+y)}\!\!]\!\!\text{}\\&\Rightarrow\text{-siny}\dfrac{\text{dy}}{\text{dx}}\text{=cos(a+y)}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(x)+x}\text{.}\dfrac{\text{d}}{\text{dx}}\text{}\!\![\!\!\text{cos(a+y)}\!\!]\!\!\text{}\\&\Rightarrow\text{-siny}\dfrac{\text{dy}}{\text{dx}}\text{=cos(a+y)+x}\text{.}\!\![\!\!\text{-sin(a+y)}\!\!]\!\!\text{}\dfrac{\text{dy}}{\text{dx}} \\ \end{align} $ 

 $ \Rightarrow \text{ }\!\![\!\!\text{ xsin(a+y)-siny }\!\!]\!\!\text{ }\dfrac{\text{dy}}{\text{dx}}\text{=cos(a+y)} $                                  …….. (1)

Since  $\text{cosy=xcos(a+y)}\Rightarrow\text{x=}\dfrac{\text{cosy}}{\text{cos(a+y)}} $ , so from the equation (1) gives

$\begin{align}&\left[\dfrac{\text{cosy}}{\text{cos(a+y)}}\text{.sin(a+y)-siny}\right]\dfrac{\text{dy}}{\text{dx}}\text{=cos(a+y)}\\&\Rightarrow\text{}\!\![\!\!\text{cosy}\text{.sin(a+y)-siny}\text{.cos(a+y)}\!\!]\!\!\text{}\text{.}\dfrac{\text{dy}}{\text{dx}}\text{=co}{{\text{s}}^{\text{2}}}\text{(a+y)}\\&\Rightarrow\text{sin(a+y-y)}\dfrac{\text{dy}}{\text{dx}}\text{=co}{{\text{s}}^{\text{2}}}\text{(a+y)} \\ \end{align} $ 

Hence, it has been proved that  $\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{co}{{\text{s}}^{\text{2}}}\text{(a+y)}}{\text{sina}} $ .


17. If   $ \mathbf{x=a(cost+tsint)} $  and  $ \mathbf{y=a(sint-tcost)} $ , find  $ \dfrac{{{\mathbf{d}}^{\mathbf{2}}}\mathbf{y}}{\mathbf{d}{{\mathbf{x}}^{\mathbf{2}}}} $ .

Ans: The given equations are

  $ \text{x=a(cost+tsint)} $                                            …… (1)

and  $ \text{y=a(sint-tcost)} $                                        …… (2)

Then, differentiating both sides of the equation (1) with respect to  $ \text{x} $  gives

$\begin{align}&\dfrac{\text{dx}}{\text{dt}}\text{=a}\left[\text{-sint+sint}\text{.}\dfrac{\text{d}}{\text{dx}}\text{(t)+t}\text{.}\dfrac{\text{d}}{\text{dt}}\text{(sint)}\right]\\&\text{=a}\left[ \text{-sint+sint+cost} \right]\text{=atcost} \\ \end{align} $ 

Again, differentiating both sides of the equation (2) with respect to  $ \text{x} $  gives

$\begin{align}&\dfrac{\text{dy}}{\text{dt}}\text{=a}\text{.}\dfrac{\text{d}}{\text{dt}}\text{(sint-tcost)}\\&\text{a}\left[\text{cost-}\left\{\text{cost}\text{.}\dfrac{\text{d}}{\text{dt}}\text{(t)+t}\text{.}\dfrac{\text{d}}{\text{dt}}\text{(cost)}\right\}\right]\\&\text{a}\!\![\!\!\text{ cost- }\!\!\{\!\!\text{ cost-tsint }\!\!\}\!\!\text{  }\!\!]\!\!\text{=atsint}\\\end{align}$ 

Therefore,

$\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\left(\dfrac{\text{dy}}{\text{dt}}\right)}{\left( \dfrac{\text{dx}}{\text{dx}} \right)}\text{=}\dfrac{\text{atsint}}{\text{atcost}}\text{=tant}$ 

Now, differentiating both sides with respect to  $ \text{x} $  gives

$\dfrac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}\dfrac{\text{d}}{\text{dx}}\left(\dfrac{\text{dy}}{\text{dx}}\right)\text{=}\dfrac{\text{d}}{\text{dx}}\text{(tant)=se}{{\text{c}}^{\text{2}}}\text{t}\text{.}\dfrac{\text{dt}}{\text{dx}}\text{=se}{{\text{c}}^{\text{2}}}\text{t}\text{.}\dfrac{\text{1}}{\text{atcost}} $       

Hence,  $\dfrac{{{\mathbf{d}}^{\mathbf{2}}}\mathbf{y}}{\mathbf{d}{{\mathbf{x}}^{\mathbf{2}}}}\text{=}\dfrac{\text{se}{{\text{c}}^{\text{3}}}\text{(t)}}{\text{at}} $ .


18. If  $ \mathbf{f(x)=}{{\left| \mathbf{x} \right|}^{\mathbf{3}}} $ , show that  $ \mathbf{{{f}'}'}\left( \mathbf{x} \right) $  exists for all real  $ \text{x} $   and find it.

Ans:  Remember that, $\left|\text{x}\right|\text{=}\left\{\begin{align}&\text{x,if}\,\text{x}\ge\text{0}\\&\text{-x,if}\,\text{x0} \\ \end{align} \right\} $ 

Therefore, if  $ \text{x}\ge \text{0,} $  then  $ \text{ f(x)=}{{\left| \text{x} \right|}^{\text{3}}}\text{=}{{\text{x}}^{\text{3}}} $ .

Then,  $ \text{{f}'}\,\text{(x)=3}{{\text{x}}^{\text{2}}} $  .

Differentiating both sides with respect to  $ \text{x} $  gives

 $ \text{{{f}'}'}\,\text{(x)=6x} $ .

Now, if $\text{x0,}$then$\text{f(x)=}{{\left|\text{x}\right|}^{\text{3}}}\text{=(-}{{\text{x}}^{\text{3}}}\text{)=}{{\text{x}}^{\text{3}}} $ .

So,  $ \text{{f}'}\,\text{(x)=3}{{\text{x}}^{\text{2}}} $ .

Therefore, differentiating both sides with respect to  $ \text{x} $  gives

 $ \text{{{f}'}'}\,\text{(x)=6x} $ .

Hence, for  $ \text{f(x)=}{{\left| \text{x} \right|}^{\text{3}}}\text{,} $   $ \text{{{f}'}'}\left( \text{x} \right) $  exists for all real values of  $ \text{x} $  and is provided as

$\text{{{f}'}'(x)=}\left\{\begin{align}&\text{6x,if}\,\text{x}\ge\text{0}\\&\text{-6x,if}\,\text{x0} \\ \end{align} \right\} $


19. Using Mathematical Induction, prove that  $\dfrac{\mathbf{d}}{\mathbf{dx}}\mathbf{(}{{\mathbf{x}}^{\mathbf{n}}}\mathbf{)=n}{{\mathbf{x}}^{\mathbf{x-1}}} $  for all positive integers  $\text{n}$ .

Ans: Let  $\text{P(n).}\,\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{x}}^{\text{n}}}\text{)=n}{{\text{x}}^{\text{x-1}}} $  for all positive integers  $ \text{n} $ .

Now, when  $ \text{n=1} $ ,

 $ \text{P(1).}\dfrac{\text{d}}{\text{dx}}\text{(x)=1=1 }\!\!\times\!\!\text{ }{{\text{x}}^{\text{1-1}}} $ 

Therefore,  $ \text{P(n}) $  is true when  $ \text{n=1} $ .

Assume that the statement  $ \text{P(k)} $  is true for some positive integer  $ \text{k} $ . 

So,  $\text{P(k).}\,\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{x}}^{\text{k}}}\text{)=k}{{\text{x}}^{\text{k-1}}} $ , 

Now, if it can be shown that the statement  $ \text{p(k+1)} $  is also true, then our proof will be complete.

Let  $\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{x}}^{\text{k+1}}}\text{)=}\dfrac{\text{d}}{\text{dx}}\text{(x }\!\!\times\!\!\text{ }{{\text{x}}^{\text{k}}}\text{)} $ 

$\begin{align}&\text{=}{{\text{x}}^{\text{k}}}\text{}\!\!\times\!\!\text{}\dfrac{\text{d}}{\text{dx}}\text{(x)+x}\!\!\times\!\!\text{}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{x}}^{\text{k}}}\text{)}\\&\text{=}{{\text{x}}^{\text{k}}}\text{}\!\!\times\!\!\text{1+x}\!\!\times\!\!\text{k}\!\!\times\!\!\text{}{{\text{x}}^{\text{k-1}}}\\&\text{=}{{\text{x}}^{\text{k}}}\text{+k}{{\text{x}}^{\text{k}}} \\ & \text{=(k+1) }\!\!\times\!\!\text{ }{{\text{x}}^{\text{k}}} \\ & \text{=(k+1) }\!\!\times\!\!\text{ }{{\text{x}}^{\text{(k+1)-1}}} \\ \end{align} $ 

So,  $ \text{P(k+1)} $  is true if  $ \text{P(k)} $  is true. 

Hence, by the principal of mathematical induction, it has been proved that the statement  $ \text{P(n)} $  is true for every positive integer  $ \text{n} $ .

That is, $\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{x}}^{\text{n}}}\text{)=n}{{\text{x}}^{\text{x-1}}} $  for all positive integers  $ \text{n} $ .


20. Using the fact that  $ \mathbf{sin(A+B)=sinAcosB+cosAsinB} $  and the differentiation, obtain the sum formula for cosines.

Ans:  The given sum formula is  $ \text{sin(A+B)=sinAcosB+cosAsinB} $ .

Now, differentiating both sides with respect to  $ \text{x} $  gives

$\dfrac{\text{d}}{\text{dx}}\left[\text{sin(A+B)}\right]\text{=}\dfrac{\text{d}}{\text{dx}}\text{(sinAcosB)+}\dfrac{\text{d}}{\text{dx}}\text{(cosAsinB)} $ 

$\begin{align}&\Rightarrow\text{cos(A+B)}\!\!\times\!\!\text{}\dfrac{\text{d}}{\text{dx}}\text{(A+B)=cosB}\!\!\times\!\!\text{}\dfrac{\text{d}}{\text{dx}}\text{(sinA)+sinA}\!\!\times\!\!\text{}\dfrac{\text{d}}{\text{dx}}\text{(cosB)+sinB}\!\!\times\!\!\text{}\dfrac{\text{d}}{\text{dx}}\text{(cosA)}\\&\text{+cosA}\!\!\times\!\!\text{}\dfrac{\text{d}}{\text{dx}}\text{(sinB)}\\&\Rightarrow\text{cos(A+B)}\!\!\times\!\!\text{}\dfrac{\text{d}}{\text{dx}}\text{(A+B)=cosB}\!\!\times\!\!\text{cosA}\dfrac{\text{d}}{\text{dx}}\text{+sinA(-sinB)}\dfrac{\text{dB}}{\text{dx}}\text{+sinB(-sinA)}\!\!\times\!\!\text{}\dfrac{\text{dA}}{\text{dx}}\\&\text{+cosAcosB}\dfrac{dB}{dx}\\&\Rightarrow\text{cos(A+B)}\left[\dfrac{\text{dA}}{\text{dx}}\text{+}\dfrac{\text{dB}}{\text{dx}}\right]\text{=(cosAcosB-sinAsinB)}\!\!\times\!\!\text{}\left[\dfrac{\text{dA}}{\text{dx}}\text{+}\dfrac{\text{dB}}{\text{dx}}\right]\\\end{align} $ 

Hence the required sum formula for cosines is  $ \text{cos(A+B)=cosAcosB-sinAsinB} $ .


21. Does there exist a function which is continuous everywhere but not differentiable at exactly two points?

Ans: Let take the function $\text{f}\left(\text{x}\right)=|\text{x}|+|\text{x}-1|$ 

Observe that, the function  $ \text{f} $  is continuous everywhere, but not differentiable at  $ \text{x}=0 $  and  $ \text{x}=1 $ .


22.If $\mathbf{y=}\left|\begin{matrix}\mathbf{f}\left(\mathbf{x}\right)&\mathbf{g}\left(\mathbf{x} \right) & \mathbf{h}\left( \mathbf{x} \right)  \\\mathbf{l} & \mathbf{m} & \mathbf{n}  \\\mathbf{a} & \mathbf{b} & \mathbf{c}  \\\end{matrix} \right| $ , prove that$\dfrac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\left|\begin{matrix}\mathbf{{f}'}\left(\mathbf{x} \right)&\mathbf{{g}'}\left(\mathbf{x}\right)&\mathbf{{h}'}\left(\mathbf{x} \right)  \\  \mathbf{l} & \mathbf{m} & \mathbf{n}  \\  \mathbf{a} & \mathbf{b} & \mathbf{c}  \\\end{matrix} \right|$  .

Ans:  The given function is  $ \text{y=}\left| \begin{align}  & \text{f(x)}\,\,\,\text{g(x)}\,\,\,\text{h(x)} \\  & \text{   l}\ \ \ \ \ \ \text{m    n} \\  & \text{   a       b    c  } \\ \end{align} \right| $ 

Evaluate the determinant.

 $ \text{y=(mc-nb)f(x)-(lc-na)g(x)+(lb-ma)h(x)} $ .

Now, differentiating both sides with respect to  $ \text{x} $  gives

 $ \begin{align}& \dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{d}}{\text{dx}}\text{ }\!\![\!\!\text{ (mc-nb)f(x) }\!\!]\!\!\text{ -}\dfrac{\text{d}}{\text{dx}}\text{ }\!\![\!\!\text{ (lc-na)g(x) }\!\!]\!\!\text{ +}\dfrac{\text{d}}{\text{dx}}\text{ }\!\![\!\!\text{ (lb-ma)h(x) }\!\!]\!\!\text{ } \\ & \text{=(mc-nb)f(x)-(lc-na)g(x)+(lb-ma)h(x)} \\  & =\left| \begin{matrix}\text{f}'\left(\text{x}\right)&\text{g}'\left(\text{x}\right)&\text{h}'\left(\text{x} \right)  \\\text{l} & \text{m} & \text{n}  \\\text{a} & \text{b} & \text{c}  \\\end{matrix} \right| \\ \end{align} $ 

Hence,  $\dfrac{\text{dy}}{\text{dx}}\text{=}\left|\begin{matrix}\text{f}'\left(\text{x}\right)&\text{g }'\left( \text{x} \right) & \text{h }'\left( \text{x} \right)  \\\text{l} & \text{m} & \text{n}  \\\text{a} & \text{b} & \text{c}  \\\end{matrix} \right| $


23.If $\mathbf{y=}{{\mathbf{e}}^{\mathbf{aco}{{\mathbf{s}}^{\mathbf{-1}}}\mathbf{x}}}\text{, -1}\le \text{x}\le 1 $ , show that $\mathbf{(1-}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{)}\dfrac{{{\mathbf{d}}^{\mathbf{2}}}\mathbf{y}}{\mathbf{d}{{\mathbf{x}}^{\mathbf{2}}}}\mathbf{-x}\dfrac{\mathbf{dy}}{\mathbf{dx}}\mathbf{-}{{\mathbf{a}}^{\mathbf{2}}}\mathbf{y=0} $ .

Ans: The given equation is

$\text{y=}{{\text{e}}^{\text{aco}{{\text{s}}^{\text{-1}}}\text{x}}} $ .

Then take logarithm both sides of the equation.

$\begin{align}&\text{logy=aco}{{\text{s}}^{\text{-1}}}\text{xloge}\\&\Rightarrow\text{logy=aco}{{\text{s}}^{\text{-1}}}\text{x} \\ \end{align} $ 

Now, differentiating both sides with respect to  $ \text{x} $  gives

$\begin{align}&\dfrac{\text{1}}{\text{y}}\dfrac{\text{dy}}{\text{dx}}\text{=ax}\dfrac{\text{1}}{\sqrt{\text{1-}{{\text{x}}^{\text{2}}}}}\\&\dfrac{\text{dy}}{\text{dx}}\text{=}\dfrac{\text{-ax}}{\sqrt{\text{1-}{{\text{x}}^{\text{2}}}}} \\ \end{align} $ 

Therefore, squaring both the sides of the equation, gives

$\begin{align}&{{\left(\dfrac{\text{dy}}{\text{dx}}\right)}^{\text{2}}}\text{=}\dfrac{{{\text{a}}^{\text{2}}}{{\text{y}}^{\text{2}}}}{\text{1-}{{\text{x}}^{\text{2}}}}\\&\Rightarrow(\text{1-}{{\text{x}}^{\text{2}}}\text{)}{{\left(\dfrac{\text{dy}}{\text{dx}}\right)}^{\text{2}}}\text{=}{{\text{a}}^{\text{2}}}{{\text{y}}^{\text{2}}}\\&\Rightarrow\text{(1-}{{\text{x}}^{\text{2}}}\text{)}{{\left(\dfrac{\text{dy}}{\text{dx}}\right)}^{\text{2}}}\text{=}{{\text{a}}^{\text{2}}}{{\text{y}}^{\text{2}}} \\ \end{align} $ 

Again, differentiating both sides with respect to  $ \text{x} $  gives  $\begin{align}&{{\left(\dfrac{\text{dy}}{\text{dx}}\right)}^{\text{2}}}\dfrac{\text{d}}{\text{dx}}\text{(1-}{{\text{x}}^{\text{2}}}\text{)+(1-}{{\text{x}}^{\text{2}}}\text{)}\!\!\times\!\!\text{}\dfrac{\text{d}}{\text{dx}}\left[{{\left(\dfrac{\text{dy}}{\text{dx}}\right)}^{\text{2}}}\right]\text{=}{{\text{a}}^{\text{2}}}\dfrac{\text{d}}{\text{dx}}\text{(}{{\text{y}}^{\text{2}}}\text{)}\\&\Rightarrow{{\left(\dfrac{\text{dy}}{\text{dx}}\right)}^{\text{2}}}\text{(-2x)+(1-}{{\text{x}}^{\text{2}}}\text{)}\!\!\times\!\!\text{2}\dfrac{\text{dy}}{\text{dx}}\text{}\!\!\times\!\!\text{}\dfrac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}{{\text{a}}^{\text{2}}}\text{}\!\!\times\!\!\text{2y}\!\!\times\!\!\text{}\dfrac{\text{dy}}{\text{dx}}\\&\Rightarrow\text{x}\dfrac{\text{dy}}{\text{dx}}\text{+(1-}{{\text{x}}^{\text{2}}}\text{)}\dfrac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}{{\text{a}}^{\text{2}}}\text{}\!\!\times\!\!\text{ y} \\ \end{align} $

Hence, it is proved that  $\text{(1-}{{\text{x}}^{\text{2}}}\text{)}\dfrac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{-x}\dfrac{\text{dy}}{\text{dx}}\text{-}{{\text{a}}^{\text{2}}}\text{y=0} $ 


NCERT Solutions for Class 12 Maths Chapter 5 - Continuity and Differentiability

NCERT Solutions for Class 12 Maths – Free PDF Download 

Do you often feel anxious when you cannot understand the questions or find the solutions for Class 12 Maths Chapter 5 Continuity and Differentiability? Here you will find the appropriate answers to Continuity and Differentiability Class 12, which are curated by experienced professionals. These solutions are created and compiled into a file and are up on the website to download. So, get started with the chapter. 


NCERT Solutions for Class 12 Maths


Class 12 Maths Chapter 5 Exercises

Chapter 5 - Continuity and Differentiability Exercises in PDF Format

Exercise 5.1

34 Questions & Solutions (10 Short Answers, 24 Long Answers)

Exercise 5.2

10 Questions & Solutions (2 Short Answers, 8 Long Answers)

Exercise 5.3

15 Questions & Solutions (9 Short Answers, 6 Long Answers)

Exercise 5.4

10 Questions & Solutions (5 Short Answers, 5 Long Answers)

Exercise 5.5

18 Questions & Solutions (4 Short Answers, 14 Long Answers)

Exercise 5.6

11 Questions & Solutions (7 Short Answers, 4 Long Answers)

Exercise 5.7

17 Questions & Solutions (10 Short Answers, 7 Long Answers)

Exercise 5.8

6 Questions & Solutions (2 Short Answers, 4 Long Answers)


Chapter 5 – Continuity and Differentiability

NCERT Solutions for Class 12 Maths Chapter 5

5.1 Introduction

In NCERT Class 12 Maths Chapter 5, you will be introduced to Continuity and differentiability and its advanced theorem and logarithms. You will learn the advanced form of what you have studied in your previous class. You will explore important concepts of Continuity, differentiability and relations between them. Moreover, you will learn the differentiation of inverse trigonometric functions. Study a new concept called exponential and logarithmic functions and powerful techniques of logarithms by geometrically distinct conditioning through differential calculus. 


5.2 Continuity

In this Continuity and Differentiability Class 12 Solutions, you will learn the rigorous formulation of the intuitive concept of function that varies with no abrupt breaks or jumps. A function is a relationship in which every value of an independent variable is associated with a dependent variable. You will engage with the topic of the algebra of continuous functions. You will study some analogous algebra of uninterrupted functions since the continuity of a function at a point is reasonable to expect results similar to the case of limits. You will learn essential definitions and theorems and how to prove the theorems. 

Exercise 5.1 – 34 Questions (10 Short Questions and 24 Long Questions)


5.3 Differentiability

In this Continuity and Differentiability Class 12 NCERT Solutions PDF, you will recall some facts you have already learnt in the previous class, such as the derivative of a real function. You will learn about the list of derivatives of the specific standard function. You will discover three theorems and how to prove them. You will also learn that the result of every differentiable function is continuous. You will learn derivatives of composite functions, derivatives of implicit functions and inverse trigonometric functions. You will learn about the Chain Rule theorem and its functionality.  

Exercise 5.2 – 10 Questions (2 Short Questions 8 Long Questions)


5.3 Differentiability

Here in this Continuity and Differentiability Class 12 NCERT PDF, you will learn in-depth about derivatives of implicit function and derivatives of an inverse trigonometric function. This section provides several examples to teach how to apply theorems while solving problems. You will learn how to prove the theorems and the functionality of the theorems.  

Exercise 5.3 – 15 Questions (9 short Questions and 6 long Questions)


5.4 Exponential and Logarithmic Functions

In this Class 12 Continuity and Differentiability, you will explore aspects of different classes of functions like polynomial functions, rational functions and trigonometric functions. You can learn about a new class of related functions and logarithmic functions. This section will emphasize on statements made that are motivational and precise proofs of theorems. You will learn new definitions of exponential function and logarithms. You will find several examples in this section, which will help you grasp and understand exponential functions and logarithmic functions.

Exercise 5.4 – 10 Questions (5 Short Questions and 5 Long Questions)


5.5 Logarithmic Differentiation

In this NCERT Solutions for Class 12 Maths Continuity and Differentiability, you will learn to differentiate the specific particular class of functions given in specific form mentioned in the textbook. This section will provide more examples than theory since the topic demands a practical approach and less of a theoretical approach. This section can give you an example of bringing out a particular class of functions provided in certain forms. 

Exercise 5.5 – 18 Questions (4 Short Questions and 14 Long Questions)


5.6 Derivatives of Functions in Parametric Forms

In NCERT Solutions Class 12 Maths CH 5, you will learn the relation between two variables is neither explicit nor implicit. You will learn how to link a third variable with each of two variables separately by establishing a relation between the first two variables. You will learn why in such a situation; the third variable is called a parameter. You will be given several examples to understand the topic better and grasp the concepts mentioned in this section. 

Exercise 5.6 – 11 Questions (1 Short Question and 1 Long Questions)


5.7 Second Order Derivative

In this NCERT Solutions for Class 12 Maths Chapter Continuity and Differentiability, you will further learn how to work with the second-order derivative. Again, this section will provide you with more examples than theory as it requires more of a practical approach than a theoretical approach. These examples will help you grasp and hold on to the theorems and concepts you have learnt in previous sections.

Exercise 5.7 – 17 Questions (10 Short Questions and 7 Long Questions)


5.8 Mean Value Theorem (Not present in current syllabus)

In this CH 5 Maths Class 12, you will learn how to state two fundamental results in Calculus without proof and you will learn the theorems' geometric interpretation. You will learn the functionality of the Mean Value Theorem and its relation to Rolle's Theorem. You will be provided with several examples that will help you engage with the theorem and grasp the whole concept of it. This theorem is slightly tricky, but if you hold on to the concept correctly and understand the examples properly, you will find it easy to apply and find solutions to the problems.


Exercise 5.8 - 6 Questions with Solutions

Key Learnings from NCERT Solutions for Class 12 Maths Chapter 5 

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability explain all the necessary concepts of continuity and differentiability of functions. Some of the important topics explained in this chapter are continuity of functions, differentiability, methods to find derivatives, exponential and logarithmic functions, logarithmic differentiation, derivatives of functions in parametric forms, second-order derivatives, mean value theorem, and derivatives of the specific standard function. Each topic is elaborated efficiently in this chapter so that students can grasp them better.


Some of the key learnings from this chapter are listed below.


  • Continuous Function: A function is called a continuous function on a given interval if it is continuous at every point of the interval.

  • Arithmetic of Continuous Functions: Arithmetic operations (sum, difference, product, and quotient of continuous functions) performed on continuous functions is continuous. For example, if g and f are continuous functions, then

(f + g) (x) = f(x) + g(x) is continuous.

(f - g) (x) = f(x) - g(x) is continuous.

(f . g) (x) = f(x) . g (x) is continuous.

(f / g ) (x) = f(x ) / g(x) (wherever g(x) ≠ 0) is continuous.


  • Differentiable: A function is called to be differentiable if the derivative of the function exists at each point in its domain. If a function f(x) is differentiable at x = a, then f′(a) exists in the domain. Some of the derivative rules to find out the derivatives of a given function are listed below:

i) (f + g)′ = f′ + g′

ii) (f - g)′ = f′ - g′

iii) (fg)′ = f′g + fg′

iv) (f/g)′ = (f′g−fg′)/g2


  • Logarithmic Functions: Logarithmic functions are the inverses of exponential functions. The logarithmic function y = logax is said to be equivalent to the exponential equation x = ay.

  • Rolle’s Theorem: If f : [a, b] → R is continuous on [a, b] and differentiable on (a, b) such that f(a) = f(b), then there exists c in (a, b) such that f ′(c) = 0.

  • Mean Value Theorem: If f : [a, b] → R is continuous on [a, b] and differentiable on (a, b), then there exists c in (a, b) such that f'(c) = (f(b) – f(a))/(b–a). (Not present in current syllabus)

Importance of Continuity and Differentiability

Continuity and Differentiability are the basic concepts in the study of calculus. Building an understanding of these basic fundamental concepts will be helpful while studying a higher branch of Mathematics called “Analysis”. Continuity and differentiability are studied in advanced Mathematics and these have multiple applications in science, engineering, economics, etc. Therefore, it is crucial for the students to gain a deep understanding of this topic.


Key Features of NCERT Solutions for Class 12 Maths Chapter 5

Students need to go thoroughly through every section of this chapter to gain better knowledge and perspective of the subject. If you go through NCERT Solution of Continuity and Differentiability that are curated by the expert professionals in this subject, you will get an even better understanding of the chapter and learn the basic and advanced concepts and theorems. The key features are:

  • You can attain more knowledge about the ideas and theorems covered in NCERT Solutions for Class 12 Maths Chapter 5.

  • You can assess your knowledge gap with the help of Chapter 5 Maths Class 12 PDF.

  • You can expand your knowledge in main topics

  • You can understand your assignments based on concepts from continuity and Continuity and Differentiability Class 12 chapters.


Conclusion

NCERT Solutions for Class 12 Maths Chapter 5 - Continuity and Differentiability of Vedantu provide a comprehensive and detailed set of solutions to all the exercises and problems in the NCERT textbook. These solutions are designed to help students develop a deep understanding of the concepts and problems related to the continuity and differentiability of functions. By using these solutions, students can practice and improve their problem-solving skills and gain the confidence they need to excel in their exams. Moreover, the solutions provide step-by-step explanations and insights into the reasoning behind each problem, which can be helpful in developing a strong foundation in mathematics. Overall, the NCERT Solutions for Class 12 Maths Continuity and Differentiability of Vedantu are an invaluable resource for students looking to excel in this subject.

FAQs on NCERT Solutions for Class 12 Maths Chapter 5 - Continuity And Differentiability

1. How many topics are there in Chapter 3 of Class 12 Mathematics NCERT textbook?

Class 12 Mathematics Chapter 3 deals with Continuity and Differentiability. It is one of most of the important chapters in Class 12 Mathematics. Before studying the chapter, every student should go through all the topics and subtopics first. That is why we have given the topics and sub-topics of the chapter here - 5.1 Introduction, 5.2 Continuity, 5.2.1 Algebra of continuous functions, 5.3. Differentiability, 5.3.1 Derivatives of composite functions, 5.3.2 Derivatives of implicit functions, 5.3.3 Derivatives of inverse trigonometric functions, 5.4 Exponential and Logarithmic Functions, 5.5. Logarithmic Differentiation, 5.6 Derivatives of Functions in Parametric Forms, 5.7 Second Order Derivative, 5.8 Mean Value Theorem.

2. Give me an overview of all the topics of Class 12 Mathematics Chapter 3.

A. Before you dig into the Class 12 Maths Chapter 3, you must learn a little about all the topics and sub-topics. 


5.1 – Introduction - You will study about the limit, differential, and continuity, their properties, methods to find derivatives by limits.


5.2 – Continuity- You will study about the continuity of a function at a given point i.e. a function is continuous if it is continuous on the whole of its domain, properties of a continuous function, types of discontinuity, intermediate value theorem, Cauchy’s definition, Heine’s definition.


5.3 – Differentiability- You will study about the concept of differentiability – it is the change of quantity with respect to another quantity, standard differentiation, fundamental rules, types of derivatives, the relationship between continuity and differentiability.


5.4 – Exponential and Logarithmic Functions- You will study the concept of exponential and logarithmic functions i.e. differentiation of the exponential function and differentiation of the logarithmic functions.


5.5 – Logarithmic Differentiation- You will study about logarithmic differentiation i.e. differentiation of the function in the form of a logarithm.


5.6 – Derivatives of Functions in Parametric Forms- You will study about derivatives of functions in parametric form.


5.7 – Second Order Derivative- You will study and learn to differentiate the functions of second and third-order.


5.8 – Mean Value Theorem- You will study about two fundamental results in calculus without proof, the geometric interpretation of theorems, Rolle’s Theorem, Mean Value Theorem. (Not present in current syllabus)

3. How many exercises are there in Class 12 Maths Chapter 3 NCERT book?

A. Class 1 Maths Chapter 3 has a total of eight exercises and each exercise consists of a bunch of questions of all types. 

  • EXERCISE 5.1 - 34 Questions

  • EXERCISE 5.2 - 10 Questions

  • EXERCISE 5.3 - 15 Questions 

  • EXERCISE 5.4 - 10 Questions

  • EXERCISE 5.5 - 18 Questions

  • EXERCISE 5.6 - 11 Questions

  • EXERCISE 5.7 - 17 Questions

  • EXERCISE 5.8 - 6 Questions

4. Why should I opt for Chapter 3 of Class 12 Mathematics NCERT Solutions?

A. There are several benefits for opting Chapter 3 of Class 12 Mathematics NCERT Solutions. Class 12 Mathematics Chapter 3 of NCERT Solutions are designed and created by the best subject matter experts from the relevant industry. All the topics covered in all the answers and each answer come with an in-depth explanation of every concept to make them understandable. These NCERT Solutions for Class 12 Chapter 3 Mathematics plays a crucial role in your preparation for various competitive exams apart from the board exams as well such as JEE Main, Olympiad etc. These solutions will be very helpful when you will study at home as per your own convenience.