NCERT Solutions 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} $ .

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)