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# NCERT Solutions Class 11 Maths Chapter 12 Limits and Derivatives

Last updated date: 05th Aug 2024
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## NCERT Solutions Class 11 Maths Chapter 12 Limits and Derivatives - FREE PDF Download

NCERT Solution for Limits And Derivatives Class 11 Chapter 12 marks the beginning of calculus, a significant branch of mathematics that deals with change and motion. Understanding Limits And Derivatives Class 11 Solutions is crucial as these concepts form the foundation for more advanced topics in calculus and have wide-ranging applications in various fields such as physics, engineering, economics, and beyond. Access the NCERT Solutions for Class 11 Maths here.

Table of Content
1. NCERT Solutions Class 11 Maths Chapter 12 Limits and Derivatives - FREE PDF Download
2. Glance on Maths Chapter 12 Class 11 - Limits and Derivatives
3. Access Exercise Wise NCERT Solutions for Chapter 12 Maths Class 11
4. Exercises Under NCERT Solutions for Class 11 Maths Chapter 12 Limits and Derivatives
5. Access NCERT Solutions for Class 11 Maths Chapter 12 – Limits and Derivatives
5.1Exercise 12.1
5.2Exercise 12.2
5.3Miscellaneous Exercise
6. Class 11 Maths Chapter 12: Exercises Breakdown
7. Other Study Material for CBSE Class 11 Maths Chapter 12
8. Chapter-Specific NCERT Solutions for Class 11 Maths
FAQs

## Glance on Maths Chapter 12 Class 11 - Limits and Derivatives

• Limits And Derivatives Class 11 PDF of NCERT Solution covers topics such as Introduction, Intuitive Idea of Derivatives, Limits, Algebra of limits, Limits of polynomials and rational functions, Limits of Trigonometric Functions, Derivatives, Algebra of derivatives of functions, Derivative of polynomials and trigonometric functions.

• Limits provide a powerful tool to answer this question, forming the foundation for understanding Calculus.

• This chapter introduces the concept of derivatives – a measure of how quickly a function changes at a specific point. Derivatives are like the speedometer for functions, telling you how fast they're changing!

• Algebra of Limits equipped with the concept of limits, Students will explore various rules and theorems to solve problems involving limits. This will be your toolkit for maneuvering through limit calculations.

• In this chapter students will be able to tackle the limits of different types of functions, including polynomials, rational functions, trigonometric functions, and more. Students will learn specific techniques for handling each type.

• This chapter lays the groundwork for the formal definition and applications of derivatives. Students will see the connection between limits and derivatives, taking your first steps into the world of Calculus.

• There are two exercises (43 fully solved questions) in class 11th maths chapter 12 Limits And Derivatives.

## Access Exercise Wise NCERT Solutions for Chapter 12 Maths Class 11

 Current Syllabus Exercises of Class 11 Maths Chapter 12 NCERT Solutions of Class 11 Maths Limits and Derivatives Exercise 12.1 NCERT Solutions of Class 11 Maths Limits and Derivatives Exercise 12.2 NCERT Solutions of Class 10 Maths Limits and Derivatives Miscellaneous Exercise
Competitive Exams after 12th Science

## Exercises Under NCERT Solutions for Class 11 Maths Chapter 12 Limits and Derivatives

• Exercise 12.1: This exercise introduces the concept of limits of a function, their properties, and different types of limits. Students will learn how to find the limit of a function using algebraic techniques and the Squeeze rule. They will also learn about left and right-hand limits and the existence of a limit.

• Exercise 12.2: In this exercise, students will learn about the concept of derivatives of a function, their geometrical interpretation, and the rules of differentiation. They will also practice finding the derivative of a function using various differentiation rules.

• Miscellaneous Exercise: This exercise includes a mix of questions covering all the concepts taught in the chapter. Students will have to apply their knowledge of limits and derivatives to solve various problems and answer questions. They will also practice finding the limit of a function using algebraic techniques and the Squeeze rule and finding the derivative of a function using various differentiation rules.

## Access NCERT Solutions for Class 11 Maths Chapter 12 – Limits and Derivatives

### Exercise 12.1

1: Evaluate the Given limit: $\mathbf{\lim _{x \rightarrow 3} x+3}$

Ans: $\lim _{x \rightarrow 3} x+3=3+3=6$

2: Evaluate the Given limit: $\mathbf{\lim _{x \rightarrow z}\left(x-\dfrac{22}{7}\right)}$

Ans: $\lim _{x \rightarrow z}\left(x-\dfrac{22}{7}\right)=\left(\pi-\dfrac{22}{7}\right)$

3: Evaluate the Given limit: $\mathbf{\lim _{r \rightarrow 1} \pi r^{2}}$

Ans: $\lim \pi r^{2}=\pi\left(1^{2}\right)=\pi$

4: Evaluate the Given limit: $\mathbf{\lim _{x \rightarrow 1} \dfrac{4 x+3}{x-2}}$

Ans: $\lim _{x \rightarrow 1} \dfrac{4 x+3}{x-2}=\dfrac{4(4)+3}{4-2}=\dfrac{16+3}{2}=\dfrac{19}{2}$

5: Evaluate the Given limit: $\mathbf{\lim _{x \rightarrow-1} \dfrac{x^{10}+x^{5}+1}{x-1}}$

Ans: $\lim _{x \rightarrow-1} \dfrac{x^{10}+x^{5}+1}{x-1}=\dfrac{(-1)^{10}+(-1)^{5}+1}{-1-1}=\dfrac{1-1+1}{-2}=-\dfrac{1}{2}$

6: Evaluate the Given limit: $\mathbf{\lim _{x \rightarrow 0} \dfrac{(x+1)^{5}-1}{x}}$

Ans: $\lim _{x \rightarrow 0} \dfrac{(x+1)^{5}-1}{x}$

Put $x+1=y s o$ that $y \rightarrow 1$ as $x \rightarrow 0$

Accordingly, $\lim _{x \rightarrow 0} \dfrac{(x+1)^{5}-1}{x}=\lim _{x \rightarrow 1} \dfrac{(y)^{5}-1}{y-1}$

5. $1^{5-1} \quad\left[\lim _{x \rightarrow a} \dfrac{x^{n}-a^{n}}{x-a}=n d^{n-1}\right]$

$=5$

$\therefore \lim _{x \rightarrow 0} \dfrac{(x+1)^{5}-1}{x}=5$

7: Evaluate the Given limit: $\mathbf{\lim _{x \rightarrow 2} \dfrac{3 x^{2}-x-10}{x^{2}-4}}$

Ans: At $x-2$, the value of the given rational function takes the form $\dfrac{0}{0}$

$\lim _{x \rightarrow 2} \dfrac{3 x^{2}-x-10}{x^{2}-4}=\lim _{x \rightarrow 2} \dfrac{(x-2)(3 x+5)}{(x-2)(x+2)}$

$\lim _{x \rightarrow 2} \dfrac{3 x+5}{x+2}$

$=\dfrac{3(2)+5}{2+2}$

$-\dfrac{11}{4}$

8: Evaluate the Given limit: $\mathbf{\lim _{x \rightarrow 3} \dfrac{x^{4}-81}{2 x^{2}-5 x-3}}$

Ans: At $x=2$, the value of the given rational function takes the form $\dfrac{0}{0}$

$\lim _{x \rightarrow 3} \dfrac{x^{4}-81}{2 x^{2}-5 x-3}=\lim _{x \rightarrow 3} \dfrac{(x-3)(x+3)\left(x^{2}+9\right)}{(x-3)(2 x+1)}$

$\lim _{x \rightarrow 3} \dfrac{(x+3)\left(x^{2}+9\right)}{(2 x+1)}$

$=\dfrac{(3+3)\left(3^{2}+9\right)}{2(3)+1}$

$=\dfrac{6 \times 18}{7}$

$=\dfrac{108}{7}$

9: Evaluate the Given limit: $\mathbf{\lim _{x \rightarrow 0} \dfrac{a x+b}{c x+1}}$

Ans:

$\lim _{x \rightarrow 0} \dfrac{a x+b}{c x+1}=\dfrac{a(0)+b}{d(0)+1}=b$

10: Evaluate the Given limit: $\mathbf{\lim _{z \rightarrow 1} \dfrac{\dfrac{1}{3^{3}}-1}{\dfrac{1}{2^{6}}-1}}$

Ans: $\lim _{z \rightarrow 1} \dfrac{z^{\dfrac{1}{3}}-1}{\dfrac{1}{z^{6}}-1}$

At $z=1$, the value of the given function takes the form $\dfrac{0}{0}$ Put $z^{\dfrac{1}{6}}=x$ so that $z \rightarrow 1$ as $x \rightarrow 1$.

Accordingly, $\lim _{x \rightarrow 1} \dfrac{\dfrac{1}{\vec{e}}-1}{z^{\dfrac{1}{t}}-1}=\lim _{x \rightarrow 1} \dfrac{x^{2}-1}{x-1}$

$=\lim _{x \rightarrow 1} \dfrac{x^{2}-1}{x-1}$

$=2.1^{2 \cdot 1} \quad\left[\lim _{x \rightarrow a} \dfrac{x^{n}-a^{n}}{x-a}=n d^{n-1}\right]$

$=2$

$\lim _{x \rightarrow 1} \dfrac{z^{\dfrac{1}{3}}-1}{\dfrac{1}{t^{6}}-1}=2$

11: Evaluate the Given limit: $\mathbf{\lim _{x \rightarrow 1} \dfrac{a x^{2}+b x+c}{c x^{2}+b x+a}, a+b+c \neq 0}$

Ans: $\lim _{x \rightarrow 1} \dfrac{a x^{2}+b x+c}{c x^{2}+b x+a}=\dfrac{a(1)^{2}+b(1)+c}{c(1)^{2}+b(1)+a}$

$=\dfrac{a+b+c}{a+b+c}$

$=1$

$[a+b+c \neq 0]$

12: Evaluate the given limit: $\mathbf{\lim _{x \rightarrow-2} \dfrac{\dfrac{1}{x}+\dfrac{1}{2}}{x+2}}$

Ans: $\lim _{x \rightarrow-2} \dfrac{\dfrac{1}{x}+\dfrac{1}{2}}{x+2}$

At $x=-2$, the value of the given function takes the form $\dfrac{0}{0}$

Now, $\lim _{x \rightarrow-2} \dfrac{\dfrac{1}{x}+\dfrac{1}{2}}{x+2}=\lim _{x \rightarrow-2} \dfrac{\left(\dfrac{2+x}{2 x}\right)}{x+2}$

$\lim _{x \rightarrow-2} \dfrac{1}{2 x}$

$\dfrac{1}{2(-2)}=\dfrac{-1}{4}$

13: Evaluate the given limit: $\mathbf{\lim _{x \rightarrow 0} \dfrac{\sin a x}{b x}}$

Ans: $\lim _{x \rightarrow 0} \dfrac{\sin a x}{b x}$

At $x=0$, the value of the given function takes the form $\dfrac{0}{0}$

Now, $\lim _{x \rightarrow 0} \dfrac{\sin a x}{b x}=\lim _{x \rightarrow 0} \dfrac{\sin a x}{a x} \times \dfrac{a x}{b x}$

$\begin{array}{l}=\lim _{x \rightarrow 0}\left(\dfrac{\sin a x}{a x}\right) \times \dfrac{a}{b}\\\dfrac{a}{b} \lim _{\operatorname{ax\rightarrow0}}\left(\dfrac{\sin a x}{a x}\right) \quad[x \rightarrow 0 \Rightarrow a x \rightarrow 0]\\=\dfrac{a}{b} \times 1\\=\left[\lim _{x \rightarrow 0}\left(\dfrac{\sin y}{y}\right)\right]\\=\dfrac{a}{b}\\\end{array}$

14: Evaluate the given limit:  $\mathbf{\lim _{x \rightarrow 0} \dfrac{\sin a x}{\sin b x}, a, b \neq 0\\}$

Ans:  $\lim _{x \rightarrow 0} \dfrac{\sin a x}{\sin b x}, a, b \neq 0\\$

At $x=0$, the value of the given function takes the form $\frac{0}{0}$

Now, $\mathop {\lim }\limits_{x \to 0} \frac{{\sin ax}}{{\sin bx}} = \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\left( {\frac{{\sin ax}}{{ax}}} \right) \times ax}}{{\left( {\frac{{\sin bx}}{{ax}}} \right) \times bx}}} \right)$

\begin{aligned}&=\frac{a}{b} \times \frac{\lim _{a x \rightarrow 0}\left(\frac{\sin a x}{a x}\right)}{\lim _{b x \rightarrow 0}\left(\frac{\sin b x}{a x}\right)} \\&x \rightarrow 0 \Rightarrow a x \rightarrow 0\end{aligned}

$\text { and } x \rightarrow 0 \Rightarrow b x \rightarrow 0$

and $x \rightarrow 0 \Rightarrow b x \rightarrow 0$

$\frac{a}{b} \times \frac{1}{1}$

$\left[\lim _{x \rightarrow 0}\left(\frac{\sin y}{y}\right)=1\right]$

15: Evaluate the given limit: $\mathbf{ \lim _{x \rightarrow z}}\mathbf{ \dfrac{\sin (\pi-x)}{\pi(\pi-x)}}$

Ans:  $\lim {x \rightarrow z}\dfrac{\sin (\pi-x)}{\pi(\pi-x)}$

It is seen that $x\rightarrow \pi \Rightarrow (\pi .x)\rightarrow 0$

$\lim_{x\rightarrow 0}(\frac{\sin ax}{ax})\times \frac{a}{b}$

$\frac{a}{b}\lim_{ax\rightarrow 0}(\frac{\sin ax}{ax})[x\rightarrow 0\Rightarrow ax\rightarrow 0]$

$=\frac{a}{b}\times 1$

$\left [ \lim_{x\rightarrow 0}(\frac{\sin y}{y}) \right ]$

$=\frac{a}{b}$

16: Evaluate the given limit: $\lim _{x \rightarrow 0} \dfrac{\cos x}{\pi-x}\\$

Ans: $\lim _{x \rightarrow 0} \dfrac{\cos x}{\pi-x}=\dfrac{\cos 0}{\pi-0}=\dfrac{1}{\pi}\\$

17: Evaluate the given limit:$\mathbf{ \lim _{x \rightarrow 0} }\mathbf{\dfrac{\cos 2 x-1}{\cos x-1}}$

Ans:

$\lim_{x\rightarrow 0}\frac{\cos2x -1}{\cos x-1}$

At, x =0  the value of given function takes a form $\frac{0}{0}$

Now, $\lim_{x\rightarrow 0}\frac{\cos 2x-1}{\cos x-1}= \lim_{x\rightarrow 0}\frac{1-\sin ^2x - 1}{1- 2 \sin ^2\frac{x}{2}-1}$

$[\cos x = 1 - 2 \sin ^2 \frac{x}{2}]$

$-\lim_{x\rightarrow 0}\frac{\sin ^2x}{\sin^2\frac{x}{2}} = \lim_{x\rightarrow 0}\frac{(\frac{\sin ^2 x}{x^2} )\times x^2}{(\frac{\sin^2 \frac{x}{2}}{(\frac{x}{2})^2})\times \frac{x^2}{4}}$

\begin{aligned}&=4 \frac{\lim _{x \rightarrow 0}\left(\frac{\sin ^{2} x}{x^{2}}\right)}{\lim _{x \rightarrow 0}\left(\frac{\sin ^{2} \frac{x}{2}}{\left(\frac{x}{2}\right)^{2}}\right)} \\&=4 \frac{\left(\lim _{x \rightarrow 0} \frac{\sin ^{2} x}{x^{2}}\right)^{2}}{\left(\lim _{x \rightarrow 0} \frac{\sin \frac{x}{2}}{\left(\frac{x}{2}\right)^{2}}\right)^{2}} \\&{\left[x \rightarrow 0 \Rightarrow \frac{x}{2} \rightarrow 0\right]} \\&=4 \frac{1^{2}}{1^{2}}\left[\lim _{y \rightarrow 0} \frac{\sin y}{y}=1\right]\end{aligned}

=8

18: Evaluate the given limit: $\mathbf{\lim _{x \rightarrow 0} } \mathbf{\dfrac{a x+x \cos x}{b \sin x}}$

Ans: $\lim _{x \rightarrow 0} \dfrac{a x+x \cos x}{b \sin x}$

At $x=0$, the value of the given function takes the form $\dfrac{0}{0}$

Now, $\lim _{x \rightarrow 0} \dfrac{a x+x \cos x}{b \sin x}=\dfrac{1}{b} \lim _{x \rightarrow 0} \dfrac{x(a+\cos x)}{\sin x}$

$\lim _{b \rightarrow 0}\left(\dfrac{x}{\sin x}\right) \times \lim _{x \rightarrow 0}(a+\cos x)$

$\dfrac{1}{b}\left(\dfrac{1}{\lim _{x \rightarrow 0}\left(\dfrac{\sin x}{x}\right)}\right) \times \lim _{x \rightarrow 0}(a+\cos x)$

$\dfrac{1}{b} \times(a+\cos 0) \quad\left[\lim _{y \rightarrow 0} \dfrac{\sin x}{x}=1\right]$

$-\dfrac{a+1}{b}$

19: Evaluate the given limit: $\mathbf{\lim _{x \rightarrow 0} x \sec x}$

Ans: $\lim _{x \rightarrow 0} x \sec x=\lim _{x \rightarrow 0} \dfrac{x}{\cos x}=\dfrac{0}{\cos 0}=\dfrac{0}{1}=0$

20: Evaluate the given limit: $\mathbf{\lim _{x \rightarrow 0}} \mathbf{\dfrac{\sin a x+b x}{a x+\sin b x} a, b, a+b \neq 0}$

Ans: At $x=0$, the value of the given function takes the form $\dfrac{0}{0}$ Now, $\lim _{x \rightarrow 0} \dfrac{\sin a x+b x}{a x+\sin b x}$

$=\lim _{x \rightarrow 0} \dfrac{\left(\dfrac{\sin a x}{a x}\right) a x+b x}{a x+b x\left(\dfrac{\sin b x}{b x}\right)}$

$=\dfrac{\left(\lim _{x \rightarrow 0} \dfrac{\sin a x}{a x}\right) \times \lim _{x \rightarrow 0}(a x)+\lim _{x \rightarrow 0}(b x)}{\lim _{x \rightarrow 0} a x+\lim _{x \rightarrow 0} b x\left(\lim _{x \rightarrow 0} \dfrac{\sin b x}{b x}\right)} \quad[$ As $x \rightarrow 0 \Rightarrow \operatorname{ax} \rightarrow 0$ and $b x \rightarrow 0]$

$=\dfrac{\lim _{x \rightarrow 0}(a x)+\lim _{x \rightarrow 0} b x}{\lim _{x \rightarrow 0} a x+\lim _{x \rightarrow 0} b x}$

$\left[\lim _{y \rightarrow 0} \dfrac{\sin x}{x}=1\right]$

$=\dfrac{\lim _{x \rightarrow 0}(a x+b x)}{\lim _{x \rightarrow 0}(a x+b x)}$

$=\lim _{x \rightarrow 0}(1)$

$=1$

21: Evaluate the given limit: $\mathbf{\lim _{x \rightarrow 0}} \mathbf{( \operatorname{cosec} x-\cot x)}$

Ans: At $x=0$, the value of the given function takes the form $\infty-\infty$ Now, $\lim _{x \rightarrow 0}(\operatorname{cosec} x-\cot x)$

$\lim _{x \rightarrow 0}\left(\dfrac{1}{\sin x}-\dfrac{\cos x}{\sin x}\right)$

$\lim _{x \rightarrow 0}\left(\dfrac{1-\cos x}{\sin x}\right)$

$=\lim _{x \rightarrow 0} \dfrac{\left(\dfrac{1-\cos x}{x}\right)}{\left(\dfrac{\sin x}{x}\right)}$

$=\dfrac{\lim _{x \rightarrow 0} \dfrac{1-\cos x}{x}}{\lim _{x \rightarrow 0} \dfrac{\sin x}{x}}$

$-\dfrac{0}{1}$

$\left[\lim _{y \rightarrow 0} \dfrac{1-\cos x}{x}=0\right.$ and $\left.\lim _{y \rightarrow 0} \dfrac{\sin x}{x}=1\right]$

$=0$

22: Evaluate the given limit: $\mathbf{\lim _{x \rightarrow \dfrac{x}{2}} \dfrac{\tan 2 x}{x-\dfrac{\pi}{2}}}$

Ans: $\lim _{x \rightarrow \dfrac{z}{2}} \dfrac{\tan 2 x}{x-\dfrac{\pi}{2}}$

At $x=\dfrac{\pi}{2}$, the value of the given function takes the form $\dfrac{0}{0}$ Now, put So that $x-\dfrac{\pi}{2}-y$ so that $x \rightarrow \dfrac{\pi}{2}, y \rightarrow 0$

$\therefore \lim _{x \rightarrow \dfrac{\pi}{2}} \dfrac{\tan 2 x}{x-\dfrac{\pi}{2}}=\lim _{y \rightarrow 0} \dfrac{\tan 2\left(y+\dfrac{\pi}{2}\right)}{y}$

$\lim _{y \rightarrow 0} \dfrac{\tan (\pi+2 y)}{y}$

$-\lim _{y \rightarrow 0} \dfrac{\tan 2 y}{y}$

$[\tan (\pi+2 y)=\tan 2 y]$

$-\lim _{y \rightarrow 0} \dfrac{\sin 2 y}{y \cos 2 y}$

$\lim _{y \rightarrow 0}\left(\dfrac{\sin 2 y}{2 y} \times \dfrac{2}{\cos 2 y}\right)$

$-\left(\lim _{y \rightarrow 0} \dfrac{\sin 2 y}{2 y}\right) \times \lim _{y \rightarrow 0}\left(\times \dfrac{2}{\cos 2 y}\right)$

$[y \rightarrow 0 \Rightarrow 2 y \rightarrow 0]$

$-1 \times \dfrac{2}{\cos 0} \quad\left[\lim _{\leftrightarrow \infty 0} \dfrac{\sin x}{x}=1\right]$

$=1 \times \dfrac{2}{1}$

$-2$

23: Find $\mathbf{\lim _{x \rightarrow 0} f(x)}$ and $\mathbf{\lim _{x \rightarrow 1} f(x)}$, where $\mathbf{f(x)=\left\{\begin{array}{ll}2 x+3, & x \leq 0 \\ 3(x+1), & x>0\end{array}\right.}$

Ans: The given function is

$f(x)=\left\{\begin{array}{ll}2 x+3, & x \leq 0 \\ 3(x+1), & x>0\end{array}\right.$

$\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0}[2 x+3]=2(0)+3-3$

$\lim _{v \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0} 3(x+1)-3(0+1)-3$

$\therefore \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0} f(x)=3$

$\lim _{x \rightarrow \pi} f(x)=\lim _{x \rightarrow 1}(x+1)=3(1+1)=6$

$\therefore \lim _{x \rightarrow+} f(x)=\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1} f(x)=6$

24: Find $\mathbf{\lim _{x \rightarrow 1} f(x)}$, when $\mathbf{f(x)=\left\{\begin{array}{ll}x^{2}-1, & x \leq 1 \\ -x-1, & x>1\end{array}\right.}$

Ans:

The given function is

$f(x)=\left\{\begin{array}{ll}x^{2}-1, & x \leq 1 \\ -x-1, & x>1\end{array}\right.$

$\therefore \lim _{x \rightarrow T} f(x)=\lim _{x \rightarrow 1}\left[x^{2}-1\right]-1^{2}-1-1-1=0$

It is observed that $\lim _{x \rightarrow \pi} f(x) \neq \lim _{x \rightarrow 1^{+}} f(x)$.

Hence, lim $f(x)$ does not exist.

25: Evaluate $\mathbf{\lim _{x \rightarrow 0} f(x)}$, where $\mathbf{f(x)=\left\{\begin{array}{ll}\dfrac{|x|}{x}, & x \neq 0 \\ 0, & x=0\end{array}\right.}$

Ans: The given function is $f(x)=\left\{\begin{array}{ll}\dfrac{|x|}{x}, & x \neq 0 \\ 0, & x=0\end{array}\right.$

$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{-}}\left[\dfrac{|x|}{x}\right]$

$-\lim _{x \rightarrow 0}\left(\dfrac{-x}{x}\right)$

(When $x$ is negative, $|x|=\cdot x$)

$-\lim _{x \rightarrow 0}(-1)$

$-\cdot 1$

$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}}\left[\dfrac{|x|}{x}\right]$

$-\lim _{x \rightarrow 0}\left(\dfrac{x}{x}\right)$

(When $x$ is positive, $|x|-x$)

$=\lim _{x \rightarrow 0}(1)$

$-1$

It is observed that $\lim _{x \rightarrow \sigma} f(x) \neq \lim _{x \rightarrow 0^{+}} f(x)$.

Hence, $\lim _{x \rightarrow 0} f(x)$ does not exist.

26: Find $\mathbf{\lim _{x \rightarrow 0} f(x)-\left\{\begin{array}{ll}\dfrac{x}{|x|}, & x \neq 0 \\ 0, & x=0\end{array}\right.}$

Ans: The given function is

$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0}\left[\dfrac{x}{|x|}\right]$

$-\lim _{x \rightarrow 0}\left(\dfrac{x}{-x}\right)$

[When $\mathrm{x}<0,|x|-\cdot \mathrm{x}]$

$-\lim _{x \rightarrow 0}(-1)$

$=\cdot 1$

$\lim _{x \rightarrow 0^{+}} f(x)=\lim _{x \rightarrow 0^{+}}\left[\dfrac{x}{|x|}\right]$

$-\lim _{x \rightarrow 0}\left(\dfrac{x}{x}\right)$

$[$ When $x>0,|x|=x]$

$-\lim _{x \rightarrow 0}(1)$

$-1$

It is observed that $\lim _{x \rightarrow \sigma} f(x) \neq \lim _{x \rightarrow 0^{+}} f(x)$.

Hence, $\lim _{x \rightarrow 0} f(x)$ does not exist.

27: Find $\mathbf{\lim _{x \rightarrow 5} f(x)}$, where $\mathbf{f(x)=|x| \cdot 5}$

Ans: The given function is $f(x)=|x| \cdot 5$

$\lim _{y \rightarrow 5^{-}} f(x)=\lim _{x \rightarrow 5}(|x|-5)$

$=\lim _{x \rightarrow 5}(x-5)$

[When $x>0,|x|-x]$

$-5 \cdot 5$

$-0$

$\lim _{x \rightarrow^{+}} f(x)=\lim _{x \rightarrow 5}(|x|-5)$

$=\lim _{x \rightarrow 5}(x-5)$

( When $x>0,|x|-x]$)

$-5-5$

$=0$

$\lim _{x \rightarrow 5^{-}} f(x)=\lim _{x \rightarrow 5^{+}} f(x)=0$

Hence, $\lim _{x \rightarrow 5} f(x)=0$

28: Suppose $\mathbf{f(x)=\left\{\begin{array}{ll}a+b x, & x<0 \\ 4, & x=1 \\ b-a x, & x>1\end{array}\right.}$ and if $\mathbf{\lim _{x \rightarrow 1} f(x)=f(1)}$ what are possible values of a and b?

Ans: The given function is

$f(x)=\left\{\begin{array}{ll}a+b x, & x<0 \\ 4, & x=1 \\ b-a x, & x>1\end{array}\right.$

$\lim _{-\pi} f(x)=\lim _{x \rightarrow 1}(a+b x)=a+b$

$\lim _{x, 1-} f(x)=\lim _{x \rightarrow 1}(b-a x)=b-a$

$f(1)=4$

It is given that $\lim _{x \rightarrow 1} f(x)=f(1)$.

$\therefore \lim _{x \rightarrow \pi} f(x)=\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1} f(x)=f(1)$

$\Rightarrow a+b-4$ and $b-a=4$

On solving these two equations, we obtain $a=0$ and $b=4$. Thus, the respective possible values of a and $\mathrm{b}$ are 0 and 4 .

29: Let $\mathbf{a_{1}, a_{2}, \ldots, a_{n}}$ be fixed real numbers and define a function

$\mathbf{f(x)=\left(x-a_{1}\right)\left(x-a_{2}\right) \ldots . .\left(x-a_{n}\right)}$

What is $\mathbf{\lim f(x) ?}$ For some $\mathbf{a \neq a_{1}, a_{2}, \ldots, a_{n}}$. Compute $\mathbf{\lim _{x \rightarrow a} f(x)}$.

Ans: The given function is $f(x)=\left(x-a_{1}\right)\left(x-a_{2}\right) \ldots . .(x-a)$ $\lim _{x \rightarrow+3} f(x)=\lim _{x \rightarrow a}\left[\left(x-a_{1}\right)\left(x-a_{2}\right) \ldots . .\left(x-a_{n}\right)\right]$

$=\left(a_{1}-a_{1}\right)\left(a_{1}-a_{2}\right) \ldots . .\left(a_{1}-a_{n}\right)=0$

$\therefore \lim _{x \rightarrow 3} f(x)=0$

Now, $\lim _{x \rightarrow a} f(x)=\lim _{x \rightarrow a}\left[\left(x-a_{1}\right)\left(x-a_{2}\right) \ldots . .\left(x-a_{n}\right)\right]$

$-\left(a-a_{1}\right)\left(a-a_{2}\right) \ldots . .(a-a)$

$\therefore \lim f(x)=\left(a-a_{1}\right)\left(a-a_{2}\right) \ldots \ldots(a-a)$

30. If f(x)=\left\{ \begin{align} & \left| x \right|+1 \\ & 0 \\ & \left| x \right|-1 \\ \end{align} \right. \begin{align} & x<0 \\ & x=0 \\ & x>1 \\ \end{align}

For what value (s) does $\mathop {\lim }\limits_{{\mathbf{x}} \to a} {\mathbf{f}}({\mathbf{x}})$ exists?

Ans: The given function is

If f(x)=\left\{ \begin{align} & \left| x \right|+1 \\ & 0 \\ & \left| x \right|-1 \\ \end{align} \right. \begin{align} & x<0 \\ & x=0 \\ & x>1 \\ \end{align}

When ${\text{a}} = 0$

$\mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to {0^ - }} (|x| + 1)$

$= \mathop {\lim }\limits_{x \to 0} ( - x + 1)$

If $x < 0,|x| = - x$

$= 0 + 1$

$= 1$

$\mathop {\lim }\limits_{x \to {0^ + }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} (|x| + 1)$

$= \mathop {\lim }\limits_{x \to 0} (x - 1)$

If $x > 0,|x| = - x$

$= 0 - 1$

$= - 1$

Here, it is observed that $\mathop {\lim }\limits_{x \to {0^ - }} f(x) \ne \mathop {\lim }\limits_{x \to {0^ + }} f(x)$.

$\therefore \mathop {\lim }\limits_{x \to 0} f(x)$ does not exist.

When $a < 0$ $\mathop {\lim }\limits_{x \to {a^ - }} f(x) = \mathop {\lim }\limits_{x \to {a^ - }} (|x| + 1)$

$= \mathop {\lim }\limits_{x \to a} ( - x + 1)$

$[x < a < 0 \Rightarrow |x| - - x]$

$= - a + 1$

$\mathop {\lim }\limits_{x \to \vec t} f(x) = \mathop {\lim }\limits_{x \to \Delta } (|x| + 1)$

$= \mathop {\lim }\limits_{x \to a} ( - x + 1)$

$[a < x < 0 \Rightarrow |x| - - x]$

$= - a$

$+ 1$

$\therefore \mathop {\lim }\limits_{x \to {a^ + }} f(x) = \mathop {\lim }\limits_{x \to {a^ + }} f(x) = - a + 1$

Thus, limit of $f(x)$ exists at $x - a$, where $a < 0$.

When ${\text{a}} > 0$

$\mathop {\lim }\limits_{x \to {a^ - }} f(x) = \mathop {\lim }\limits_{x \to {a^ - }} (|x| + 1)$

$= \mathop {\lim }\limits_{x \to a} ( - x - 1)$

$[0 < x < a \Rightarrow |x| - x]$

$= a - 1$

$\mathop {\lim }\limits_{x{ \to ^2}} f(x) = \mathop {\lim }\limits_{x \to \Delta } (|x| - 1)$

$= \mathop {\lim }\limits_{x \to a} ( - x - 1)$

$[0 < x < a \Rightarrow |x| = x]$

$= a - 1$

$\therefore \mathop {\lim }\limits_{x \to \pi } f(x) = \mathop {\lim }\limits_{x{ \to ^ + }} f(x) = a - 1$

Thus, limit of $f(x)$ exists at $x = a$, where $a > 0$ Thus, $\mathop {\lim }\limits_{x \to a} f(x)$ exists for all $a \ne 0$.

31: If the function f(x) satisfies, $\mathbf{\lim _{x \rightarrow 1} \dfrac{f(x)-2}{x^{2}-1}=\pi}$, evaluate $\mathbf{\lim _{x \rightarrow 1} f(x)}$

Ans: $\lim _{x \rightarrow 1} \dfrac{f(x)-2}{x^{2}-1}=\pi$

$\Rightarrow \dfrac{\lim _{x \rightarrow 1}(f(x)-2)}{\lim _{x \rightarrow 1}\left(x^{2}-1\right)}=\pi$

$\Rightarrow \lim _{x \rightarrow 1}(f(x)-2)=\pi \lim _{x \rightarrow 1}\left(x^{2}-1\right)$

$\Rightarrow \lim _{x \rightarrow 1}(f(x)-2)=\pi\left(1^{2}-1\right)$

$\Rightarrow \lim _{x \rightarrow 1}(f(x)-2)=0$

$\Rightarrow \lim _{x \rightarrow 1} f(x)-\lim _{x \rightarrow 1} 2=0$

$\Rightarrow \lim _{x \rightarrow 1} f(x)-2=0$

$\therefore \lim _{x \rightarrow 1} f(x)=2$

32: If $\mathbf{f(x)=\left\{\begin{array}{ll}m x^{2}+n, & x<0 \\ n x+m, & 0 \leq x \leq 1 \\ n x^{3}+m, & x>1\end{array}\right.}$

For what integers m and n does $\mathbf{\lim _{x \rightarrow 0} f(x)}$ and $\mathbf{\lim _{x \rightarrow 1} f(x)}$ exist?

Ans: $f(x)=\left\{\begin{array}{ll}m x^{2}+n, & x<0 \\ n x+m, & 0 \leq x \leq 1 . \\ n x^{3}+m, & x>1\end{array}\right.$

$\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0}\left(m x^{2}+n\right)$

$=m(0)^{2}+n$

$=n$

$=n(0)+m$

$=m$

Thus, $\lim _{x \rightarrow 0^{+}} f(x)$ exists if $\mathrm{m}=\mathrm{n}$.

$\lim _{x \rightarrow \sqrt{-}} f(x)=\lim _{x \rightarrow 1}(n x+m)$

$=n(1)+m$

$=m+n$

$\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1}\left(n x^{3}+m\right)$

$=n(1)^{3}+m$

$=m+n$

$\therefore \lim _{x \rightarrow \sqrt{ }} f(x)=\lim _{x \rightarrow+} f(x)=\lim _{x \rightarrow 1} f(x) .$

Thus, $\lim _{u \rightarrow 1} f(x)$ exists for any internal value of $\mathrm{m}$ and $\mathrm{n}$.

### Exercise 12.2

1: Find the derivative of $\mathbf{x^{2}-2}$ at $\mathbf{x=10}$.

Ans: Let $f(x)=x^{2}-2 .$ Accordingly. $f^{\prime}(10)=\lim _{h \rightarrow 0} \dfrac{f(10+h)-f(10)}{h}$

$=\lim _{h \rightarrow 0} \dfrac{\left[(10+h)^{2}-2\right]-\left(10^{2}-2\right)}{h}$

$=\lim _{h \rightarrow 0} \dfrac{10^{2}+2 \cdot 10 \cdot h+h^{2}-2-10^{2}+2}{h}$

$=\lim _{h \rightarrow 0} \dfrac{20 h+h^{2}}{h}$

$=\lim _{h \rightarrow 0}(20+h)=20+0=20$

Thus, the derivative of $x^{2}-2$ at $x-10$ is 20 .

2. Find the derivative of x at x=1.

Ans: Let $f(x)=x .$ Accordingly.

$f^{\prime}(1)=\lim _{n \rightarrow 0} \dfrac{f(1+h)-f(1)}{h}$

$=\lim _{h \rightarrow 0} \dfrac{(1+h)-1}{h}$

$=\lim _{h \rightarrow 0} \dfrac{h}{h}$

$=\lim _{h \rightarrow 0}(1)=1$

Thus, the derivative of $x$ at $x=1$ is 1 .

3: Find the derivative of $\mathbf{99 \mathrm{\mathbf{x}}}$ at $\mathrm{\mathbf{x}}$-100.

Ans: Let $f(x)=99 x$. Accordingly,

$f^{\prime}(100)=\lim _{h \rightarrow 0} \dfrac{f(100+h)-f(100)}{h}$

$=\lim _{n \rightarrow 0} \dfrac{99(100+h)-99(100)}{h}$

$=\lim _{h \rightarrow 0} \dfrac{99 \times 100+99 h-99 \times 100}{h}$

$=\lim _{h \rightarrow 0} \dfrac{99 h}{h}$

$=\lim _{h \rightarrow 0}(99)=99$

Thus, the derivative of $99 x$ at $x=100$ is 99 .

4: Find the derivative of the following functions from first principles.

(i) $\mathbf{x^{3}-27}$

(ii) $\mathbf{(x-1)(x-2)}$

(iii) $\mathbf{\dfrac{1}{x^{2}}}$

(iv) $\mathbf{\dfrac{x+1}{x-1}}$

Ans: (i) Let $f(x)=x^{3}-27$. Aocordingly, from the first principle,

$f^{\prime}(x)=\lim _{n \rightarrow 0} \dfrac{f(x+h)-f(x)}{h}$

$=\lim _{h \rightarrow 0} \dfrac{\left[(x+h)^{3}-27\right]-\left(x^{3}-27\right)}{h}$

$=\lim _{h \rightarrow 0} \dfrac{x^{3}+h^{3}+3 x^{2} h+3 x t^{2}-x^{3}}{h}$

$=\lim _{h \rightarrow 0} \dfrac{h^{3}+3 x^{2} h+3 x h^{2}}{h}$

$=\lim _{h \rightarrow 0}\left(h^{3}+3 x^{2} h+3 x h^{2}\right)$

$=0+3 x^{2}+0=3 x^{2}$

(ii) Let $f(x)=(x-1)(x-2)$. Accordingly, from the first principle,

$f^{\prime}(x)=\lim _{n \rightarrow 0} \dfrac{f(x+h)-f(x)}{h}$

$=\lim _{h \rightarrow 0} \dfrac{(x+h-1)(x+h-2)-(x-1)(x-2)}{h}$

$=\lim _{n \rightarrow 0} \dfrac{\left(x^{2}+h x-2 x+h x+t^{2}-2 h-x-h+2\right)-\left(x^{2}-2 x-x+2\right)}{h}$

$=\lim _{h \rightarrow 0} \dfrac{\left(h x+h x+h^{2}-2 h-h\right)}{h}$

$=\lim _{h \rightarrow 0} \dfrac{2 h x+h^{2}-3 h}{h}$

$=\lim _{n \rightarrow 0}(2 x+h-3)$

$-2 x-3$

(iii) Let $\mathrm{f}(\mathrm{x})=\dfrac{1}{x^{2}}$. Accordingly, from the first principle,

$f^{\prime}(x)=\lim _{n \rightarrow 0} \dfrac{f(x+h)-f(x)}{h}$

$=\lim _{h \rightarrow 0} \dfrac{\dfrac{1}{(x+h)^{2}}-\dfrac{1}{x^{2}}}{h}$

$=\lim _{h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{x^{2}-(x+h)^{2}}{x^{2}(x+h)^{2}}\right]$

$=\lim _{h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{x^{2}-x^{2}-2 h x-h^{2}}{x^{2}(x+h)^{2}}\right]$

$=\lim _{n \rightarrow 0} \dfrac{1}{h}\left[\dfrac{-A-2 h x}{x^{2}(x+h)^{2}}\right]$

$=\lim _{h \rightarrow 0}\left[\dfrac{-h^{2}-2 x}{x^{2}(x+h)^{2}}\right]$

$=\dfrac{0-2 x}{x^{2}(x+0)^{2}}=\dfrac{-2}{x^{3}}$

(iv) Let $f(x)=\dfrac{x+1}{x-1}$. Accordingly, from the frst principle,

$f^{\prime}(x)=\lim _{n \rightarrow 0} \dfrac{f(x+h)-f(x)}{h}$

$=\lim _{h \rightarrow 0} \dfrac{\left(\dfrac{x+h+1}{x+h-1}-\dfrac{x+1}{x-1}\right)}{h}$

$=\lim _{n \rightarrow 0} \dfrac{1}{h}\left[\dfrac{(x-1)(x+h+1)-(x+1)(x+h-1)}{(x-1)(x+h-1)}\right]$

$=\lim _{n \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\left(x^{2}+h x+x-x-h-1\right)-\left(x^{2}+h x-x+x+h-1\right)}{(x-1)(x+h-1)}\right]$

$=\lim _{n \rightarrow 0} \dfrac{1}{h}\left[\dfrac{-2 h}{(x-1)(x+h-1)}\right]$

$=\lim _{n \rightarrow 0}\left[\dfrac{-2}{(x-1)(x+h-1)}\right]$

$=\dfrac{-2}{(x-1)(x-1)}=\dfrac{-2}{(x-1)^{2}}$

5: For the function $\mathbf{F(x)=\dfrac{x^{10}}{100}+\dfrac{x^{59}}{99}+\cdots+\dfrac{x^{2}}{2}+x+1}$

Prove that $\mathbf{f(1)=100 f^{\prime}(0)}$

Ans: The given function is

$F(x)=\dfrac{x^{\mathrm{m}}}{100}+\dfrac{x^{59}}{99}+\cdots+\dfrac{x^{2}}{2}+x+1$

$\dfrac{d}{d x} f(x)=\dfrac{d}{d x}\left[\dfrac{x^{1 \mathbf{m}}}{100}+\dfrac{x^{m}}{99}+\cdots+\dfrac{x^{2}}{2}+x+1\right]$

$\dfrac{d}{d x} f(x)=\dfrac{d}{d x}\left(\dfrac{x^{100}}{100}\right)+\dfrac{d}{d x}\left(\dfrac{x^{99}}{99}\right)+\cdots+\dfrac{d}{d x}\left(\dfrac{x^{2}}{2}\right)+\dfrac{d}{d x}(x)+\dfrac{d}{d x}(1)$

On using theorem $\dfrac{d}{d x}\left(x^{n}\right)=n x^{n-1}$, we obtain

$\dfrac{d}{d x} f(x)=\dfrac{100 x^{99}}{100}+\dfrac{99 x^{88}}{99}+\cdots+\dfrac{2 x}{2}+1+0$

$=x^{99}+x^{88}+\cdots+x+1$

$\therefore f^{\prime}(x)=x^{99}+x^{18}+\cdots+x+1$

At $x=0$

$f^{\prime}(0)=1$

At $x=1$,

Thus, $\mathrm{f}(1)=100 \mathrm{f}(0)$

6: Find the derivative of $\mathbf{x^{n}+a x^{n-1}+a^{2} x^{n-2}+\cdots+a^{n-1} \chi+a^{n}}$ for some fixed real number a.

Ans: Let $f(x)=x^{n}+a x^{n-1}+a^{2} x^{n-2}+\cdots+a^{n-1} x+a^{n}$

$\dfrac{d}{d x} f(x)=\dfrac{d}{d x}\left(x^{n}+a x^{n-1}+a^{2} x^{n-2}+\cdots+a^{n-1} x+d^{n}\right)$

$=\dfrac{d}{d x}\left(x^{n}\right)+a \dfrac{d}{d x}\left(x^{n-1}\right)+a^{2} \dfrac{d}{d x}\left(x^{n-2}\right)+\cdots+a^{n-1} \dfrac{d}{d x}(x)+a^{n} \dfrac{d}{d x}(1)$

On using theorem $\dfrac{d}{d x}\left(x^{n}\right)=n^{n-1}$, we obtain

$f^{\prime}(x)=n x^{n-1}+a(n-1) x^{n-2}+a^{2}(n-2) x^{n-3}+\cdots+a^{n-1}+a^{n}(0)$

$\therefore f^{\prime}(x)=n x^{n-1}+a(n-1) x^{n-2}+a^{2}(n-2) x^{n-3}+\cdots+a^{n-1}$

7: For some constants a and $\mathrm{b}$, find the derivative of

(i) $\mathbf{(x-a)(x-b)}$

(ii) $\mathbf{\left(a x^{2}+b\right)^{2}}$

(iii) $\mathbf{\dfrac{x-a}{x-b}}$

Ans: (i) Let $f(x)=(x-a)(x-b)$

$\Rightarrow f(x)=x^{2}-(a+b) x+a b$

$\therefore f^{\prime}(x)=\dfrac{d}{d x}\left(x^{2}-(a+b) x+a b\right)$

$=\dfrac{d}{d x}\left(x^{2}\right)-(a+b) \dfrac{d}{d x}(x)+\dfrac{d}{d x}(a b)$

On using theorem $\dfrac{d}{d x}\left(x^{n}\right)=n x^{n-1}$, we obtain

$f^{\prime}(x)=2 x-(a+b)+0$

$=2 x-a-b$

(ii) Let $f(x)-\left(a x^{2}+b\right)^{2}$

$\Rightarrow f(x)=a^{2} x^{4}+2 a b x^{2}+b^{2}$

$\therefore f(x)=\dfrac{d}{d x}\left(a^{2} x^{4}+2 a b x^{2}+b^{2}\right)$

$=a^{2} \dfrac{d}{d x}\left(x^{4}\right)+2 a b \dfrac{d}{d x}\left(x^{2}\right)+\dfrac{d}{d x} b^{2}$

On using theorem $\dfrac{d}{d x}\left(x^{n}\right)=n x^{n-1}$, we obtain

$f^{\prime}(x)=a^{2}\left(4 x^{3}\right)+2 a b(2 x)+b^{2}(0)$

$=4 a^{2} x^{3}+4 a b x$

$=4 a x\left(a x^{2}+b\right)$

(iii) Let $f(x)=\dfrac{x-a}{x-b}$

$\Rightarrow f^{\prime}(x)=\dfrac{d}{d x}\left(\dfrac{x-a}{x-b}\right)$

By quotient rule,

$f^{\prime}(x)=\dfrac{(x-b) \dfrac{d}{d x}(x-a)-(x-a) \dfrac{d}{d x}(x-b)}{(x-b)^{2}}$

$=\dfrac{(x-b)(1)-(x-a)(1)}{(x-b)^{2}}$

$=\dfrac{x-b-x+a}{(x-b)^{2}}$

$=\dfrac{a-b}{(x-b)^{2}}$

8: Find the derivative of $\mathbf{\dfrac{x^{n}-a^{n}}{x-a}}$ for some constant a.

Ans: Let $f(x)=\dfrac{x^{n}-d^{n}}{x-a}$

$\Rightarrow f^{\prime}(x)=\dfrac{d}{d x}\left(\dfrac{x^{n}-a^{n}}{x-a}\right)$

By quotient rule, $f^{\prime}(x)=\dfrac{(x-a) \dfrac{d}{d x}\left(x^{n}-a^{n}\right)-\left(x^{n}-a^{n}\right) \dfrac{d}{d x}(x-a)}{(x-a)^{2}}$

$=\dfrac{(x-a)\left(n x^{n-1}-0\right)-\left(x^{n}-a^{n}\right)}{(x-a)^{2}}$

$=\dfrac{n x^{n}-a n x^{n-1}-x^{n}+a^{n}}{(x-a)^{2}}$

9: Find the derivative of

(i) $\mathbf{2 x-\dfrac{3}{4}}$

(ii) $\mathbf{\left(5 x^{3}+3 x-1\right)(x-1)}$

(iii) $\mathbf{x^{-3}(5+3 x)}$

(iv) $\mathbf{x^{5}\left(3-6 x^{-9}\right)}$

(v) $\mathbf{x^{-4}\left(3-4 x^{-5}\right)}$

(vi) $\mathbf{\dfrac{2}{x+1}-\dfrac{x^{2}}{3 x-1}}$

Ans: (i) Let $f(x)=2 x-\dfrac{3}{4}$

$f^{\prime}(x)=\dfrac{d}{d x}\left(2 x-\dfrac{3}{4}\right)$

$=2 \dfrac{d}{d x}(x)-\dfrac{d}{d x}\left(\dfrac{3}{4}\right)$

$-2-0$

(ii) Let $f(x)=\left(5 x^{3}+3 x-1\right)(x-1)$

By Leibnitz product rule,

$f^{\prime}(x)=\left(5 x^{3}+3 x-1\right) \dfrac{d}{d x}(x-1)+(x-1) \dfrac{d}{d x}\left(5 x^{3}+3 x-1\right)$

$-\left(5 x^{3}+3 x-1\right)(1)+(x-1)\left(5.3 x^{2}+3-0\right)$

$-\left(5 x^{3}+3 x-1\right)+(x-1)\left(15 x^{2}+3\right)$

$-5 x^{3}+3 x-1+15 x^{3}+3 x-15 x^{2}-3$

$=20 x^{3}-15 x^{2}+6 x-4$

(iii) Let $f(x)=x^{-3}(5+3 x)$

By Leibnitz product rule,

$f^{\prime}(x)=x^{-3} \dfrac{d}{d x}(5+3 x)+(5+3 x) \dfrac{d}{d x}\left(x^{-3}\right)$

$=x^{-3}(0+3)+(5+3 x)\left(3 x^{-3-1}\right)$

$=x^{-3}(3)+(5+3 x)\left(3 x^{-4}\right)$

$=3 x^{-3}-15 x^{-4}-9 x^{-3}$

$=-6 x^{3}-15 x^{4}$

$=-3 x^{-3}\left(2+\dfrac{5}{x}\right)$

$=\dfrac{-3 x^{-3}}{x}(2 x+5)$

$=\dfrac{-3}{x^{4}}(5+2 x)$

(iv) Let $f(x)=x^{5}\left(3-6 x^{-9}\right)$

By Leibnitz product rule,

$f^{\prime}(x)=x^{5} \dfrac{d}{d x}\left(3-6 x^{9}\right)+\left(3-6 x^{-9}\right) \dfrac{d}{d x}\left(x^{5}\right)$

$=x^{5}\left\{0-6(-9) x^{-2-1}\right\}+\left(3-6 x^{9}\right)\left(5 x^{4}\right)$

$=x^{5}\left(54 x^{-10}\right)+15 x^{4}-30 x^{-5}$

$=54 x^{5}+15 x^{4}-30 x^{5}$

$=24 x^{5}+15 x^{4}$

$=15 x^{4}+\dfrac{24}{x^{5}}$

(v) Let $f(x)=x^{-4}\left(3-4 x^{5}\right)$

By Leibnitz product rule,

$f^{\prime}(x)=x^{-4} \dfrac{d}{d x}\left(3-4 x^{-5}\right)+\left(3-4 x^{-5}\right) \dfrac{d}{d x}\left(x^{-4}\right)$

$=x^{-4}\left\{0-4(-5) x^{-5-1}\right\}+\left(3-4 x^{5}\right)(-4) x^{-4-1}$

$=x^{-1}\left(20 x^{6}\right)+\left(3-4 x^{5}\right)\left(-4 x^{-5}\right)$

$=20 x^{10}-12 x^{-5}+16 x^{-0}$

$=36 x^{-10}-12 x^{-6}$

$=\dfrac{12}{x^{5}}+\dfrac{36}{x^{20}}$

(vi) Let $f(x)=\dfrac{2}{x+1}-\dfrac{x^{2}}{3 x-1}$

$f^{\prime}(x)=\dfrac{d}{d x}\left(\dfrac{2}{x+1}\right)-\dfrac{d}{d x}\left(\dfrac{x^{2}}{3 x-1}\right)$

By quotient rule,

$f^{\prime}(x)=\left[\dfrac{(x+1) \dfrac{d}{d x}(2)-2 \dfrac{d}{d x}(x+1)}{(x+1)^{2}}\right]-\left[\dfrac{(3 x-1) \dfrac{d}{d x}\left(x^{2}\right)-x^{2} \dfrac{d}{d x}(3 x-1)}{(3 x-1)^{2}}\right]$

$=\left[\dfrac{(x+1)(0)-2(0)}{(x+1)^{2}}\right]-\left[\dfrac{(3 x-1)(2 x)-x^{2}(3)}{(3 x-1)^{2}}\right]$

$=\dfrac{-2}{(x+1)^{2}}-\left[\dfrac{6 x^{2}-2 x-3 x^{2}}{(3 x-1)^{2}}\right]$

$=\dfrac{-2}{(x+1)^{2}}-\left[\dfrac{3 x^{2}-2 x^{2}}{(3 x-1)^{2}}\right]$

$=\dfrac{-2}{(x+1)^{2}}-\dfrac{x(3 x-2)}{(3 x-1)^{2}}$

10: Find the derivative of cos $\mathrm{\mathbf{x}}$ from first principle.

Ans: Let $\mathrm{f}(\mathrm{x})-\cos \mathrm{x}$. Accordingly, from the first principle, $f^{\prime}(x)=\lim _{n \rightarrow 0} \dfrac{f(x+h)-f(x)}{h}$

$=\lim _{n \rightarrow 0}\left[\dfrac{\cos (x+h)-\cos (x)}{h}\right]$

$=\lim _{n \rightarrow 0}\left[\dfrac{\cos x \cos h-\sin x \sin h-\cos x}{h}\right]$

$=\lim _{n \rightarrow 0}\left[\dfrac{-\cos x(1-\cos h)}{h} \dfrac{-\sin x \sin h}{h}\right]$

$=-\cos x\left[\lim _{n \rightarrow 0}\left(\dfrac{1-\cos h}{h}\right)\right]-\sin x\left[\lim _{n \rightarrow 0}\left(\dfrac{\sin h}{h}\right)\right]$

$=-\cos x(0)-\sin x(1) \quad\left[\lim _{n \rightarrow 0} \dfrac{1-\cos h}{h}=0\right.$ and $\left.\lim _{n \rightarrow 0} \dfrac{\sin h}{h}=1\right]$

$\therefore f^{\prime}(x)=-\sin x$

11: Find the derivative of the following functions:

(i) $\mathbf{\sin x \cos x}$

(ii) $\mathbf{\sec x}$

(iii) $\mathbf{5 \sec x+4 \cos x}$

(iv) $\mathbf{\operatorname{cosec} x}$

(v) $\mathbf{3 \cot x+5 \operatorname{cosec} x}$

(vi) $\mathbf{5 \sin x-6 \cos x+7}$

(vii) $\mathbf{2 \tan x-7 \sec x}$

Ans: (i) Let $f(x)=\sin x \cos x .$ Accordingly, from the first principle,

$f^{\prime}(x)=\lim _{n \rightarrow 0} \dfrac{f(x+h)-f(x)}{h}$

$=\lim _{n \rightarrow 0} \dfrac{\sin (x+h) \cos (x+h)-\sin x \cos x}{h}$

$=\lim _{h \rightarrow 0} \dfrac{1}{2 h}[2 \sin (x+h) \cos (x+h)-2 \sin x \cos x]$

$=\lim _{n \rightarrow 0} \dfrac{1}{2 h}[\sin 2(x+h)-\sin 2 x]$

$=\lim _{h \rightarrow 0} \dfrac{1}{2 h}\left[2 \cos \dfrac{2 x+2 h+2 x}{2} \cdot \sin \dfrac{2 x+2 h-2 x}{2}\right]$

$=\lim _{h \rightarrow 0} \dfrac{1}{2 h}\left[2 \cos \dfrac{4 x+2 h}{2} \cdot \sin \dfrac{2 h}{2}\right]$

$=\lim _{n \rightarrow 0} \dfrac{1}{2 h}[\cos (2 x+h) \sin h]$

$=\lim _{n \rightarrow 0} \cos (2 x+h) \cdot \lim _{n \rightarrow 0} \dfrac{\sin h}{h}$

$=\cos (2 x+h) \cdot 1$

$=\cos 2 x$

(ii) Let $f(x)=\sec x .$ Accordingly, from the first principle,

$\begin{array}{l}f^{\prime}(x)=\lim _{n \rightarrow 0} \dfrac{f(x+h)-f(x)}{h}\\=\lim _{n \rightarrow 0} \dfrac{\sec (x+h)-\sec x}{h}\\=\lim _{h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{1}{\cos (x+h)}-\dfrac{1}{\cos x}\right]\\=\lim _{n \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\cos x-\cos (x+h)}{\cos x \cos (x+h)}\right]\\=\dfrac{1}{\cos x} \cdot \lim _{n \rightarrow 0} \dfrac{1}{h}\left[\dfrac{-2 \sin \left(\dfrac{x+x+h}{2}\right) \sin \left(\dfrac{x-x-h}{2}\right)}{\cos (x+h)}\right]\\=\dfrac{1}{\cos x} \cdot \lim _{h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{-2 \sin \left(\dfrac{2 x+h}{2}\right) \sin \left(\dfrac{-h}{2}\right)}{\cos (x+h)}\right]\\=\dfrac{1}{\cos x} \cdot \lim _{h \rightarrow 0} \dfrac{1}{2 h} \dfrac{\left.-2 \sin \left(\dfrac{2 x+h}{2}\right) \dfrac{\sin \left(\dfrac{-h}{2}\right)}{\left(\dfrac{h}{2}\right)}\right]}{\cos (x+h)}\\=\dfrac{1}{\cos x} \cdot \lim _{i \rightarrow 0} \dfrac{\sin \left(\dfrac{h}{2}\right)}{\left(\dfrac{h}{2}\right)} \cdot \lim _{i \rightarrow 0} \dfrac{\sin \left(\dfrac{2 x+h}{2}\right)}{\cos (x+h)}\\=\dfrac{1}{\cos x} \cdot 1 \cdot \dfrac{\sin x}{\cos x}\\=\sec x \tan x\\ \\ \text { (iii) Let } f(x)=5 \sec x+4 \cos x \text {. Accordingly, from the first principle, }\\f^{\prime}(x)=\lim _{h \rightarrow 0} \dfrac{f(x+h)-f(x)}{h}\\=\lim _{n \rightarrow 0} \dfrac{5 \sec (x+h)+4 \cos (x+h)-[5 \sec x+4 \cos x]}{h}\end{array}$

$\begin{array}{l}=5 \lim _{n \rightarrow 0} \dfrac{[\sec (x+h)-\sec x]}{h}+4 \lim _{n \rightarrow 0} \dfrac{[\cos (x+h)-\cos x]}{h}\\=5 \lim _{n \rightarrow 0} \dfrac{1}{h}\left[\dfrac{1}{\cos (x+h)}-\dfrac{1}{\cos x}\right]+4 \lim _{n \rightarrow 0} \dfrac{1}{h}[\cos (x+h)-\cos x]\\=5 \lim _{n \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\cos x-\cos (x+h)}{\cos x \cos (x+h)}\right]+4 \lim _{n \rightarrow 0} \dfrac{1}{h}[\cos x \cos h-\sin x \sin h-\cos x]\\=\dfrac{5}{\cos x} \cdot \lim _{h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{-2 \sin \left(\dfrac{2 x+h}{2}\right) \sin \left(\dfrac{-h}{2}\right)}{\cos (x+h)}\right]+4\left[-\cos x \lim _{h \rightarrow 0} \dfrac{(1-\cos x)}{h}-\sin x \lim _{h \rightarrow 0} \dfrac{\sin h}{h}\right]\\=\dfrac{5}{\cos x} \cdot \lim _{h \rightarrow 0} \dfrac{\left.\sin \left(\dfrac{2 x+h}{2}\right) \dfrac{\sin \left(\dfrac{-h}{2}\right)}{\left(\dfrac{h}{2}\right)}\right]}{\cos (x+h)}+4[-\cos x(0)-\sin x(1)]\\=\dfrac{5}{\cos x} \cdot\left[\lim _{n \rightarrow 0} \dfrac{\sin \left(\dfrac{h}{2}\right)}{\left(\dfrac{h}{2}\right)} \cdot \lim _{n \rightarrow 0} \dfrac{\sin \left(\dfrac{2 x+h}{2}\right)}{\cos (x+h)}\right]-4 \sin x\\=\dfrac{5}{\cos x} \cdot \dfrac{\sin x}{\cos x} \cdot 1-4 \sin x\\=5 \sec x \tan x-4 \sin x\\ \\ \text { (iv) Let } f(x)=\operatorname{cosec} x \text {. Accordingly, from the first principle. }\\f^{\prime}(x)=\lim _{n \rightarrow 0} \dfrac{f(x+h)-f(x)}{h}\\=\lim _{n \rightarrow 0} \dfrac{1}{h}[\operatorname{cosec}(x+h)-\operatorname{cosec} x]\\=\lim _{n \rightarrow 0} \dfrac{1}{h}\left[\dfrac{1}{\sin (x+h)}-\dfrac{1}{\sin x}\right]\\=\lim _{n \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\sin x-\sin (x+h)}{\sin x \sin (x+h)}\right]\end{array}$

$=\lim _{h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{2 \cos \left(\dfrac{x+x+h}{2}\right) \cdot \sin \left(\dfrac{x-x-h}{2}\right)}{\sin x \sin (x+h)}\right]$

$=\lim _{h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{2 \cos \left(\dfrac{2 x+h}{2}\right) \cdot \sin \left(\dfrac{-h}{2}\right)}{\sin x \sin (x+h)}\right]$

$=\lim _{h \rightarrow 0} \dfrac{ \left.-\cos \left(\dfrac{2 x+h}{2}\right) \dfrac{\sin \left(\dfrac{-h}{2}\right)}{\left(\dfrac{h}{2}\right)}\right]}{\sin x \sin (x+h)}$

$=\lim _{h \rightarrow 0}\left(\dfrac{-\cos \left(\dfrac{2 x+h}{2}\right)}{\sin x \sin (x+h)}\right) \cdot \lim _{\dfrac{h}{2} \rightarrow 0} \dfrac{\sin \left(\dfrac{h}{2}\right)}{\left(\dfrac{h}{2}\right)}$

$=\left(\dfrac{-\cos x}{\sin x \sin x}\right) \cdot 1$

$=-\operatorname{cosec} x \cot x$

(v) Let $f(x)=3 \cot x+5 \operatorname{cosec} x$. Accordingly, from the first principle, $f^{\prime}(x)=\lim _{n \rightarrow 0} \dfrac{f(x+h)-f(x)}{h}$

$=\lim _{n \rightarrow 0} \dfrac{1}{h}[3 \cot (x+h)+5 \operatorname{cosec}(x+h)-3 \cot x-5 \operatorname{cosec} x]$

$=3 \lim _{n \rightarrow 0} \dfrac{1}{h}[\cot (x+h)-\cot x]+5 \lim _{n \rightarrow 0} \dfrac{1}{h}[\operatorname{cosec}(x+h)-\operatorname{cosec} x]$

$\ldots .$

Now, $\lim _{n \rightarrow 0} \dfrac{1}{h}[\cot (x+h)-\cot x]=\lim _{n \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\cos (x+h)}{\sin (x+h)}-\dfrac{\cos x}{\sin x}\right]$

$=\lim _{n \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\cos (x+h) \sin x-\cos x \sin (x+h)}{\sin x \sin (x+h)}\right]$

$=\lim _{n \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\sin (x-x-h)}{\sin x \sin (x+h)}\right]$

$\begin{array}{l}=\lim _{n \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\sin (-h)}{\sin x \sin (x+h)}\right]\\=\lim _{h \rightarrow 0} \dfrac{\sin h}{h} \cdot \lim _{n \rightarrow 0}\left[\dfrac{1}{\sin x \sin (x+h)}\right]\\=-1 \cdot \dfrac{1}{\sin x \sin (x+h)}=\dfrac{-1}{\sin ^{2} x}=-\operatorname{cosec}^{2} x \quad \ldots .(2)\\\lim _{n \rightarrow 0} \dfrac{1}{h}[\operatorname{cosec}(x+h)-\operatorname{cosec} x]=\lim _{n \rightarrow 0} \dfrac{1}{h}\left[\dfrac{1}{\sin (x+h)}-\dfrac{1}{\sin x}\right]\\=\lim _{n \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\sin x-\sin (x+h)}{\sin x \sin (x+h)}\right]\\=\lim _{h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{2 \cos \left(\dfrac{x+x+h}{2}\right) \cdot \sin \left(\dfrac{x-x-h}{2}\right)}{\sin x \sin (x+h)}\right]\\=\lim _{h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{2 \cos \left(\dfrac{2 x+h}{2}\right) \cdot \sin \left(\dfrac{-h}{2}\right)}{\sin x \sin (x+h)}\right]\\=\lim _{h \rightarrow 0} \dfrac{-\cos \left(\dfrac{2 x+h}{2}\right) \cdot \dfrac{\sin \left(\dfrac{-h}{2}\right)}{\left(\dfrac{h}{2}\right)}}{\sin x \sin (x+h)}\\=\lim _{h \rightarrow 0}\left(\dfrac{-\cos \left(\dfrac{2 x+h}{2}\right)}{\sin x \sin (x+h)}\right) \cdot \lim _{n \rightarrow 0} \dfrac{\sin \left(\dfrac{h}{2}\right)}{\left(\dfrac{h}{2}\right)}\\=\left(\dfrac{-\cos x}{\sin x \sin x}\right) \cdot 1\\=-\operatorname{cosec} x \cot x\\\text { From (1), (2), and (3), we obtain }\\f^{\prime}(x)=-3 \operatorname{cosec}^{2} x-5 \operatorname{cosec} x \cot x\end{array}$

(vi) Let $\mathrm{f}(\mathrm{x})=5 \sin \mathrm{x}-6 \cos \mathrm{x}+7$. Accordingly, from the first principle, $f^{\prime}(x)=\lim _{n \rightarrow 0} \dfrac{f(x+h)-f(x)}{h}$

$=\lim _{h \rightarrow 0} \dfrac{1}{h}[5 \sin (x+h)-6 \cos (x+h)+7-5 \sin x+6 \cos x-7]$

$=5 \lim _{n \rightarrow 0} \dfrac{1}{h}[\sin (x+h)-\sin x]-6 \lim _{n \rightarrow 0} \dfrac{1}{h}[\cos (x+h)-\cos x]$

$=5 \lim _{n \rightarrow 0} \dfrac{1}{h}\left[2 \cos \left(\dfrac{x+h+x}{2}\right) \cdot \sin \left(\dfrac{x+h-x}{2}\right)\right]-6 \lim _{n \rightarrow 0} \dfrac{\cos x \cos h-\sin x \sinh -\cos x}{h}$

$=5 \lim _{n \rightarrow 0} \dfrac{1}{h}\left[2 \cos \left(\dfrac{2 x+h}{2}\right) \cdot \sin \left(\dfrac{h}{2}\right)\right]-6 \lim _{n \rightarrow 0}\left[\dfrac{-\cos x(1-\cos h)-\sin x \sin h}{h}\right]$

$=5 \lim _{n \rightarrow 0} \dfrac{1}{h}\left[\cos \left(\dfrac{2 x+h}{2}\right) \cdot \dfrac{\sin \left(\dfrac{h}{2}\right)}{\left(\dfrac{h}{2}\right)}\right]-6 \lim _{n \rightarrow 0}\left[\dfrac{-\cos x(1-\cos h)}{h}-\dfrac{\sin x \sin h}{h}\right]$

$=5\left[\lim _{n \rightarrow 0} \cos \left(\dfrac{2 x+h}{2}\right)\right]\left[\lim _{h \rightarrow 0} \dfrac{\sin \left(\dfrac{h}{2}\right)}{\left(\dfrac{h}{2}\right)}\right]-6\left[-\cos x\left(\lim _{n \rightarrow 0} \dfrac{1-\cos h}{h}\right)-\sin x\left(\lim _{n \rightarrow 0} \dfrac{\sin h}{h}\right)\right]$

$=5 \cos x \cdot 1-6[(-\cos x) \cdot(0)-\sin x \cdot 1]$

$=5 \cos x+6 \sin x$

(vii) Let $f(x)=2 \tan x-7 \sec x .$ Accordingly, from the first principle, $f^{\prime}(x)=\lim _{n \rightarrow 0} \dfrac{f(x+h)-f(x)}{h}$

$=\lim _{n \rightarrow 0} \dfrac{1}{h}[2 \tan (x+h)-7 \sec (x+h)-2 \tan x+7 \sec x]$

$=2 \lim _{n \rightarrow 0} \dfrac{1}{h}[\tan (x+h)-\tan x]-7 \lim _{n \rightarrow 0} \dfrac{1}{h}[\sec (x+h)-\sec x]$

$=2 \lim _{h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\sin (x+h)}{\cos (x+h)}-\dfrac{\sin x}{\cos x}\right]-7 \lim _{n \rightarrow 0} \dfrac{1}{h}\left[\dfrac{1}{\operatorname{cosec}(x+h)}-\dfrac{1}{\operatorname{cosec} x}\right]$

$=2 \lim _{n \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\cos x \sin (x+h)-\sin x \cos (x+h)}{\cos x \cos (x+h)}\right]-7 \lim _{n \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\cos x-\cos (x+h)}{\cos x \cos (x+h)}\right]$

$=2 \lim _{h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\sin x+h-x}{\cos x \cos (x+h)}\right]-7 \lim _{n \rightarrow 0} \dfrac{1}{h}\left[\dfrac{-2 \sin \left(\dfrac{x+x+h}{2}\right) \sin \left(\dfrac{x-x-h}{2}\right)}{\cos x \cos (x+h)}\right]$

$=2\left[\lim _{n \rightarrow 0}\left(\dfrac{\sin h}{h}\right) \dfrac{1}{\cos x \cos (x+h)}\right]-7 \lim _{n \rightarrow 0} \dfrac{1}{h}\left[\dfrac{-2 \sin \left(\dfrac{2 x+h}{2}\right) \sin \left(\dfrac{-h}{2}\right)}{\cos x \cos (x+h)}\right]$

$=2\left(\lim _{h \rightarrow 0} \dfrac{\sin h}{h}\right)\left[\lim _{h \rightarrow 0} \dfrac{1}{\cos x \cos (x+h)}\right]-7\left(\lim _{h \rightarrow 0} \dfrac{\sin \left(\dfrac{h}{2}\right)}{\dfrac{h}{2}}\right)\left(\lim _{h \rightarrow 0} \dfrac{\sin \left(\dfrac{2 x+h}{2}\right)}{\cos x \cos (x+h)}\right)$

$=2 \cdot 1 \cdot 1 \dfrac{1}{\cos x \cos x}-7 \cdot 1\left(\dfrac{\sin x}{\cos x \cos x}\right)$

$=2 \sec ^{2} x-7 \sec x \tan x$

### Miscellaneous Exercise

1: Find the derivative of the following functions from first principle:

(i) -x

(ii) $\mathbf{(-x)^{-1}}$

(iii) $\mathbf{\sin (x+1)}$

(iv) $\mathbf{\cos \left(x-\dfrac{\pi}{8}\right)}$

Ans: (i) Let $f(x)=-x$. Accordingly, $f(x+h)=-(x+h)$

By first principle, $f^{\prime}(x)=\lim _{n \rightarrow 0} \dfrac{f(x+h)-f(x)}{h}$

$=\lim _{h \rightarrow 0} \dfrac{-(x+h)-(-x)}{h}$

$=\lim _{n \rightarrow 0} \dfrac{-x-h+x}{h}$

$=\lim _{n \rightarrow 0} \dfrac{-h}{h}$

$=\lim _{h \rightarrow 0}(-1)=-1$

(ii) Let $f(x)=(-x)^{-1}=\dfrac{1}{-x}=\dfrac{-1}{x} .$ Accordingly, $f(x+h)=\dfrac{-1}{(x+h)}$

By first principle,

$f^{\prime}(x)=\lim _{n \rightarrow 0} \dfrac{f(x+h)-f(x)}{h}$

$=\lim _{h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{-1}{(x+h)}-\left(\dfrac{-1}{x}\right)\right]$

$=\lim _{n \rightarrow 0} \dfrac{1}{h}\left[\dfrac{-x+(x+h)}{x(x+h)}\right]$

$=\lim _{h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{h}{x(x+h)}\right]$

$=\lim _{n \rightarrow 0} \dfrac{1}{x(x+h)}$

$=\dfrac{1}{x \cdot X}=\dfrac{1}{x^{2}}$

(iii) Let $f(x)=\sin (x+1)$. Accordingly, $f(x+h)=\sin (x+h+1)$

By first principle,

$f^{\prime}(x)=\lim _{n \rightarrow 0} \dfrac{f(x+h)-f(x)}{h}$

$=\lim _{n \rightarrow 0} \dfrac{1}{h}[\sin (x+h+1)-\sin (x+1)]$

$=\lim _{n \rightarrow 0} \dfrac{1}{h}\left[2 \cos \left(\dfrac{x+h+1+x+1}{2}\right) \sin \left(\dfrac{x+h+1-x-1}{2}\right)\right]$

$=\lim _{n \rightarrow 0} \dfrac{1}{h}\left[2 \cos \left(\dfrac{2 x+h+2}{2}\right) \sin \left(\dfrac{h}{2}\right)\right]$

$=\lim _{n \rightarrow 0}\left[\cos \left(\dfrac{2 x+h+2}{2}\right) \cdot \dfrac{\sin \left(\dfrac{h}{2}\right)}{\left(\dfrac{h}{2}\right)}\right]$

$\begin{array}{l}=\lim _{n \rightarrow 0} \dfrac{1}{h} \cos \left(\dfrac{2 x+h+2}{2}\right) \cdot \lim _{\dfrac{b}{2} \rightarrow} \dfrac{\sin \left(\dfrac{h}{2}\right)}{h} \dfrac{h}{\left.\dfrac{h}{2}\right)} \quad\left[\text { As } h \rightarrow 0 \Rightarrow \dfrac{h}{2} \rightarrow 0\right]\\=\cos \left(\dfrac{2 x+0+2}{2}\right) \cdot 1 \quad\left[\lim _{h \rightarrow 0} \dfrac{\sin x}{x}=1\right]\\=\cos (x+1)\\\text { (iv) Let } f(x)=\cos \left(x-\dfrac{\pi}{8}\right) \text {. Accordingly, } f(x+h)-\cos \left(x+h-\dfrac{\pi}{8}\right)\\\text { By first principle, }\\f^{\prime}(x)=\lim _{n \rightarrow 0} \dfrac{f(x+f)-f(x)}{h}\\=\lim _{h \rightarrow 0} \dfrac{1}{h}\left[\cos \left(x+h-\dfrac{\pi}{8}\right)-\cos \left(x-\dfrac{\pi}{8}\right)\right]\\=\lim _{n \rightarrow 0} \dfrac{1}{h}\left[-2 \sin \dfrac{\left(x+h-\dfrac{\pi}{8}+x-\dfrac{\pi}{8}\right)}{2} \sin \left(\dfrac{x+h-\dfrac{\pi}{8}-x+\dfrac{\pi}{8}}{2}\right)\right]\\=\lim _{n \rightarrow 0} \dfrac{1}{h}\left[-2 \sin \left(\dfrac{2 x+h-\dfrac{\pi}{4}}{2}\right) \sin \left(\dfrac{h}{2}\right)\right]\\=\lim _{n \rightarrow 0}\left[-\sin \left(\dfrac{2 x+h-\dfrac{\pi}{4}}{2}\right) \dfrac{\sin \left(\dfrac{h}{2}\right)}{\left(\dfrac{h}{2}\right)}\right]\\\left.=\lim _{n \rightarrow 0}\left[-\sin \left(\dfrac{2 x+h-\dfrac{\pi}{4}}{2}\right)\right] \cdot \lim _{\dfrac{\pi}{2} \rightarrow 0} \dfrac{\sin \left(\dfrac{h}{2}\right)}{\left(\dfrac{h}{2}\right)} \quad \text { [As } h \rightarrow 0 \Rightarrow \dfrac{h}{2} \rightarrow 0\right]\\=-\sin \left(\dfrac{2 x+0-\dfrac{\pi}{4}}{2}\right) \cdot 1\end{array}$

$=-\sin \left(x-\dfrac{\pi}{8}\right)$

2: Find the derivative of the following functions (it is to be understood that a, b, c, d. p, q, r and s are fixed non-zero constants and $m$ and $n$ are integers): $(x+a)$

Ans: Let $f(x)=x+a$. Accordingly. $f(x+h)-x+h+a$ By first principle,

$f^{\prime}(x)=\lim _{n \rightarrow 0} \dfrac{f(x+h)-f(x)}{h}$

$\lim _{n \rightarrow 0} \dfrac{x+h+a-x-a}{h}$

$\lim _{n \rightarrow 0}\left(\dfrac{h}{h}\right)$

$-\lim _{n \rightarrow 0}(1)$

$=1$

3: Find the derivative of the following functions (it is to be understood that $a, b, c$. d, $p, q, r$ and s are fixed non- zero constants and $m$ and $n$ are integers): $\mathbf{(p x+q)\left(\dfrac{r}{x}+s\right)}$

Ans: Let $f(x)=(p x+q)\left(\dfrac{r}{x}+s\right)$

By Leibnitz product rule.

$f^{\prime}(x)=(p x+q)\left(\dfrac{r}{x}+s\right)^{\prime}+\left(\dfrac{r}{x}+s\right)(p x+q)^{\prime}$

$-(p x+q)\left(r x^{-1}+s\right)^{\prime}+\left(\dfrac{r}{x}+s\right)(p)$

$-(p x+q)\left(-n x^{2}\right)+\left(\dfrac{r}{x}+s\right) p$

$=(p x+q)\left(\dfrac{-r}{x^{2}}\right)+\left(\dfrac{r}{x}+s\right) p$

$=\dfrac{-p x}{x}-\dfrac{q r}{x^{2}}+\dfrac{p r}{x}+p s$

$p s-\dfrac{q r}{x^{2}}$

4: Find the derivative of the following functions (it is to be understood that $a, b, c$, d, $p, q, r$ and $s$ are fixed non-zero constants and $m$ and $n$ are integers): $\mathbf{(a x+b)(c x+d)^{2}}$

Ans: Let $f^{\prime}(x)=(a x+b)(c x+d)^{2}$

By Leibnitz product rule,

$f^{\prime}(x)=(a x+b) \dfrac{d}{d x}(c x+d)^{2} \dfrac{d}{d x}(a x+b)$

$(a x+b) \dfrac{d}{d x}\left(c^{2} x^{2}+2 c d x^{2}\right)+(c x+d)^{2} \dfrac{d}{d x}(a x+b)$

$(a x+b)\left[\dfrac{d}{d x}\left(c^{2} x^{2}\right)+\dfrac{d}{d x}(2 c d x)+\dfrac{d}{d x} d^{2}\right]+(c x+d)^{2}\left[\dfrac{d}{d x} a x+\dfrac{d}{d x} b\right]$

$=(a x+b)\left(2 c^{2} x+2 c d\right)+(c x+d)^{2} a$

$-2 c(a x+b)(c x+d)+a(c x+d)^{2}$

5: Find the derivative of the following functions (it is to be understood that $a, b, c$, d, $p, q, r$ and $s$ are fixed non zero constants and $m$ and $n$ are integers): $\mathbf{\dfrac{a x+b}{c x+d}}$

Ans: Let $f(x)=\dfrac{a x+b}{c x+d}$

By quotient rule,

$f(x)=\dfrac{(c x+d) \dfrac{d}{d x}(a x+b)-(a x+b) \dfrac{d}{d x}(c x+d)}{(c x+d)^{2}}$

$=\dfrac{(c x+d)(a)-(a x+d)(c)}{(c x+d)^{2}}$

$\dfrac{a c x+a d-a c x-b c}{(c x+d)^{2}}$

$\dfrac{a d-b c}{(c x+d)^{2}}$

6: Find the derivative of the following functions (it is to be understood that $a, b, c$, d, $p, q, r$ and $s$ are fixed non-zero constants and $m$ and $n$ are integers): $\mathbf{\dfrac{1+\dfrac{1}{x}}{1-\dfrac{1}{x}}}$

Ans: Let$f(x)=\dfrac{1+\dfrac{1}{x}}{1-\dfrac{1}{x}}=\dfrac{\dfrac{x+1}{x}}{\dfrac{x-1}{x}}=\dfrac{x+1}{x-1}$, where $x \neq 0$

By quotient rule, $f^{\prime}(x)=\dfrac{(x-1) \dfrac{d}{d x}(x-1)-(x+1) \dfrac{d}{d x}(x-1)}{(x-1)^{2}}, x \neq 0,1$

$=\dfrac{(x-1)(1)-(x+1)(1)}{(x-1)^{2}}, x \neq 0,1$

$\dfrac{x-1-x-1}{(x-1)^{2}}, x \neq 0,1$

$\dfrac{-2}{(x-1)^{2}}, x \neq 0,1$

7: Find the derivative of the following functions (it is to be understood that $a, b, c$, d, $p, q, r$ and s are fixed non-zero constants and $m$ and $n$ are integers) $\mathbf{: \dfrac{1}{a x^{2}+b x+c}}$

Ans: Let $f(x)=\dfrac{1}{a x^{2}+b x+c}$

By quotient rule,

$f^{\prime}(x)=\dfrac{\left(a x^{2}+b x+c\right) \dfrac{d}{d x}(1)-\dfrac{d}{d x}\left(a x^{2}+b x+c\right)}{\left(a x^{2}+b x+c\right)^{2}}$

$\dfrac{\left(a x^{2}+b x+c\right)(0)-(2 a x+b)}{\left(a x^{2}+b x+c\right)^{2}}$

$\dfrac{-(2 a x+b)}{\left(a x^{2}+b x+c\right)^{2}}$

8: Find the derivative of the following functions (it is to be understood that $a, b, c$ d, p, q, rand s are fixed non-zero constants and m and $n$ are integers): $\mathbf{\dfrac{a x+b}{p x^{2}+q x+r}}$

Ans: Let $f(x)=\dfrac{a x+b}{p x^{2}+q x+r}$

By quotient rule,

$f^{\prime}(x)=\dfrac{\left(p x^{2}+q x+r\right) \dfrac{d}{d x}(a x+b)-(a x+b) \dfrac{d}{d x}\left(p x^{2}+q x+r\right)}{\left(p x^{2}+q x+r\right)^{2}}$

$=\dfrac{\left(p x^{2}+q x+r\right)(a)-(a x+b)(2 p x+q)}{\left(p x^{2}+q x+r\right)^{2}}$

$=\dfrac{a p x^{2}+a q x+a r-a q x+2 n p x+b q}{\left(p x^{2}+q x+r\right)^{2}}$

$\dfrac{-a p x^{2}+2 b p x+a r-b q}{\left(p x^{2}+q x+r\right)^{2}}$

9: Find the derivative of the following functions (it is to be understood that $a, b, c$. d, $p$, q, $r$ and s are fixed non-zero constants and $m$ and $n$ are integers): $\mathbf{\dfrac{p x^{2}+q x+r}{a x+b}}$

Ans: Let $f(x)=\dfrac{p x^{2}+q x+r}{a x+b}$

By quotient rule,

$\dot{f}(x)=\dfrac{(a x+b) \dfrac{d}{d x}\left(p x^{2}+q x+\eta\right)-\left(p x^{2}+q x+r\right) \dfrac{d}{d x}(a x+b)}{(a x+b)^{2}}$

$\dfrac{(a x+b)(2 p x+q)-\left(p x^{2}+q x+r\right)(a)}{(a x+b)^{2}}$

$=\dfrac{2 a p x^{2}+a q x+2 b p x+b q-a q x^{2}-a q x-a r}{(a x+b)^{2}}$

$\dfrac{a p x^{2}+2 b p x+b q-a r}{(a x+b)^{2}}$

10: Find the derivative of the following functions (it is to be understood that $a, b, c$, d, $p$, q, $r$ and $s$ are fixed non-zero constants and $m$ and $n$ are integers): $\mathbf{\dfrac{a}{x^{4}}-\dfrac{b}{x^{2}}+\cos x}$

Ans: Let $f(x)=\dfrac{a}{x^{4}}-\dfrac{b}{x^{2}}+\cos x$

$f^{\prime}(x)=\dfrac{d}{d x}\left(\dfrac{a}{x^{4}}\right)-\dfrac{d}{d x}\left(\dfrac{a}{x^{2}}\right)+\dfrac{d}{d x}(\cos x)$

$a \dfrac{d}{d x}\left(x^{-4}\right)-b \dfrac{d}{d x}\left(x^{2}\right)+\dfrac{d}{d x}(\cos x)$

$-a\left(-4 x^{-5}\right)-b\left(-2 x^{3}\right)+(-\sin x) \quad\left[\dfrac{d}{d x}\left(x^{n}\right)=n x^{n-1}\right.$ and $\left.\dfrac{d}{d x}(\cos x)=-\sin x\right]$

$\dfrac{-4 a}{x^{5}}+\dfrac{2 b}{x^{3}}-\sin x$

11: Find the derivative of the following functions (it is to be understood that $a, b, c$, d, $p, q, r$ and $s$ are fixed nonzero constants and $m$ and $n$ are integers): $\mathbf{4 \sqrt{x}-2}$

Ans: Let $f(x)=4 \sqrt{x}-2$

$f^{\prime}(x)=\dfrac{d}{d x}(4 \sqrt{x}-2)=\dfrac{d}{d x}(4 \sqrt{x})-\dfrac{d}{d x}(2)$

$=4 \dfrac{d}{d x}\left(x^{\dfrac{1}{2}}\right)-0=4\left(\dfrac{1}{2} x^{\dfrac{1}{2}}\right)$

$=\left(2 x^{-\dfrac{1}{2}}\right)=\dfrac{2}{\sqrt{x}}$

12: Find the derivative of the following functions (it is to be understood that $a, b, c$, d, $p, q, r$ and $s$ are fixed non-zero constants and $m$ and $n$ are integers): $\mathbf{(a x+b)^{n}}$

Ans: Let $f(x)=(a x+b)^{n} .$ Accordingly, $f(x+h)-\{a(x+h)+b\}^{n}-(a x+a h+b)^{n}$

By first principle,

$f^{\prime}(x)=\lim _{n \rightarrow 0} \dfrac{f(x+h)-f(x)}{h}$

$=\lim _{h \rightarrow 0} \dfrac{(a x+a h+b)-(a x+b)^{n}}{h}$

$=\lim _{h \rightarrow 0} \dfrac{(a x+b)^{n}\left(1+\dfrac{a h}{a x+b}\right)^{n}-(a x+b)^{n}}{h}$

$=(a x+b)^{n} \lim _{n \rightarrow 0} \dfrac{1}{h}\left[\left\{1+n\left(\dfrac{a h}{a x+b}\right)+\dfrac{n(n-1)}{2}\left(\dfrac{a h}{a x+b}\right)^{2}+\cdots\right\}-1\right] \quad$ (using binomial theorem)

$=(a x+b)^{n} \lim _{n \rightarrow 0} \dfrac{1}{h}\left[\left(\dfrac{a h}{a x+b}\right)+\dfrac{n(n-1) a^{2} h^{2}}{2(a x+b)^{2}}+\cdots\right.$ (Terms cortaining higher degrees of $\left.\left.h\right)\right]$

$=(a x+b)^{n} \lim _{n \rightarrow 0}\left[\dfrac{n a}{(a x+b)}+\dfrac{n(n-1) \nexists^{7} h^{2}}{2(a x+b)^{2}}+\cdots\right]$

$=(a x+b)^{n}\left[\dfrac{n a}{(a x+b)}+0\right]$

$=n a \dfrac{(a x+b)^{n}}{a x+b}$

$-n a(a x+b)^{n-1}$

13: Find the derivative of the following functions (it is to be understood that $a, b, c$, d, $p, q, r$ and $s$ are fixed non-zero constants and $m$ and $n$ are integers): $\mathbf{(a x+b)^{n}(c x+d)^{m}}$

Ans: Let $f(x)=(a x+b)^{n}(c x+d)^{m}$

By Leibnitz product rule,

$f^{\prime}(x)=(a x+b)^{n} \dfrac{d}{d x}(c x+d)^{m}+(c x+d)^{m} \dfrac{d}{d x}(a x+b)^{n}$

Now let $f_{1}(x)=(c x+d)^{m}$

$f_{1}(x+h)=(c x+c h+d)^{m}$

$f_{1}^{\prime}(x)=\lim _{n \rightarrow 0} \dfrac{f_{1}(x+h)-f_{1}(x)}{h}$

$=\lim _{n \rightarrow 0} \dfrac{(c x+c h+d)^{m}-(c x+d)^{m}}{h}$

$=(c x+d)^{m} \lim _{n \rightarrow 0} \dfrac{1}{h}\left[\left(1+\dfrac{c h}{c x+d}\right)^{m}-1\right]$

$=(c x+d)^{m} \lim _{h \rightarrow 0} \dfrac{1}{h}\left[\left(1+\dfrac{m c h}{(c x+d)}+\dfrac{m(m-1)}{2} \dfrac{c^{2} h^{2}}{(c x+d)^{2}}+\cdots\right)^{m}-1\right]$

$=(c x+d)^{m} \lim _{n \rightarrow 0} \dfrac{1}{h}\left[\dfrac{m c h}{(c x+d)}+\dfrac{m(m-1) c^{2} h^{2}}{2(c x+d)^{2}}+\cdots\right.$ (Terms containing higher degree oh $\left.\left.h\right)\right]$

$=(c x+d)^{m} \lim _{h \rightarrow 0}\left[\dfrac{m c}{(c x+d)}+\dfrac{m(m-1) c^{2} h^{2}}{2(c x+d)^{2}}+\cdots\right]$

$=(C x+a)^{m}\left[\dfrac{m c h}{(c x+d)}+0\right]$

$=\dfrac{m c(c x+d)^{m}}{(c x+d)}$

$=m c(c x+d)^{m-1}$

$\dfrac{d}{d x}(c x+d)^{m}=m d(x+d)^{m-1}$

Similarly, $\dfrac{d}{d x}(a x+b)^{n}=n a(a x+b)^{n-1}$

... (3)

Therefore, from (1), (2), and (3), we obtain

$f^{\prime}(x)=(a x+b)^{n}\left\{m c(c x+d)^{m-1}\right\}+(c+d)^{m}\left\{n a(a x+b)^{n-1}\right\}$

$=(a x+b)^{n-1}(c x+d)^{m-1}[m c(a x+b)+n a(c x+d)]$

14: Find the derivative of the following functions (it is to be understood that $a, b, c$. d, $p, q, r$ and s are fixed non-zero constants and $m$ and $n$ are integers): $\mathbf{\sin (x+a)}$

Ans: Let, $f(x)=\sin (x+a)$

$f(x+h)=\sin (x+h+a)$

By first principle, $f^{\prime}(x)=\lim _{n \rightarrow 0} \dfrac{f(x+h)-f(x)}{h}$

$=\lim _{n \rightarrow 0} \dfrac{\sin (x+h+a)-\sin (x+a)}{h}$

$=\lim _{n \rightarrow 0} \dfrac{1}{h}\left[2 \cos \left(\dfrac{x+h+a+x+a}{2}\right) \sin \left(\dfrac{x+h+a-x-a}{2}\right)\right]$

$=\lim _{n \rightarrow 0} \dfrac{1}{h}\left[2 \cos \left(\dfrac{2 x+2 a+h}{2}\right) \sin \left(\dfrac{h}{2}\right)\right]$

$=\lim _{h \rightarrow 0}\left[\cos \left(\dfrac{2 x+2 a+h}{2}\right)\left[\dfrac{\sin \left(\dfrac{h}{2}\right)}{\left(\dfrac{h}{2}\right)}\right]\right]$

$=\lim _{h \rightarrow 0} \cos \left(\dfrac{2 x+2 a+h}{2}\right) \cdot \lim _{\dfrac{h}{2} \rightarrow 0}\left[\dfrac{\sin \left(\dfrac{h}{2}\right)}{\left(\dfrac{h}{2}\right)}\right] \quad\left[\right.$ As $\left.h \rightarrow 0 \Rightarrow \dfrac{h}{2} \rightarrow 0\right]$

$=\cos \left(\dfrac{2 x+2 a}{2}\right) \times 1 \quad\left[\lim _{n \rightarrow 0} \dfrac{\sin x}{x}=1\right]$

$=\cos (x+a)$

15: Find the derivative of the following functions (it is to be understood that $a, b, c$, d, $p, q, r$ and $s$ are fixed non-zero constants and $m$ and $n$ are integers): $\mathbf{\operatorname{cosec} x \cot x}$

Ans: Let $f(x)=\operatorname{cosec} x \cot x$

By Leibnitz product rule,

$f^{\prime}(x)=\operatorname{cosec} x(\cot x)^{\prime}+\cot x(\operatorname{cosec} x)^{\prime} \ldots .(1)$

Let $f_{1}(x)=\cot x .$ Accordingly, $f_{1}(x+h)=\cot (x+h)$

By first principle,

$f^{\prime \prime}(x)=\lim _{n \rightarrow 0} \dfrac{f(x+h)-f(x)}{h}$

$=\lim _{h \rightarrow 0} \dfrac{\cot (x+h)-\cot (x)}{h}$

$=\lim _{h \rightarrow 0} \dfrac{1}{h}\left(\dfrac{\cos (x+h)}{\sin (x+h)}-\dfrac{\cos (x)}{\sin x}\right)$

$=\lim _{n \rightarrow 0} \dfrac{1}{h}\left(\dfrac{\sin x \cos (x+h)-\cos x \sin (x+h)}{\sin x \sin (x+h)}\right)$

$=\lim _{h \rightarrow 0} \dfrac{1}{h}\left(\dfrac{\sin (x-x+h)}{\sin x \sin (x+h)}\right)$

$=\dfrac{1}{\sin x^{n \rightarrow 0}} \lim _{h} \dfrac{1}{h}\left[\dfrac{\sin (-h)}{\sin (x+h)}\right]$

$=\dfrac{-1}{\sin x}\left(\lim _{n \rightarrow 0} \dfrac{\sin h}{h}\right)\left(\lim _{n \rightarrow 0} \dfrac{1}{\sin (x+h)}\right)$

$=\dfrac{-1}{\sin x} \cdot 1 \cdot\left(\lim _{n \rightarrow 0} \dfrac{1}{\sin (x+0)}\right)$

$=\dfrac{-1}{\sin ^{2} x}$

$=-\operatorname{cosec}^{2} x$

$\therefore(\cot x)^{\prime}=-\operatorname{cosec}^{2} x \quad \ldots$ (2)

Now, let $f_{2}(x)=\operatorname{cosec} x .$ Accordingly, $f_{2}(x+h)=\operatorname{cosec}(x+h)$

By first principle, $f_{2}(x)=\lim _{n \rightarrow 0} \dfrac{f_{2}(x+h)-f_{2}(x)}{h}$

$=\lim _{h \rightarrow 0} \dfrac{1}{h}[\operatorname{cosec}(x+h)-\operatorname{cosec}(x)]$

$=\lim _{n \rightarrow 0} \dfrac{1}{h}\left(\dfrac{1}{\sin (x+h)}-\dfrac{1}{\sin x}\right)$

$=\lim _{n \rightarrow 0} \dfrac{1}{h}\left(\dfrac{\sin x-\sin (x+h)}{\sin x \sin (x+h)}\right)$

$=\dfrac{1}{\sin x} \cdot \lim _{h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{2 \cos \left(\dfrac{x+x+h}{2}\right) \sin \left(\dfrac{x-x-h}{2}\right)}{\sin (x+h)}\right]$

$=\dfrac{1}{\sin x} \cdot \lim _{n \rightarrow 0} \dfrac{1}{h}\left[\dfrac{2 \cos \left(\dfrac{2 x+h}{2}\right) \sin \left(\dfrac{-h}{2}\right)}{\sin (x+h)}\right]$

$=\dfrac{1}{\sin x} \cdot \lim _{h \rightarrow 0}\left[\dfrac{-\sin \left(\dfrac{h}{2}\right)}{\left(\dfrac{h}{2}\right)} \cdot \dfrac{\cos \left(\dfrac{2 x+h}{2}\right)}{\sin (x+h)}\right]$

$=\dfrac{-1}{\sin x} \cdot \lim _{h \rightarrow 0} \dfrac{\sin \left(\dfrac{h}{2}\right)}{\left(\dfrac{h}{2}\right)} \cdot \lim _{n \rightarrow 0} \dfrac{\cos \left(\dfrac{2 x+h}{2}\right)}{\sin (x+h)}$

$=\dfrac{-1}{\sin x} \cdot 1 \cdot \dfrac{\cos \left(\dfrac{2 x+h}{2}\right)}{\sin (x+0)}$

$=\dfrac{-1}{\sin x} \cdot \dfrac{\cos x}{\sin x}$

$=-\operatorname{cosec} x \cdot \cot x$

$\therefore(\operatorname{cosec} x)^{\prime}=-\operatorname{cosec} x \cdot \cot x$

From (1), (2), and (3), we obtain

$f^{\prime}(x)=\operatorname{cosec} x\left(-\operatorname{cosec}^{2} x\right)+\cot x(-\operatorname{cosec} x \cot x)$

$=-\operatorname{cosec}^{3} x-\cot ^{2} x \operatorname{cosec} x$

16: Find the derivative of the following functions (it is to be understood that $a, b, c$, d, $p, q, r$ and $s$ are fixed non-zero constants and $m$ and $n$ are integers): $\mathbf{\dfrac{\cos x}{1+\sin x}}$

Ans: Let $f(x)=\dfrac{\cos x}{1+\sin x}$

By quotient rule,

$f^{\prime}(x)=\dfrac{(1+\sin x) \dfrac{d}{d x}(\cos x)-(\cos x) \dfrac{d}{d x}(1+\sin x)}{(1+\sin x)^{2}}$

$=\dfrac{(1+\sin x)(-\sin x)-(\cos x)(\cos x)}{(1+\sin x)^{2}}$

$=\dfrac{-\sin x-\sin ^{2} x-\cos ^{2} x}{(1+\sin x)^{2}}$

$=\dfrac{-\sin x-\left(\sin ^{2} x+\cos ^{2} x\right)}{(1+\sin x)^{2}}$

$=\dfrac{-\sin x-1}{(1+\sin x)^{2}}$

$=\dfrac{-(1-\sin x)}{(1+\sin x)^{2}}$

$=\dfrac{-1}{(1+\sin x)^{2}}$

17: Find the derivative of the following functions (it is to be understood that $a, b, c$, d, $p, q, r$ and $s$ are fixed non zero constants and m and n are integers): $\mathbf{\dfrac{\sin x+\cos x}{\sin x-\cos x}}$

Ans:17: Let $f(x)=\dfrac{\sin x+\cos x}{\sin x-\cos x}$

By quotient rule,

$f^{\prime \prime}(x)=\dfrac{(\sin x-\cos x) \dfrac{d}{d x}(\sin x+\cos x)-(\sin x+\cos x) \dfrac{d}{d x}(\sin x-\cos x)}{(\sin x+\cos x)^{2}}$

$=\dfrac{(\sin x-\cos x)(\cos x-\sin x)-(\sin x+\cos x)(\cos x+\sin x)}{(\sin x+\cos x)^{2}}$

$=\dfrac{-(\sin x-\cos x)^{2}-(\sin x+\cos x)^{2}}{(\sin x+\cos x)^{2}}$

$=\dfrac{-\left[\sin ^{2} x+\cos ^{2} x-2 \sin x \cos x+\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x\right]}{(\sin x+\cos x)^{2}}$

$=\dfrac{-[1+1]}{(\sin x-\cos x)^{2}}$

$=\dfrac{-2}{(\sin x-\cos x)^{2}}$

18: Find the derivative of the following functions (it is to be understood that $a, b, c$, d. $p, q, r$ and $s$ are fixed non-zero constants and $m$ and $n$ are integers): $\mathbf{\dfrac{\sec x-1}{\sec x+1}}$

Ans: Let $f(x)=\dfrac{\sec x-1}{\sec x+1}$

$f(x)=\dfrac{\dfrac{1}{\cos x}-1}{\dfrac{1}{\cos x}+1}=\dfrac{1-\cos x}{1+\cos x}$

By quotient rule,

$f^{\prime}(x)=\dfrac{(1+\cos x) \dfrac{d}{d x}(1-\cos x)-(1-\cos x) \dfrac{d}{d x}(1+\cos x)}{(1+\cos x)^{2}}$

$=\dfrac{(1+\cos x)(\sin x)-(1-\cos x)(-\sin x)}{(1+\cos x)^{2}}$

$=\dfrac{\sin x+\cos x \sin x+\sin x-\sin x \cos x}{(1+\cos x)^{2}}$

$=\dfrac{2 \sin x}{(1+\cos x)^{2}}$

$=\dfrac{2 \sin x}{\left(1+\dfrac{1}{\sec x}\right)^{2}}=\dfrac{2 \sin x}{\dfrac{(\sec x+1)^{2}}{\sec ^{2} x}}$

$=\dfrac{2 \sin x \sec ^{2} x}{(\sec x+1)^{2}}$

$=\dfrac{\dfrac{2 \sin x}{\cos x} \sec x}{(\sec x+1)^{2}}$

$=\dfrac{2 \sec x \tan x}{(\sec x+1)^{2}}$

19: Find the derivative of the following functions (it is to be understood that $a, b, c$, d. $p, q, r$ and s are fixed non-zero constants and $m$ and $n$ are integers): $\mathbf{\sin ^{n} x}$

Ans: Let $y=\sin ^{n} x$

Accordingly, for $n=1, y=\sin x$

$\therefore \dfrac{d y}{d x}=\cos x$, i.e., $\dfrac{d}{d x} \sin x=\cos x$

For $n=2, y=\sin ^{2} x$.

$\therefore \dfrac{d y}{d x}=\dfrac{d}{d x}(\sin x \sin x)$

$=(\sin x)^{\prime}\left(\sin x+\sin x(\sin x)^{\prime} \quad\right.$ (By Leibnitz product rule)

$=\cos x \sin x+\sin x \cos x$

$=2 \sin x \cos x$

$\ldots .(1)$

For $n=3, y=\sin ^{3} x$

$\therefore \dfrac{d y}{d x}=\dfrac{d}{d x}\left(\sin x \sin ^{2} x\right)$

$=(\sin x)^{\prime} \sin ^{2} x+\sin x(\sin x)^{\prime}$

(By Leibnitz product rule)

$-\cos x \sin ^{2} x+\sin x(2 \sin x \cos x) \quad[$ Using $(1)]$

$=\cos x \sin ^{2} x+\sin ^{2} x \cos x$

$=3 \sin ^{2} x \cos x$

We assert that $\dfrac{d}{d x}\left(\sin ^{n} x\right)=n \sin ^{(n-1)} x \cos x$

Let our assertion be true for $n=k$.

i.e., $\dfrac{d}{d x}\left(\sin ^{k} x\right)=k \sin ^{(k-1)} x \cos x \quad \ldots .$ (2)

Corsider

$\dfrac{d}{d x}\left(\sin ^{k+1} x\right)=\dfrac{d}{d x}\left(\sin x \sin ^{(k)} x\right)$

$=(\sin x)^{\prime} \sin ^{k} x+\sin x\left(\sin ^{k} x\right)^{n}$

(By Leibnitz product rule)

$=\cos x \sin ^{k} x+\sin x\left(k \sin ^{k-1} \cos x\right) \quad[$ Using $(2)]$

$=\cos x \sin ^{k} x+2 \sin ^{k} x \cos x$

$-(k+1) \sin ^{k} x \cos x$

Thus, our assertion is true for $n=k+1$.

Hence, by mathematical induction, $\dfrac{d}{d x}\left(\sin ^{n} x\right)=n \sin ^{(n-1)} x \cos x$

20: Find the derivative of the following functions (it is to be understood that $a, b, c$, d, $p, q, r$ and s are fixed non-zero constants and $m$ and n are integers): $\mathbf{\dfrac{a+b \sin x}{c+d \cos x}}$

Ans: Let $f(x)=\dfrac{a+b \sin x}{c+d \cos x}$

By quotient rule,

$f^{\prime}(x)=\dfrac{(c+d \cos x) \dfrac{d}{d x}(a+b \sin x)-(a+b \sin x) \dfrac{d}{d x}(c+d \cos x)}{(c+d \cos x)^{2}}$

$=\dfrac{(c+d \cos x)(b \cos x)-(a+b \sin x)(-d \sin x)}{(c+d \cos x)^{2}}$

$=\dfrac{c b \cos x+b d \cos ^{2} x+a d \sin x+b d \sin ^{2} x}{(c+d \cos x)^{2}}$

$=\dfrac{b c \cos x+a d \sin x+b d\left(\cos ^{2} x+\sin ^{2} x\right)}{(C+d \cos x)^{2}}$

$=\dfrac{b c \cos x+a d \sin x+b d}{(c+d \cos x)^{2}}$

21: Find the derivative of the following functions (it is to be understood that $a, b, c$, d, $p, q, r$ and s are fixed non-zero constants and m and $n$ are integers): $\mathbf{\dfrac{\sin (x+a)}{\cos x}}$

Ans: Let $f(x)=\dfrac{\sin (x+a)}{\cos x}$

By quotient rule,

$f^{\prime}(x)=\dfrac{\cos x \dfrac{d}{d x}[\sin (x+a)]-\sin (x+a) \dfrac{d}{d x} \cos x}{\cos ^{2} x}$

$f^{\prime}(x)=\dfrac{\cos x \dfrac{d}{d x}[\sin (x+a)]-\sin (x+a) \dfrac{d}{d x}(-\sin x)}{\cos ^{2} x}$

Let $g(x)-\sin (x+a) .$ Accordingly,$g(x+h)=\sin (x+h+a)$

By first principle,

$g^{\prime}(x)=\lim _{h \rightarrow 0} \dfrac{g(x+h)-g(x)}{h}$

$=\lim _{n \rightarrow 0} \dfrac{1}{h}[\sin (x+h+a)-\sin (x+a)]$

$=\lim _{n \rightarrow 0} \dfrac{1}{h}\left[2 \cos \left(\dfrac{x+h+a+x+a}{2}\right) \sin \left(\dfrac{x+h+a-x-a}{2}\right)\right]$

$=\lim _{n \rightarrow 0} \dfrac{1}{h}\left[2 \cos \left(\dfrac{2 x+2 a+h}{2}\right) \sin \left(\dfrac{h}{2}\right)\right]$

$=\lim _{n \rightarrow 0}\left[\cos \left(\dfrac{2 x+2 a+h}{h}\right)\left\{\dfrac{\sin \left(\dfrac{h}{2}\right)}{\left(\dfrac{h}{2}\right)}\right\}\right]$

$=\lim _{n \rightarrow 0} \cos \left(\dfrac{2 x+2 a+h}{h}\right) \cdot \lim _{n \rightarrow 0}\left\{\dfrac{\sin \left(\dfrac{h}{2}\right)}{\left(\dfrac{h}{2}\right)}\right\} \quad\left[\right.$ As $\left.h \rightarrow 0 \Rightarrow \dfrac{h}{2} \rightarrow 0\right]$

$=\left(\cos \dfrac{2 x+2 a}{2}\right) \times 1 \quad\left[\lim _{n \rightarrow 0} \dfrac{\sin h}{h}=1\right]$

$=\cos (x+a) \quad \ldots$ (ii)

From (i) and (ii), we obtain $f^{\prime}(x)=\dfrac{\cos x \cos (x+a)+\sin x \sin (x+a)}{\cos ^{2} x}$

$=\dfrac{\cos (x+a-x)}{\cos ^{2} x}$

$=\dfrac{\cos a}{\cos ^{2} x}$

22: Find the derivative of the following functions (it is to be understood that $a, b, c$, d. $p, q, r$ and $s$ are fixed non-zero constants and $m$ and $n$ are integers) $\mathbf{: x^{4}(5 \sin x-3 \cos x)}$

Ans: Let $f(x)=x^{4}(5 \sin x-3 \cos x)$

Byproduct rule.

$f^{\prime}(x)=x^{4} \dfrac{d}{d x}(5 \sin x-3 \cos x)+(5 \sin x-3 \cos x) \dfrac{d}{d x}\left(x^{4}\right)$

$=x^{4}\left[5 \dfrac{d}{d x}(\sin x)-3 \dfrac{d}{d x}(\cos x)\right]+(5 \sin x-3 \cos x) \dfrac{d}{d x}\left(x^{4}\right)$

$=x^{4}[5 \cos x-3(-\sin x)]+(5 \sin x-3 \cos x)\left(4 x^{3}\right)$

$=x^{3}[5 x \cos x+3 x \sin x+20 \sin x-12 \cos x]$

23: Find the derivative of the following functions (it is to be understood that $a, b, c$, d, p, q, r and s are fixed non-zero constants and m and n are integers): $\mathbf{\left(x^{2}+1\right) \cos x}$

Ans: Let $f(x)=\left(x^{2}+1\right) \cos x$

By product rule.

$f^{\prime}(x)=\left(x^{2}+1\right) \dfrac{d}{d x}(\cos x)+\cos x \dfrac{d}{d x}\left(x^{2}+1\right)$

$=\left(x^{2}+1\right)(-\sin x)+\cos x(2 x)$

$=-x^{2} \sin x-\sin x+2 x \cos x$

24: Find the derivative of the following functions (it is to be understood that $a, b, c$, d, $p, q, r$ and s are fixed non-zero constants and $m$ and $n$ are integers): $\mathbf{\left(a x^{2}+\sin x\right)(p+q) \cos x)}$

Ans: Let $f(x)=\left(a x^{2}+\sin x\right)(p+q \cos x)$

By product rule.

$f^{\prime}(x)=\left(a x^{2}+\sin x\right) \dfrac{d}{d x}(p+q \cos x)+(p+q \cos x) \dfrac{d}{d x}\left(a x^{2}+\sin x\right)$

$=\left(a x^{2}+\sin x\right)(-q \sin x)+(p+q \cos x)(2 a x+\cos x)$

$=-q \sin x\left(a x^{2}+\sin x\right)+(p+q \cos x)(2 a x+\cos x)$

25: Find the derivative of the following functions (it is to be understood that $a, b, c$, d. $p, q, r$ and $s$ are fixed non-zero constants and $m$ and $n$ are integers): $\mathbf{(x+\cos x)(x-\tan x)}$

Ans: Let $f(x)=(x+\cos x)(x-\tan x)$

By product rule,

$f^{\prime}(x)=(x+\cos x) \dfrac{d}{d x}(x-\tan x)+(x-\tan x) \dfrac{d}{d x}(x+\cos x)$

$=(x+\cos x)\left[\dfrac{d}{d x}(x)-\dfrac{d}{d x}(\tan x)\right]+(x-\tan x)(1-\sin x)$

$=(x+\cos x)\left[1-\dfrac{d}{d x}(\tan x)\right]+(x-\tan x)(1-\sin x)$

Let $g(x)=\tan x .$ Accordingly,$g(x+h)=\tan (x+h)$

By first principle,

$g^{\prime \prime}(x)=\lim _{n \rightarrow 0} \dfrac{g(x+h)-g(x)}{h}$

$=\lim _{h \rightarrow 0} \dfrac{\tan (x+h)-\tan (x)}{h}$

$=\lim _{n \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\sin (x+h)}{\cos (x+h)}-\dfrac{\sin x}{\cos x}\right]$

$=\lim _{n \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\sin (x+h) \cos x-\sin x \cos (x+h)}{\cos x \cos (x+h)}\right]$

$=\dfrac{1}{\cos x^{n \rightarrow 0}} \lim _{h} \dfrac{1}{h}\left[\dfrac{\sin (x+h-x)}{\cos (x+h)}\right]$

$=\dfrac{1}{\cos x^{h \rightarrow 0}} \lim _{h} \dfrac{1}{h}\left[\dfrac{\sin h}{\cos (x+h)}\right]$

$=\dfrac{1}{\cos x}\left(\lim _{n \rightarrow 0} \dfrac{\sin h}{h}\right)\left(\lim _{n \rightarrow 0} \dfrac{1}{\cos (x+h)}\right)$

$=\dfrac{1}{\cos x} \cdot \cdot\left(\dfrac{1}{\cos (x+0)}\right)$

$=\dfrac{1}{\cos ^{2} x}$

$=\sec ^{2} x \quad$... (ii)

Therefore, from (i) and (ii). We obtain

$f^{\prime}(x)=(x+\cos x)\left(1-\sec ^{2} x\right)+(x-\tan x)(1-\sin x)$

$=(x+\cos x)\left(-\tan ^{2} x\right)+(x-\tan x)(1-\sin x)$

$=-\tan ^{2} x(x+\cos x)+(x-\tan x)(1-\sin x)$

26: Find the derivative of the following functions (it is to be understood that $a, b, c$, d, $p, q, r$ and s are fixed non-zero constants and m and $n$ are integers): $\mathbf{\dfrac{4 x+5 \sin x}{3 x+7 \cos x}}$

Ans: Let $f(x)=\dfrac{4 x+5 \sin x}{3 x+7 \cos x}$

Quotient rule,

$f^{\prime}(x)=\dfrac{(3 x+7 \cos x) \dfrac{d}{d x}(4 x+5 \sin x)-(4 x+5 \sin x) \dfrac{d}{d x}(3 x+7 \cos x)}{(3 x+7 \cos x)^{2}}$

$=\dfrac{(3 x+7 \cos x)\left[4 \dfrac{d}{d x}(x)+5 \dfrac{d}{d x}(\sin x)\right]-(4 x+5 \sin x)\left[3 \dfrac{d}{d x}(x)+7 \dfrac{d}{d x}(\cos x)\right]}{(3 x+7 \cos x)^{2}}$

$=\dfrac{(3 x+7 \cos x)[4 x+5 \cos x]-(4 x+5 \sin x)[3-7 \sin x]}{(3 x+7 \cos x)^{2}}$

$=\dfrac{12 x+15 x \cos x+28 x \cos x+35 \cos ^{2} x-12 x+28 x \sin x-15 \sin x+35\left(\cos ^{2} x+\sin ^{2} x\right)}{(3 x+7 \cos x)^{2}}$

$=\dfrac{15 x \cos x+28 \cos x+28 x \sin x-15 \sin x+35\left(\cos ^{2} x+\sin ^{2} x\right)}{(3 x+7 \cos x)^{2}}$

$=\dfrac{35+15 x \cos x+28 \cos x+28 x \sin x-15 \sin x}{(3 x+7 \cos x)^{2}}$

27: Find the derivative of the following functions (it is to be understood that $a, b, c$, d, $p, q, r$ and $s$ are fixed non-zero constants and $m$ and $n$ are integers): $\mathbf{\dfrac{x^{2} \cos \left(\dfrac{\pi}{4}\right)}{\sin x}}$

Ans: Let $\mathrm{f}(\mathrm{x})=\dfrac{x^{2} \cos \left(\dfrac{\pi}{4}\right)}{\sin x}$

By quotient rule, $f^{\prime}(x)=\cos \left(\dfrac{\pi}{4}\right)\left[\dfrac{\sin x \dfrac{d}{d x}\left(x^{2}\right)-x^{2} \dfrac{d}{d x}(\sin x)}{\sin ^{2} x}\right]$

$=\cos \left(\dfrac{\pi}{4}\right)\left[\dfrac{\sin x(2 x)-x^{2}(\cos x)}{\sin ^{2} x}\right]$

$=\dfrac{x \cos \dfrac{\pi}{4}[2 \sin x-x \cos x]}{\sin ^{2} x}$

28: Find the derivative of the following functions (it is to be understood that $a, b, c$. d, p, q, r and s are fixed non-zero constants and $m$ and $n$ are integers): $\mathbf{\dfrac{x}{1+\tan x}}$

Ans: Let $f(x)=\dfrac{x}{1+\tan x}$

$f(x)=\dfrac{(1+\tan x) \dfrac{d}{d x}(x)-(x) \dfrac{d}{d x}(1+\tan x)}{(1+\tan x)^{2}}$

$f^{\prime}(x)=\dfrac{(1+\tan x)-x \dfrac{d}{d x}(1+\tan x)}{(1+\tan x)^{2}}$

Let $g(x)=1+\tan x .$ Accordingly $g(x+h)=1+\tan (x+h)$.

By first principle, $\dot{g}(x)=\lim _{n \rightarrow 0} \dfrac{g(x+h)-g(x)}{h}$

$\lim _{h \rightarrow 0}\left[\dfrac{1+\tan (x+h)-1-\tan (x)}{h}\right]$

$\lim _{n \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\sin (x+h)}{\cos (x+h)}-\dfrac{\sin x}{\cos x}\right]$

$\lim _{n \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\sin (x+h) \cos x-\sin x \cos x(x+h)}{\cos (x+h) \cos x}\right]$

$\lim _{n \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\sin (x+h-x)}{\cos (x+h) \cos x}\right]$

$\lim _{h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\sinh }{\cos (x+h) \cos x}\right]$

$-\left(\lim _{n \rightarrow 0} \dfrac{\sinh }{h}\right) \cdot\left(\lim _{n \rightarrow 0} \dfrac{1}{\cos (x+h) \cos x}\right)$

$-1 \times \dfrac{1}{\cos ^{2}}=\sec ^{2} x$

$\Rightarrow \dfrac{d}{d x}\left(1+\tan ^{2} x\right)=\sec ^{2} x$

From (i) and (ii), we obtain

$\dot{f}(x)=\dfrac{1+\tan x-x \sec ^{2} x}{(1+\tan x)^{2}}$

29: Find the derivative of the following functions (it is to be understood that $\mathrm{a}, \mathrm{b}, \mathrm{c}$, d, p, q, r and s are fixed non-zero constants and m and n are integers): $\mathbf{(x+\sec x)(x-\tan x)}$

Ans: Let $f(x)=(x+\sec x)(x-\tan x)$

By product rule.

$f(x)=(x+\sec x) \dfrac{d}{d x}(x-\tan x)+(x-\tan x) \dfrac{d}{d x}(x+\sec x)$

$-(x+\sec x)\left[\dfrac{d}{d x}(x)-\dfrac{d}{d x} \tan x\right]+(x-\tan x)\left[\dfrac{d}{d x}(x)-\dfrac{d}{d x} \sec x\right]$

$\left.-f(x+\sec x)\left[1-\dfrac{d}{d x} \tan x\right)\right]+(x-\tan x)\left[1+\dfrac{d}{d x} \sec x\right]$

$\ldots(\mathrm{i})$

Let $f_{1}(x)=\tan x, f_{2}(x)=\sec x$

Accordingly, $f_{1}(x+h) \cdot \tan (x+h)$ and $f_{2}(x+h)-\sec (x+h)$

$f_{1}^{\prime}(x)=\lim _{n \rightarrow 0}\left(\dfrac{f_{1}(x+h)-f_{1}(x)}{h}\right)$

$=\lim _{h \rightarrow 0}\left[\dfrac{\tan (x+h)-\tan (x)}{h}\right]$

$\begin{array}{l}-\lim _{n \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\sin (x+h)}{\cos (x+h)}-\dfrac{\sin x}{\cos x}\right]\\-\lim _{n \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\sin (x+h) \cos x-\sin x \cos x(x+h)}{\cos (x+h) \cos x}\right]\\-\lim _{n \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\sin (x+h-x)}{\cos (x+h) \cos x}\right]\\-\lim _{n \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\sinh }{\cos (x+h) \cos x}\right]\\-\left(\lim _{n \rightarrow 0} \dfrac{\sinh }{h}\right) \cdot\left(\lim _{n \rightarrow 0} \dfrac{1}{\cos (x+h) \cos x}\right)\\-1 \times \dfrac{1}{\cos ^{2}}=\sec ^{2} x\\\Rightarrow \dfrac{d}{d x}\left(1+\tan ^{2} x\right)=\sec ^{2} x\\f_{2}^{\prime}(x)=\lim _{n \rightarrow 0}\left(\dfrac{f_{2}+(x+h)-f_{2}(x)}{h}\right)\\=\lim _{h \rightarrow 0}\left(\dfrac{\sec (x+h)-\sec (x)}{h}\right)\\=\lim _{n \rightarrow 0} \dfrac{1}{h}\left(\dfrac{1}{\cos (x+h)}-\dfrac{1}{\cos x}\right)\\=\lim _{n \rightarrow 0} \dfrac{1}{h}\left(\dfrac{\cos x-\cos (x+h)}{\cos (x+h) \cos x}\right)\\=\dfrac{1}{\cos x^{\prime \rightarrow 0}} \lim _{h} \dfrac{1}{h}\left[\dfrac{-2 \sin \left(\dfrac{x+x+h}{2}\right) \cdot \sin \left(\dfrac{x-x-h}{2}\right)}{\cos (x+h)}\right]\\=\dfrac{1}{\cos x} \lim _{h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{-2 \sin \left(\dfrac{2 x+h}{2}\right) \cdot \sin \left(\dfrac{-h}{2}\right)}{\cos (x+h)}\right]\end{array}$

$=\dfrac{1}{\cos x} \lim _{h \rightarrow 0} \dfrac{1}{h}\left[\dfrac{\sin \left(\dfrac{2 x+h}{2}\right)\left\{\dfrac{\sin\left(\dfrac{h}{2}\right)}{\left(\dfrac{h}{2}\right)}\right\}}{\cos (x+h)}\right]$

$=\sec x \dfrac{\left\{\lim _{n \rightarrow 0} \sin \left(\dfrac{2 x+h}{2}\right)\right\}\left\{\lim _{\dfrac{h}{2} \rightarrow 0} \dfrac{\sin \left(\dfrac{h}{2}\right)}{\left(\dfrac{h}{2}\right)}\right\}}{\lim _{n \rightarrow 0} \cos (x+h)}$

$=\sec x \cdot \dfrac{\sin x \cdot 1}{\cos x}$

$\Rightarrow \dfrac{d}{d x} \sec x=\sec x \tan x$

From (i). (ii), and (iii), we obtain

$f^{\prime}(x)=(x+\sec x)\left(1-\sec ^{2} x\right)+(x-\tan x)(1+\sec x \tan x)$

30: Find the derivative of the following functions (it is to be understood that $a, b, c$, d, $p, q, r$ and s are fixed non-zero constants and $m$ and $n$ are integers): $\mathbf{\dfrac{x}{\sin ^{n} x}}$

Ans: Let $\mathrm{f}(\mathrm{x})=\dfrac{x}{\sin ^{n} x}$

By quotient rule, $f^{\prime}(x)=\dfrac{\sin ^{n} x \dfrac{d}{d x} x-x \dfrac{d}{d x} \sin ^{n} x}{\sin ^{2 n} x}$

It can be easily shown that $\dfrac{d}{d x} \sin ^{n} x=n \sin ^{n-1} x \cos x$

Therefore,

$f^{\prime}(x)=\dfrac{\sin ^{n} \times \dfrac{d}{d x} x-x \dfrac{d}{d x} \sin ^{n} x}{\sin ^{2 n} x}$

$=\dfrac{\sin ^{n} x \cdot 1-x\left(\operatorname{nin}^{n-1} x \cos x\right)}{\sin ^{2 n} x}$

$=\dfrac{\sin ^{n-1} x(\sin x-n x \cos x)}{\sin ^{2 n} x}$

$=\dfrac{\sin x-n x \cos x}{\sin ^{n+1} x}$

## Class 11 Maths Chapter 12: Exercises Breakdown

 Exercise Number of Questions Exercise 12.1 32 Questions and solutions Exercise 12.2 11 Questions and solutions Miscellaneous Exercise 30 Questions and Solutions

## Conclusion

The Class 11 Limits and Derivatives Solutions provided by Vedantu, is a valuable tool for 11th-grade students. It helps introduce mathematical concepts in an accessible manner. The provided solutions and explanations simplify complex ideas, making it easier for 11th-grade students to understand the material. By using Vedantu's resources, students can develop a deeper understanding of NCERT concepts. Class 11 Maths Ch Limits And Derivatives  are a helpful aid for 11th-grade students, empowering them to excel in their studies and develop a genuine appreciation for Limits And Derivatives. In previous years' exams, this chapter typically features around 5-7 questions.

## Other Study Material for CBSE Class 11 Maths Chapter 12

 S. No Important Links for Chapter 12 Limits And Derivatives 1 Class 11 Limits and Derivatives Important Questions 2 Class 11 Limits and Derivatives Revision Notes 3 Class 11 Limits and Derivatives Important Formulas 4 Class 11 Limits and Derivatives NCERT Exemplar Solution

## Chapter-Specific NCERT Solutions for Class 11 Maths

Given below are the chapter-wise NCERT Solutions for Class 11 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions Class 11 Maths Chapter 12 Limits and Derivatives

1. What is the first principle of Class 11 Maths Chapter Limits And Derivatives NCERT Solutions?

The first principle of derivatives is the foundation for defining the derivative of a function. It essentially defines the derivative based on the concept of limits.

Imagine you have a function, f(x), that represents some quantity changing with respect to its input, x. The derivative of this function at a specific point, x=a, tells you the instantaneous rate of change of the function at that point.

The first principle states that the derivative of f(x) at x=a can be calculated as the limit of the following expression as h approaches zero:

lim(h -> 0) [(f(a + h) - f(a)) / h]

2. What is the purpose of limits and derivatives class 11?

Here we will understand limits and derivatives with examples for each of them:

Limits: Imagine a car speeding up. Limits help us understand what speed the car will eventually reach (if it keeps accelerating). They tell us if a value gets closer and closer to a specific number as we keep changing the input.

Derivatives: They're like the car's speedometer. They tell us how fast the car's speed is changing at any given moment.A high derivative means the speed is changing rapidly (like accelerating fast), while a low derivative means the speed isn't changing much (like maintaining a steady speed).

3. What is the relationship between limits and derivatives class 11?

The relationship between limits and derivatives is like that of bricks and a building. Limits are the essential building blocks, and derivatives are the structure built upon them.

• Limits define derivatives: The concept of a derivative is built on the foundation of limits. We define the derivative of a function at a point as the limit of a specific expression involving that function and the point. This expression essentially calculates the average rate of change of the function around that point. The derivative itself is the instantaneous rate of change, which is the limit of that average rate of change as we get infinitely close to the point.

• Limits help analyze derivative behavior: Since derivatives are defined using limits, we can use limit properties to analyze the behavior of derivatives. For example, we can use limits to determine if a derivative exists at a particular point or if it approaches positive or negative infinity.

4. When to differentiate limits in class 11th limits and derivatives?

• L'Hôpital's Rule: This rule applies in specific scenarios where the original limit expression results in an indeterminate form (0/0 or ∞/∞). Under these conditions, L'Hôpital's rule allows you to replace the original expression with another involving the derivative of the numerator and denominator (under certain conditions). This essentially utilizes differentiation to transform the limit problem into a more solvable form.

• While differentiation is a valuable tool in mathematics, it's generally not the first approach to solving limit problems. Limits are more versatile and can handle a wider range of scenarios. However, L'Hôpital's rule provides an exception where differentiation can be used strategically to transform a specific type of limit problem into a solvable form.

5. What is the basic rule of derivatives in class 11 maths ch 12?

The absolute foundation for most differentiation rules Students will encounter in Calculus is the limit definition of the derivative. It establishes the connection between derivatives and limits, and serves as the building block for understanding how derivatives are calculated.

Given a function f(x), the derivative of f(x) at a point x=a is defined as:

lim(h -> 0) [(f(a + h) - f(a)) / h]

• f(x): This represents the function whose derivative we're trying to find.

• a: This is the specific point (x-value) where we want to determine the derivative.

• h: This is a small, positive number. We imagine letting h approach zero to get an infinitely small change in the input (x-value).

• f(a + h): This represents the value of the function f(x) when the input is shifted by h units to the right of the point x=a (a+h).

• f(a): This is the value of the function f(x) at the original point x=a.

• (f(a + h) - f(a)) / h: This part calculates the average rate of change of the function f(x) between the points x=a and x=a+h. Imagine the slope of a secant line that intersects the curve of the function at these two points.

7. Why is zero of such significance?

Modern electronics would cease to exist without zero. There exists no Calculus without zero, which implies no contemporary engineering as well. Much of our modern world actually breaks apart without zero. However, humanity did not grasp the number zero for the overwhelming bulk of our history. Zero is an indispensable part of mathematics and its importance in most cases has been underrated.