NCERT Solutions For Class 11 Maths Chapter 12 Limits and Derivatives - 2025-26

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

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

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

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

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

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.

Important Related Links for CBSE Class 11 Maths