NCERT Solutions for Class 11 Maths Chapter 13 - Limits And Derivatives

Exercise 13.1: 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 13.2: 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 Mathematics  Chapter 13 – Limits and Derivatives

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, $\lim _{x \rightarrow 0} \frac{\sin a x}{\sin b x}=\lim _{x \rightarrow 0}\left(\frac{\left.\frac{\sin a x}{a x}\right) \times a x}{\left(\frac{\sin b x}{a x}\right) \times b x}\right.$

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

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

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 $\mathbf{f(x)=\left\{\begin{array}{ll}|x|+1, & x<0 \\ 0, & x=0 \\ |x|-1, & x>1\end{array}\right.}$ For what value (s) does $\mathbf{\lim _{x \rightarrow 3} f(x)}$ exists?

Ans: The given function is

If $f(x)=\left\{\begin{array}{ll}|x|+1, & x<0 \\ 0, & x=0\end{array} .\right.$

When $\mathrm{a}=0$

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

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

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

$=0+1$

$-1$

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

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

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

$=0-1$

$=-1$

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

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

When $a<0$ $\lim _{x \rightarrow \tau} f(x)=\lim _{x \rightarrow a^{-}}(|x|+1)$

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

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

$=-a+1$

$\lim _{x \rightarrow \vec{t}} f(x)=\lim _{x \rightarrow \Delta}(|x|+1)$

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

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

$=-a$

$+1$

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

Thus, limit of $f(x)$ exists at $x-a$, where $a<0$. When $\mathrm{a}>0$

$\lim _{x \rightarrow \mathbb{Z}} f(x)=\lim _{x \rightarrow a}(|x|+1)$

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

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

$=a-1$

$\lim _{x \rightarrow^{2}} f(x)=\lim _{x \rightarrow \Delta}(|x|-1)$

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

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

$=a-1$

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

Thus, limit of $f(x)$ exists at $x=a$, where $a>0$ Thus, $\lim _{x \rightarrow a} f(x)$ exists for all $a \neq 0$.

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 13.2

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 $\mathbf{99 \mathrm{\mathbf{x}}}$ 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 .

3: 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 .

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,

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

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

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

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

Importance of Limits and Derivatives Class 11

Calculus is the study of continuity and the rate of change. "Limits and Derivatives" make up a substantial part of the calculus. You'll get better at this chapter if you practise it a lot. Practicing with exercise solutions, sample papers, and question papers is important. Understanding how mathematical concepts are applied in real-life situations helps to understand the chapter well. Limits and derivatives are useful in advanced mathematics, physics, and engineering. They can be used to determine electric and magnetic fields, calculate the rate of change of a quantity, and study temperature changes.

NCERT Solutions for Class 11 Maths Chapter 13 - Limits and Derivatives

NCERT Solutions for Class 11 Maths Chapter 13 Limits and Derivatives elaborates all the foundational concepts. The solutions are made available for everyone on the Vedantu app. Anyone can do NCERT Solutions for Class 11 Maths Chapter 13 PDF download for free.

We Cover all Exercises in the Chapter Given Below:-

Chapter 13 Maths Class 11

Limits and Derivatives Class 11 covers several sub-topics that acts as the foundation for advanced Calculus. Following is an indicative list of such topics –

• Definition of Limits

The limit is essentially the assignment of values to specific functions at such points where no previous values have been defined.

• Limits of 0/0 Form

It indicates an ‘Indeterminant' where after taking the limit, the answer stands at 0/0. Example of undefined number comes to infinity or 1/0.

• Limits of $x_n$ Formula

The concept relates to the power rule for derivatives.

• Limits of Trigonometric Functions

Trigonometric functions such as cosine and sine have important limit properties.

• Derivatives by the First Principle at a Point

This concept relates to the limit of slopes of secant line or difference quotient.

• Derivatives by the First Principle at a General Point

The first principle derivative leads to the usage of algebra to determine a general expression for a curve's slope.

• Derivatives by $x_n$ Formula

It is based on the principle that derivative formulas may be utilised in such instances where n is a positive integer.

• Derivatives by Sin & Cos

For a sine function derivative, the rate of change of sin(x) at a given angle x is indicated by the cosine of such angle.

• Derivatives by Another Trigonometric Formula

There are derivatives of four other types of trigonometric functions. 

Why are Limit and Derivatives Class 11 NCERT Solution a Must-Read for Students?

Limits Class 11, no doubt, makes for a very important topic for students. Owing to the nature of the chapter, a lot of deliberation is required from students’ end apart from rigorous practice. Having a good grasp over Calculus is no mean feat. To that effect, NCERT solutions for Class 11 Maths Chapter 13 Limits and Derivatives would aid students to prepare effectively. Following are a few ways in which solutions are helpful –

• There are a lot of practice sums for students to solve. The more practice one does, the better is the preparation.

• A variety of questions are included in the solution. It helps students to understand the different types of questions that may come in the examination.

• Format of the solutions is in accordance with the CBSE syllabus. A lot of questions are asked directly or indirectly from the NCERT book in the examination. Studying from Limits and Derivatives Class 11 solutions allows students to attempt such questions successfully.

Due focus is given to the basic concepts of Calculus apart from discussing the advanced ones. The formulas are collated in such a manner that it becomes easier for students to understand and memorise it. The lucid language of ch 13 Maths Class 11 ensures that students face no difficulty in comprehending complex concepts.

Benefits of Using Vedantu for Class 11 Chapter 13 - Limits and Derivatives 

Key Features of NCERT Solutions, These solutions are designed to help students achieve proficiency in their studies. They are crafted by experienced educators who excel in teaching Class 11 Maths. Some of the features include:

• Comprehensive explanations for each exercise and questions, promoting a deeper understanding of the subject.

• Clear and structured presentation for easy comprehension.

• Accurate answers aligned with the curriculum, boosting students' confidence in their knowledge.

• Visual aids like diagrams and illustrations to simplify complex concepts.

• Additional tips and insights to enhance students' performance.

• Chapter summaries for quick revision.

• Online accessibility and downloadable resources for flexible study and revision.

Conclusion

The NCERT Solutions for Class 11 Maths Chapter 13 - Limits And Derivatives, 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. These solutions are a helpful aid for 11th-grade students, empowering them to excel in their studies and develop a genuine appreciation for Limits And Derivatives.

NCERT Solutions for Class 11 Maths Chapters

• Chapter 1 - Sets

• Chapter 2 - Relations and Functions

• Chapter 3 - Trigonometric Functions

• Chapter 4 - Principle of Mathematical Induction

• Chapter 5 - Complex Numbers and Quadratic Equations

• Chapter 6 - Linear Inequalities

• Chapter 7 - Permutations and Combinations

• Chapter 8 - Binomial Theorem

• Chapter 9 - Sequences and Series

• Chapter 10 - Straight Lines

• Chapter 11 - Conic Sections

• Chapter 12 - Introduction to Three Dimensional Geometry

• Chapter 14 - Mathematical Reasoning

• Chapter 15 - Statistics

• Chapter 16 - Probability