Answer :

$\lim _{x \to 3} x+3=3+3=6$

Answer :

$\lim _{x \to \pi}(x-\frac{22}{7})=(\pi-\frac{22}{7})$

Answer :

$\lim _{r \to 1} \pi r^{2}=\pi(1)^{2}=\pi$

Answer :

$\lim _{x \to 4} \frac{4 x+3}{x-2}=\frac{4(4)+3}{4-2}=\frac{16+3}{2}=\frac{19}{2}$

Answer :

$\lim _{x \to-1} \frac{x^{10}+x^{5}+1}{x-1}=\frac{(-1)^{10}+(-1)^{5}+1}{-1-1}=\frac{1-1+1}{-2}=-\frac{1}{2}$

Answer :

$\lim _{x \to 0} \frac{(x+1)^{5}-1}{x}$

Put $x+1=y$ so that $y^5 - $ 1 as $x -0 $.

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

$=\lim _{y \to 1} \frac{y^{5}-1^{5}}{y-1}$

$=5 \cdot 1^{5-1}$

$[\lim _{x \to a} \frac{x^{n}-a^{n}}{x-a}=n a^{n-1}]$

$=5$

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

Answer :

At $x=2$, the value of the given rational function takes the form $\frac{0}{0}$.

$ \begin{aligned} \therefore \lim _{x \to 2} \frac{3 x^{2}-x-10}{x^{2}-4} & =\lim _{x \to 2} \frac{(x-2)(3 x+5)}{(x-2)(x+2)} \\ & =\lim _{x \to 2} \frac{3 x+5}{x+2} \\ & =\frac{3(2)+5}{2+2} \\ & =\frac{11}{4} \end{aligned} $

Answer :

At $x=2$, the value of the given rational function takes the form $\frac{0}{0}$.

$ \begin{aligned} \therefore \lim _{x \to 3} \frac{x^{4}-81}{2 x^{2}-5 x-3} & =\lim _{x \to 3} \frac{(x-3)(x+3)(x^{2}+9)}{(x-3)(2 x+1)} \\ & =\lim _{x \to 3} \frac{(x+3)(x^{2}+9)}{2 x+1} \\ & =\frac{(3+3)(3^{2}+9)}{2(3)+1} \\ & =\frac{6 \times 18}{7} \\ & =\frac{108}{7} \end{aligned} $

Answer :

$\lim _{x \to 0} \frac{a x+b}{c x+1}=\frac{a(0)+b}{c(0)+1}=b$

Answer :

$\lim _{z \to 1} \frac{z^{\frac{1}{3}}-1}{z^{\frac{1}{6}}-1}$

At $z=1$, the value of the given function takes the form $\frac{0}{0}$.

Put $z^{\frac{1}{6}}=x$ so that $z ^{\frac{1}{6}}- 1 \text{ as } x-1 $ .

Accordingly, $\lim _{z \to 1} \frac{z^{\frac{1}{3}}-1}{z^{\frac{1}{6}}-1}=\lim _{x \to 1} \frac{x^{2}-1}{x-1}$

$ =\lim _{x \to 1} \frac{x^{2}-1^{2}}{x-1} $

$ =2.1^{2-1} \quad[\lim _{x \to a} \frac{x^{n}-a^{n}}{x-a}=n a^{n-1}] $

$=2$

$\therefore \lim _{z \to 1} \frac{z^{\frac{1}{3}}-1}{z^{\frac{1}{6}}-1}=2$

Answer :

$\lim _{x \to 1} \frac{a x^{2}+b x+c}{c x^{2}+b x+a}=\frac{a(1)^{2}+b(1)+c}{c(1)^{2}+b(1)+a}$

$ \begin{aligned} & =\frac{a+b+c}{a+b+c} \\ & =1 \quad[a+b+c \neq 0] \end{aligned} $

Answer :

$\lim _{x \to-2} \frac{\frac{1}{x}+\frac{1}{2}}{x+2}$

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

Now, $\lim _{x \to-2} \frac{\frac{1}{x}+\frac{1}{2}}{x+2}=\lim _{x \to-2} \frac{(\frac{2+x}{2 x})}{x+2}$

$ \begin{aligned} & =\lim _{x \to-2} \frac{1}{2 x} \\ & =\frac{1}{2(-2)}=\frac{-1}{4} \end{aligned} $

Answer :

$\lim _{x \to 0} \frac{\sin a x}{b x}$

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

Now, $\lim _{x \to 0} \frac{\sin a x}{b x}=\lim _{x \to 0} \frac{\sin a x}{a x} \times \frac{a x}{b x}$

$=\lim _{x \to 0}(\frac{\sin a x}{a x}) \times(\frac{a}{b})$

$=\frac{a}{b} \lim _{a x \to 0}(\frac{\sin a x}{a x}) \quad[x \to 0 \Rightarrow a x \to 0]$

$=\frac{a}{b} \times 1 \quad[\lim _{y \to 0} \frac{\sin y}{y}=1]$

$=\frac{a}{b}$

Answer :

$\lim _{x \to 0} \frac{\sin a x}{\sin b x}, a, b \neq 0$

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

Now, $\lim _{x \to 0} \frac{\sin a x}{\sin b x}=\lim _{x \to 0} \frac{(\frac{\sin a x}{a x}) \times a x}{(\frac{\sin b x}{b x}) \times b x}$

$ \begin{matrix} =(\frac{a}{b}) \times \frac{\lim _{a x \to 0}(\frac{\sin a x}{a x})}{\lim _{b x \to 0}(\frac{\sin b x}{b x})} & { \begin{bmatrix} x \to 0 \Rightarrow a x \to 0 \\ \text{ and } x \to 0 \Rightarrow b x \to 0 \end{bmatrix} } \\ =(\frac{a}{b}) \times \frac{1}{1} & {[\lim _{y \to 0} \frac{\sin y}{y}=1]} \\ =\frac{a}{b} & \end{matrix} $

Answer :

$\lim _{x \to \pi} \frac{\sin (\pi-x)}{\pi(\pi-x)}$

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

$ \begin{aligned} \therefore \lim _{x \to \pi} \frac{\sin (\pi-x)}{\pi(\pi-x)} & =\frac{1}{\pi} \lim _{(\pi-x) \to 0} \frac{\sin (\pi-x)}{(\pi-x)} \\ & =\frac{1}{\pi} \times 1 \quad[\lim _{y \to 0} \frac{\sin y}{y}=1] \\ & =\frac{1}{\pi} \end{aligned} $

Answer :

$\lim _{x \to 0} \frac{\cos x}{\pi-x}=\frac{\cos 0}{\pi-0}=\frac{1}{\pi}$

Answer :

$\lim _{x \to 0} \frac{\cos 2 x-1}{\cos x-1}$

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

Now,

$ \begin{aligned} & \lim _{x \to 0} \frac{\cos 2 x-1}{\cos x-1}=\lim _{x \to 0} \frac{1-2 \sin ^{2} x-1}{1-2 \sin ^{2} \frac{x}{2}-1} \quad[\cos x=1-2 \sin ^{2} \frac{x}{2}] \\ & =\lim _{x \to 0} \frac{\sin ^{2} x}{\sin ^{2} \frac{x}{2}}=\lim _{x \to 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}} \\ & =4 \frac{\lim _{x \to 0}(\frac{\sin ^{2} x}{x^{2}})}{\lim _{x \to 0}(\frac{\sin ^{2} \frac{x}{2}}{(\frac{x}{2})^{2}})} \\ & =4 \frac{(\lim _{x \to 0} \frac{\sin x}{x})^{2}}{(\lim _{\frac{x}{2} \to 0} \frac{\sin \frac{x}{2}}{\frac{x}{2}})^{2}} \quad[x \to 0 \Rightarrow \frac{x}{2} \to 0] \\ & =4 \frac{1^{2}}{1^{2}} \quad[\lim _{y \to \mid} \frac{\sin y}{y}=1] \\ & =4 \end{aligned} $

Answer :

$\lim _{x \to 0} \frac{a x+x \cos x}{b \sin x}$

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

Now,

$ \begin{aligned} \lim _{x \to 0} \frac{a x+x \cos x}{b \sin x} & =\frac{1}{b} \lim _{x \to 0} \frac{x(a+\cos x)}{\sin x} \\ & =\frac{1}{b} \lim _{x \to 0}(\frac{x}{\sin x}) \times \lim _{x \to 0}(a+\cos x) \\ & =\frac{1}{b} \times \frac{1}{(\lim _{x \to 0} \frac{\sin x}{x})} \times \lim _{x \to 0}(a+\cos x) \\ & =\frac{1}{b} \times(a+\cos 0) \quad[\lim _{x \to 0} \frac{\sin x}{x}=1] \\ & =\frac{a+1}{b} \end{aligned} $

Answer :

$\lim _{x \to 0} x \sec x=\lim _{x \to 0} \frac{x}{\cos x}=\frac{0}{\cos 0}=\frac{0}{1}=0$

Answer :

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

Now,

$ \begin{aligned} & \lim _{x \to 0} \frac{\sin a x+b x}{a x+\sin b x} \\ & =\lim _{x \to 0} \frac{(\frac{\sin a x}{a x}) a x+b x}{a x+b x(\frac{\sin b x}{b x})} \\ & =\frac{(\lim _{a x \to 0} \frac{\sin a x}{a x}) \times \lim _{x \to 0}(a x)+\lim _{x \to 0} b x}{\sin b x)} \quad[\text{ As } x \to 0 \Rightarrow a x \to 0 \text{ and } b x \to 0 \text{ ] } \\ & \lim _{x \to 0} a x+\lim _{x \to 0} b x(\lim _{b x \to 0} \frac{\sin b x}{b x}) \\ & =\frac{\lim _{x \to 0}(a x)+\lim _{x \to 0} b x}{\lim _{x \to 0} a x+\lim _{x \to 0} b x} \quad[\lim _{x \to 0} \frac{\sin x}{x}=1] \\ & =\frac{\lim _{x \to 0}(a x+b x)}{\lim _{x \to 0}(a x+b x)} \\ & =\lim _{x \to 0}(1) \\ & =1 \end{aligned} $

Answer :

At $x=0$, the value of the given function takes the form $\infty-\infty$.

Now,

$ \begin{aligned} & \lim _{x \to 0}(cosec x-\cot x) \\ & =\lim _{x \to 0}(\frac{1}{\sin x}-\frac{\cos x}{\sin x}) \\ & =\lim _{x \to 0}(\frac{1-\cos x}{\sin x}) \\ & =\lim _{x \to 0} \frac{(\frac{1-\cos x}{x})}{.\frac{\sin x}{x})} \\ & =\frac{\lim _{x \to 0} \frac{1-\cos x}{x}}{\lim _{x \to 0} \frac{\sin x}{x}} \quad[\lim _{x \to 0} \frac{1-\cos x}{x}=0 \text{ and } \lim _{x \to 0} \frac{\sin x}{x}=1] \\ & =\frac{0}{1} \\ & =0 \end{aligned} $

Answer :

$\lim _{x \to \frac{\pi}{2}} \frac{\tan 2 x}{x-\frac{\pi}{2}}$

At $x=\frac{\pi}{2}$, the value of the given function takes the form $\frac{0}{0}$.

Now, put $x-\frac{\pi}{2}=y \quad x \to \frac{\pi}{2}, y \to 0$

$ \begin{matrix} \therefore \lim _{x \to \frac{\pi}{2}} \frac{\tan 2 x}{x-\frac{\pi}{2}} & =\lim _{y \to 0} \frac{\tan 2(y+\frac{\pi}{2})}{y} & \\ & =\lim _{y \to 0} \frac{\tan (\pi+2 y)}{y} & \quad[\tan (\pi+2 y)=\tan 2 y] \\ & =\lim _{y \to 0} \frac{\tan 2 y}{y} & \\ & =\lim _{y \to 0} \frac{\sin 2 y}{\cos 2 y} \\ & =\lim _{y \to 0}(\frac{\sin 2 y}{2 y} \times \frac{2}{\cos 2 y}) & \\ & =(\lim _{2 y \to 0} \frac{\sin 2 y}{2 y}) \times \lim _{y \to 0}(\frac{2}{\cos 2 y}) \quad[\lim _{x \to 0} \frac{\sin x}{x}=1] \\ & =1 \times \frac{2}{\cos 0} & \\ & =1 \times \frac{2}{1} & \\ & =2 & \end{matrix} $

Answer :

The given function is

$f(x)= \begin{cases}2 x+3, & x \leq 0 \\ 3(x+1), & x>0\end{cases}$

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

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

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

$\lim _{x \to 1^{-}} f(x)=\lim _{x \to 1} 3(x+1)=3(1+1)=6$ $\lim _{x \to 1^{+}} f(x)=\lim _{x \to 1} 3(x+1)=3(1+1)=6$

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

Answer :

The given function is

$f(x)=\begin{cases} x^{2}-1, x \leq 1 \\ -x^{2}-1, x>1 \end{cases} .$

$\lim _{x \to 1^{-}} f(x)=\lim _{x \to 1}[x^{2}-1]=1^{2}-1=1-1=0$

$\lim _{x \to 1^{+}} f(x)=\lim _{x \to 1}[-x^{2}-1]=-1^{2}-1=-1-1=-2$

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

Hence, $\lim _{x \to 1} f(x)$ does not exist.

Answer :

The given function is

$f(x)= \begin{cases}\frac{|x|}{x}, & x \neq 0 \\ 0, & x=0\end{cases}$

$ \begin{matrix} \lim _{x \to 0^{-}} f(x) & =\lim _{x \to 0^{-}}[\frac{|x|}{x}] & \\ & =\lim _{x \to 0}(\frac{-x}{x}) \quad & \\ & =\lim _{x \to 0}(-1) \\ & =-1 \\ \lim _{x \to 0^{+}} f(x) & =\lim _{x \to 0^{+}}[\frac{|x|}{x}] & \\ & =\lim _{x \to 0}[\frac{x}{x}] \\ & =\lim _{x \to 0}(1) \\ & =1 \end{matrix} $

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

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

Answer :

The given function is

$ \begin{aligned} & f(x)= \begin{cases}\frac{x}{|x|}, & x \neq 0 \\ 0, & x=0\end{cases} \\ & \lim _{x \to 0^{-}} f(x)=\lim _{x \to 0^{-}}[\frac{x}{|x|}] \\ & =\lim _{x \to 0}[\frac{x}{-x}] \quad[\text{ When } x<0,|x|=-x] \\ & =\lim _{x \to 0}(-1) \\ & =-1 \\ & \lim _{x \to 0^{+}} f(x)=\lim _{x \to 0^{+}}[\frac{x}{|x|}] \\ & =\lim _{x \to 0}[\frac{x}{x}] \quad[\text{ When } x>0,|x|=x] \\ & =\lim _{x \to 0}(1) \\ & =1 \end{aligned} $

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

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

Answer :

The given function is $f(x)=|x|-5$.

$ \begin{aligned} & \begin{aligned} \lim _{x \to 5^{-}} f(x) & =\lim _{x \to 5^{-}}[|x|-5] \\ & =\lim _{x \to 5}(x-5) \quad \\ & =5-5 \\ & =0 \end{aligned} \\ & \begin{aligned} \lim _{x \to 5^{+}} f(x) & =\lim _{x \to 5^{+}}(|x|-5) \\ & =\lim _{x \to 5}(x-5) \quad \\ & =5-5 \\ & =0 \end{aligned} \\ & \therefore \lim _{x \to 5^{5}} f(x)=\lim _{x \to 5^{+}} f(x)=0 \end{aligned} $

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

Answer :

The given function is

$f(x)= \begin{cases}a+b x, & x<1 \\ 4, & x=1 \\ b-a x & x>1\end{cases}$

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

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

$f(1)=4$

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

$\therefore \lim _{.x \to|^{-}} f(x)=\lim _{x \to 1^{+}} f(x)=\lim _{x \to 1} f(x)=f(1)$

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

On solving these two equations, we obtain $a=0$ and $b=4$.

Thus, the respective possible values of $a$ and $b$ are 0 and 4 .

Answer :

The given function is $f(x)=(x-a_1)(x-a_2) \ldots(x-a_n)$

$ \begin{aligned} \lim _{x \to a_1} f(x) & =\lim _{x \to a_1}[(x-a_1)(x-a_2) \ldots(x-a_n)] \\ & =[\lim _{x \to a_1}(x-a_1)][\lim _{x \to a_1}(x-a_2)] \ldots[\lim _{x \to a_1}(x-a_n)] \\ & =(a_1-a_1)(a_1-a_2) \ldots(a_1-a_n)=0 \end{aligned} $

$\therefore \lim _{x \to a_1} f(x)=0$

Now, $\lim _{x \to a} f(x)=\lim _{x \to a}[(x-a_1)(x-a_2) \ldots(x-a_n)]$

$ \begin{aligned} & =[\lim _{x \to a}(x-a_1)][\lim _{x \to a}(x-a_2)] \ldots[\lim _{x \to a}(x-a_n)] \\ & =(a-a_1)(a-a_2) \ldots .(a-a_n) \end{aligned} $

$\therefore \lim _{x \to a} f(x)=(a-a_1)(a-a_2) \ldots(a-a_n)$

Answer :

The given function is $f(x)= \begin{cases}|x|+1, & x<0 \\ 0, & x=0 \\ |x|-1, & x>0\end{cases}$

When $a=0$,

$ \begin{aligned} \lim _{x \to 0^{-}} f(x) & =\lim _{x \to 0^{-}}(|x|+1) \\ & =\lim _{x \to 0}(-x+1) \quad[\text{ If } x<0,|x|=-x] \\ & =-0+1 \\ & =1 \end{aligned} $

$ \begin{matrix} \lim _{x \to 0^{+}} f(x) & =\lim _{x \to 0^{+}}(|x|-1) & \\ & =\lim _{x \to 0}(x-1) \quad[\text{ If } x>0,|x|=x] \\ & =0-1 & \\ & =-1 & \end{matrix} $

Here, it is observed that $\lim _{x \to 00^{-}} f(x) \neq \lim _{x \to 0^{+}} f(x)$.

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

When $a<0$,

$ \begin{aligned} & \begin{aligned} \lim _{x \to a^{-}} f(x) & =\lim _{x \to a^{-}}(|x|+1) \\ & =\lim _{x \to a}(-x+1) \quad[x<a<0 \Rightarrow|x|=-x] \\ & =-a+1 \end{aligned} \\ & \begin{aligned} \lim _{x \to a^{+}} f(x) & =\lim _{x \to a^{+}}(|x|+1) \\ & =\lim _{x \to a}(-x+1) \quad[a<x<0 \Rightarrow|x|=-x] \\ & =-a+1 \end{aligned} \\ & \therefore \lim _{x \to a^{-}} f(x)=\lim _{x \to a^{+}} f(x)=-a+1 \end{aligned} $

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

When $a>0$

$ \begin{aligned} & \lim _{x \to a^{-}} f(x)=\lim _{x \to a^{-}}(|x|-1) \\ & =\lim _{x \to a}(x-1) \quad[0<x<a \Rightarrow|x|=x] \\ & =a-1 \\ & \lim _{x \to a^{+}} f(x)=\lim _{x \to a^{+}}(|x|-1) \\ & =\lim _{x \to a}(x-1) \quad[0<a<x \Rightarrow|x|=x] \\ & =a-1 \\ & \therefore \lim _{x \to a^{-}} f(x)=\lim _{x \to a^{+}} f(x)=a-1 \end{aligned} $

Thus, limit of $f(x)$ exists at $x=a$, where $a>0$.

Thus, $\lim _{x \to a} f(x)$ exists for all $a \neq 0$.

Answer :

$\lim _{x \to 1} \frac{f(x)-2}{x^{2}-1}=\pi$

$\Rightarrow \frac{\lim _{x \to 1}(f(x)-2)}{\lim _{x \to 1}(x^{2}-1)}=\pi$

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

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

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

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

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

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

$ f(x)= \begin{cases}m x^{2}+n, & x<0 \\ n x+m, & 0 \leq x \leq 1 \\ n x^{3}+m, & x>1 \quad \text{. For what integers } m \text{ and } n \text{ does } \lim _{x \to 0} f(x) \quad \lim _{x \to 1} f(x) \text{ exist? }\end{cases} $

Answer :

The given function is

$ \begin{aligned} & f(x)= \begin{cases}m x^{2}+n, & x<0 \\ n x+m, & 0 \leq x \leq 1 \\ n x^{3}+m, & x>1\end{cases} \\ & \lim _{x \to 0^{-}} f(x)=\lim _{x \to 0}(m x^{2}+n) \\ & =m(0)^{2}+n \\ & =n \\ & \lim _{x \to 0^{+}} f(x)=\lim _{x \to 0}(n x+m) \\ & =n(0)+m \\ & =m . \end{aligned} $

Thus, $\lim _{x \to 0} f(x)$ exists if $m=n$.

$ \begin{aligned} \lim _{x \to 1^{-}} f(x) & =\lim _{x \to 1}(n x+m) \\ & =n(1)+m \\ & =m+n \end{aligned} $

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

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

$ =m+n $

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

Thus, $\lim _{x \to 1} f(x)$ exists for any integral value of $m$ and $n$.

Answer :

Let $f(x)=x^{2} - 2$. Accordingly,

$ \begin{aligned} f^{\prime}(10) & =\lim _{h \to 0} \frac{f(10+h)-f(10)}{h} \\ & =\lim _{h \to 0} \frac{[(10+h)^{2}-2]-(10^{2}-2)}{h} \\ & =\lim _{h \to 0} \frac{10^{2}+2 \cdot 10 \cdot h+h^{2}-2-10^{2}+2}{h} \\ & =\lim _{h \to 0} \frac{20 h+h^{2}}{h} \\ & =\lim _{h \to 0}(20+h)=(20+0)=20 \end{aligned} $

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

Answer :

Letf $(x)=x$. Accordingly,

$ \begin{aligned} f^{\prime}(1) & =\lim _{h \to 0} \frac{f(1+h)-f(1)}{h} \\ & =\lim _{h \to 0} \frac{(1+h)-1}{h} \\ & =\lim _{h \to 0} \frac{h}{h} \\ & =\lim _{h \to 0}(1) \\ & =1 \end{aligned} $

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

Answer :

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

$ \begin{aligned} f^{\prime}(100) & =\lim _{h \to 0} \frac{f(100+h)-f(100)}{h} \\ & =\lim _{h \to 0} \frac{99(100+h)-99(100)}{h} \\ & =\lim _{h \to 0} \frac{99 \times 100+99 h-99 \times 100}{h} \\ & =\lim _{h \to 0} \frac{99 h}{h} \\ & =\lim _{h \to 0}(99)=99 \end{aligned} $

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

Answer :

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

$ \begin{aligned} f^{\prime}(x) & =\lim _{h \to 0} \frac{f(x+h)-f(x)}{h} \\ & =\lim _{h \to 0} \frac{[(x+h)^{3}-27]-(x^{3}-27)}{h} \\ & =\lim _{h \to 0} \frac{x^{3}+h^{3}+3 x^{2} h+3 x h^{2}-x^{3}}{h} \\ & =\lim _{h \to 0} \frac{h^{3}+3 x^{2} h+3 x h^{2}}{h} \\ & =\lim _{h \to 0}(h^{2}+3 x^{2}+3 x h) \\ & =0+3 x^{2}+0=3 x^{2} \end{aligned} $

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

$ \begin{aligned} & f^{\prime}(x)=\lim _{h \to 0} \frac{f(x+h)-f(x)}{h} \\ & =\lim _{h \to 0} \frac{(x+h-1)(x+h-2)-(x-1)(x-2)}{h} \\ & =\lim _{h \to 0} \frac{(x^{2}+h x-2 x+h x+h^{2}-2 h-x-h+2)-(x^{2}-2 x-x+2)}{h} \\ & =\lim _{h \to 0} \frac{(h x+h x+h^{2}-2 h-h)}{h} \\ & =\lim _{h \to 0} \frac{2 h x+h^{2}-3 h}{h} \\ & =\lim _{h \to 0}(2 x+h-3) \\ & =(2 x+0-3) \\ & =2 x-3 \\ & \begin{matrix} f(x)=\frac{1}{x^{2}} \text{. Accordingly, } f \\ \text{ (iii) Let } \\ f^{\prime}(x)=\lim _{h \to 0} \frac{f(x+h)-f(x)}{h} \end{matrix} \\ & =\lim _{h \to 0} \frac{\frac{1}{(x+h)^{2}}-\frac{1}{x^{2}}}{h} \\ & =\lim _{h \to 0} \frac{1}{h}[\frac{x^{2}-(x+h)^{2}}{x^{2}(x+h)^{2}}] \\ & =\lim _{h \to 0} \frac{1}{h}[\frac{x^{2}-x^{2}-h^{2}-2 h x}{x^{2}(x+h)^{2}}] \\ & =\lim _{h \to 0} \frac{1}{h}[\frac{-h^{2}-2 h x}{x^{2}(x+h)^{2}}] \\ & =\lim _{h \to 0}[\frac{-h-2 x}{x^{2}(x+h)^{2}}] \\ & =\frac{0-2 x}{x^{2}(x+0)^{2}}=\frac{-2}{x^{3}} \\ & f(x)=\frac{x+1}{x-1} \\ & \text{ (iv) Let } \quad x-1 \text{. Accordingly, from the first principle, } \end{aligned} $

$ \begin{aligned} & f^{\prime}(x)=\lim _{h \to 0} \frac{f(x+h)-f(x)}{h} \\ & =\lim _{h \to 0} \frac{(\frac{x+h+1}{x+h-1}-\frac{x+1}{x-1})}{h} \\ & =\lim _{h \to 0} \frac{1}{h}[\frac{(x-1)(x+h+1)-(x+1)(x+h-1)}{(x-1)(x+h-1)}] \\ & =\lim _{h \to 0} \frac{1}{h}[\frac{(x^{2}+h x+x-x-h-1)-(x^{2}+h x-x+x+h-1)}{(x-1)(x+h-1)}] \\ & =\lim _{h \to 0} \frac{1}{h}[\frac{-2 h}{(x-1)(x+h-1)}] \\ & =\lim _{h \to 0}[\frac{-2}{(x-1)(x+h-1)}] \\ & =\frac{-2}{(x-1)(x-1)}=\frac{-2}{(x-1)^{2}} \end{aligned} $

Answer :

The given function is $f(x)=\frac{x^{100}}{100}+\frac{x^{99}}{99}+\ldots+\frac{x^{2}}{2}+x+1$

$\frac{d}{d x} f(x)=\frac{d}{d x}[\frac{x^{100}}{100}+\frac{x^{99}}{99}+\ldots+\frac{x^{2}}{2}+x+1]$

$\frac{d}{d x} f(x)=\frac{d}{d x}(\frac{x^{100}}{100})+\frac{d}{d x}(\frac{x^{99}}{99})+\ldots+\frac{d}{d x}(\frac{x^{2}}{2})+\frac{d}{d x}(x)+\frac{d}{d x}(1)$

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

$\frac{d}{d x} f(x)=\frac{100 x^{99}}{100}+\frac{99 x^{98}}{99}+\ldots+\frac{2 x}{2}+1+0$

$ =x^{99}+x^{98}+\ldots+x+1 $

$\therefore f^{\prime}(x)=x^{99}+x^{98}+\ldots+x+1$

At $x=0$,

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

At $x=1$,

$f^{\prime}(1)=1^{99}+1^{98}+\ldots+1+1=[1+1+\ldots+1+1] _{\text{to terms }}=1 \times 100=100$

Thus, $f^{\prime}(1)=100 \times f^{1}(0)$

Answer :

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

$\therefore f^{\prime}(x)=\frac{d}{d x}(x^{n}+a x^{n-1}+a^{2} x^{n-2}+\ldots+a^{n-1} x+a^{n})$

$=\frac{d}{d x}(x^{\prime \prime})+a \frac{d}{d x}(x^{n-1})+a^{2} \frac{d}{d x}(x^{n-2})+\ldots+a^{n-1} \frac{d}{d x}(x)+a^{n} \frac{d}{d x}(1)$

On using theorem $\frac{d}{d x} x^{n}=n x^{n-1}$, we obtain

$ \begin{aligned} f^{\prime}(x) & =n x^{n-1}+a(n-1) x^{n-2}+a^{2}(n-2) x^{n-3}+\ldots+a^{n-1}+a^{n}(0) \\ & =n x^{n-1}+a(n-1) x^{n-2}+a^{2}(n-2) x^{n-3}+\ldots+a^{n-1} \end{aligned} $

Answer :

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

$ \begin{aligned} \Rightarrow f(x) & =x^{2}-(a+b) x+a b \\ \therefore f^{\prime}(x) & =\frac{d}{d x}(x^{2}-(a+b) x+a b) \\ & =\frac{d}{d x}(x^{2})-(a+b) \frac{d}{d x}(x)+\frac{d}{d x}(a b) \end{aligned} $

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

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

(ii) Let $f(x)=(a x^{2}+b)^{2}$

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

$\therefore f^{\prime}(x)=\frac{d}{d x}(a^{2} x^{4}+2 a b x^{2}+b^{2})=a^{2} \frac{d}{d x}(x^{4})+2 a b \frac{d}{d x}(x^{2})+\frac{d}{d x}(b^{2})$

On using theorem $\frac{d}{d x} x^{n}=n x^{n-1}$, we obtain

$ \begin{aligned} f^{\prime}(x) & =a^{2}(4 x^{3})+2 a b(2 x)+b^{2}(0) \\ & =4 a^{2} x^{3}+4 a b x \\ & =4 a x(a x^{2}+b) \end{aligned} $

(iii)

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

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

By quotient rule,

$ \begin{aligned} f^{\prime}(x) & =\frac{(x-b) \frac{d}{d x}(x-a)-(x-a) \frac{d}{d x}(x-b)}{(x-b)^{2}} \\ & =\frac{(x-b)(1)-(x-a)(1)}{(x-b)^{2}} \\ & =\frac{x-b-x+a}{(x-b)^{2}} \\ & =\frac{a-b}{(x-b)^{2}} \end{aligned} $

Answer :

Let $f(x)=\frac{x^{n}-a^{n}}{x-a}$

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

By quotient rule,

$ \begin{aligned} f^{\prime}(x) & =\frac{(x-a) \frac{d}{d x}(x^{n}-a^{n})-(x^{n}-a^{n}) \frac{d}{d x}(x-a)}{(x-a)^{2}} \\ & =\frac{(x-a)(n x^{n-1}-0)-(x^{n}-a^{n})}{(x-a)^{2}} \\ & =\frac{n x^{n}-a n x^{n-1}-x^{n}+a^{n}}{(x-a)^{2}} \end{aligned} $

Answer :

(i) Let $f(x)=2 x-\frac{3}{4}$

$ \begin{aligned} f^{\prime}(x) & =\frac{d}{d x}(2 x-\frac{3}{4}) \\ & =2 \frac{d}{d x}(x)-\frac{d}{d x}(\frac{3}{4}) \\ & =2-0 \\ & =2 \end{aligned} $

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

By Leibnitz product rule,

$ \begin{aligned} f^{\prime}(x) & =(5 x^{3}+3 x-1) \frac{d}{d x}(x-1)+(x-1) \frac{d}{d x}(5 x^{3}+3 x-1) \\ & =(5 x^{3}+3 x-1)(1)+(x-1)(5.3 x^{2}+3-0) \\ & =(5 x^{3}+3 x-1)+(x-1)(15 x^{2}+3) \\ & =5 x^{3}+3 x-1+15 x^{3}+3 x-15 x^{2}-3 \\ & =20 x^{3}-15 x^{2}+6 x-4 \end{aligned} $

(iii) Letf $(x)=x^{3 e^{-3}}(5+3 x)$

By Leibnitz product rule,

$ \begin{aligned} f^{\prime}(x) & =x^{-3} \frac{d}{d x}(5+3 x)+(5+3 x) \frac{d}{d x}(x^{-3}) \\ & =x^{-3}(0+3)+(5+3 x)(-3 x^{-3-1}) \\ & =x^{-3}(3)+(5+3 x)(-3 x^{-4}) \\ & =3 x^{-3}-15 x^{-4}-9 x^{-3} \\ & =-6 x^{-3}-15 x^{-4} \\ & =-3 x^{-3}(2+\frac{5}{x}) \\ & =\frac{-3 x^{-3}}{x}(2 x+5) \\ & =\frac{-3}{x^{4}}(5+2 x) \end{aligned} $

(iv) Let $f(x)=x^{5}(3 - 6 x^{\mathbf{a} \in 9})$

By Leibnitz product rule,

$ \begin{aligned} f^{\prime}(x) & =x^{5} \frac{d}{d x}(3-6 x^{-9})+(3-6 x^{-9}) \frac{d}{d x}(x^{5}) \\ & =x^{5}{0-6(-9) x^{-9-1}}+(3-6 x^{-9})(5 x^{4}) \\ & =x^{5}(54 x^{-10})+15 x^{4}-30 x^{-5} \\ & =54 x^{-5}+15 x^{4}-30 x^{-5} \\ & =24 x^{-5}+15 x^{4} \\ & =15 x^{4}+\frac{24}{x^{5}} \end{aligned} $

(v) Let $f(x)=x^{4 \epsilon^{4} \cdot 4}(3 - 4 x^{3 \epsilon^{\prime} 5})$

By Leibnitz product rule,

$ \begin{aligned} & f^{\prime}(x)=x^{-4} \frac{d}{d x}(3-4 x^{-5})+(3-4 x^{-5}) \frac{d}{d x}(x^{-4}) \\ & =x^{-4}{0-4(-5) x^{-5-1}}+(3-4 x^{-5})(-4) x^{-4-1} \\ & =x^{-4}(20 x^{-6})+(3-4 x^{-5})(-4 x^{-5}) \\ & =20 x^{-10}-12 x^{-5}+16 x^{-10} \\ & =36 x^{-10}-12 x^{-5} \\ & =-\frac{12}{x^{5}}+\frac{36}{x^{10}} \end{aligned} $

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

By quotient rule,

$ \begin{aligned} f^{\prime}(x) & =[\frac{(x+1) \frac{d}{d x}(2)-2 \frac{d}{d x}(x+1)}{(x+1)^{2}}]-[\frac{(3 x-1) \frac{d}{d x}(x^{2})-x^{2} \frac{d}{d x}(3 x-1)}{(3 x-1)^{2}}] \\ & =[\frac{(x+1)(0)-2(1)}{(x+1)^{2}}]-[\frac{(3 x-1)(2 x)-(x^{2})(3)}{(3 x-1)^{2}}] \\ & =\frac{-2}{(x+1)^{2}}-[\frac{6 x^{2}-2 x-3 x^{2}}{(3 x-1)^{2}}] \\ & =\frac{-2}{(x+1)^{2}}-[\frac{3 x^{2}-2 x^{2}}{(3 x-1)^{2}}] \\ & =\frac{-2}{(x+1)^{2}}-\frac{x(3 x-2)}{(3 x-1)^{2}} \end{aligned} $

Answer :

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

$ \begin{aligned} f^{\prime}(x) & =\lim _{h \to 0} \frac{f(x+h)-f(x)}{h} \\ & =\lim _{h \to 0} \frac{\cos (x+h)-\cos x}{h} \end{aligned} $

$ \begin{aligned} & =\lim _{h \to 0}[\frac{\cos x \cos h-\sin x \sin h-\cos x}{h}] \\ & =\lim _{h \to 0}[\frac{-\cos x(1-\cos h)-\sin x \sin h}{h}] \\ & =\lim _{h \to 0}[\frac{-\cos x(1-\cos h)}{h}-\frac{\sin x \sin h}{h}] \\ & =-\cos x(\lim _{h \to 0} \frac{1-\cos h}{h})-\sin x \lim _{h \to 0}(\frac{\sin h}{h}) \\ & =-\cos x(0)-\sin x(1) \\ & =-\sin x \\ & \therefore f^{\prime}(x)=-\sin x \end{aligned} $

Answer :

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

$ \begin{aligned} f^{\prime}(x) & =\lim _{h \to 0} \frac{f(x+h)-f(x)}{h} \\ & =\lim _{h \to 0} \frac{\sin (x+h) \cos (x+h)-\sin x \cos x}{h} \\ & =\lim _{h \to 0} \frac{1}{2 h}[2 \sin (x+h) \cos (x+h)-2 \sin x \cos x] \\ & =\lim _{h \to 0} \frac{1}{2 h}[\sin 2(x+h)-\sin 2 x] \\ & =\lim _{h \to 0} \frac{1}{2 h}[2 \cos \frac{2 x+2 h+2 x}{2} \cdot \sin \frac{2 x+2 h-2 x}{2}] \\ & =\lim _{h \to 0} \frac{1}{h}[\cos \frac{4 x+2 h}{2} \sin \frac{2 h}{2}] \\ & =\lim _{h \to 0} \frac{1}{h}[\cos (2 x+h) \sin h] \\ & =\lim _{h \to 0} \cos (2 x+h) \cdot \lim _{h \to 0} \frac{\sin h}{h} \\ & =\cos (2 x+0) \cdot 1 \\ & =\cos 2 x \end{aligned} $

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

$ \begin{aligned} & f^{\prime}(x)=\lim _{h \to 0} \frac{f(x+h)-f(x)}{h} \\ & =\lim _{h \to 0} \frac{\sec (x+h)-\sec x}{h} \\ & =\lim _{h \to 0} \frac{1}{h}[\frac{1}{\cos (x+h)}-\frac{1}{\cos x}] \\ & =\lim _{h \to 0} \frac{1}{h}[\frac{\cos x-\cos (x+h)}{\cos x \cos (x+h)}] \\ & =\frac{1}{\cos x} \cdot \lim _{h \to 0} \frac{1}{h}[\frac{-2 \sin (\frac{x+x+h}{2}) \sin (\frac{x-x-h}{2})}{\cos (x+h)}] \\ & =\frac{1}{\cos x} \cdot \lim _{h \to 0} \frac{1}{h}[\frac{-2 \sin (\frac{2 x+h}{2}) \sin (-\frac{h}{2})}{\cos (x+h)}] \\ & =\frac{1}{\cos x} \lim _{h \to 0} \frac{[\sin (\frac{2 x+h}{2}) \frac{\sin (\frac{h}{2})}{(\frac{h}{2})}]}{\cos (x+h)} \\ & =\frac{1}{\cos x} \cdot \lim _{\frac{h}{2} \to 0} \frac{\sin (\frac{h}{2})}{(\frac{h}{2})} \cdot \lim _{h \to 0} \frac{\sin (\frac{2 x+h}{2})}{\cos (x+h)} \\ & =\frac{1}{\cos x} \cdot 1 \frac{\sin x}{\cos x} \\ & =\sec x \tan x \end{aligned} $

(iii) Letf $(x)=5 \sec x+4 \cos x$. Accordingly, from the first principle,

$ \begin{aligned} & f^{\prime}(x)=\lim _{h \to 0} \frac{f(x+h)-f(x)}{h} \\ & =\lim _{h \to 0} \frac{5 \sec (x+h)+4 \cos (x+h)-[5 \sec x+4 \cos x]}{h} \\ & =5 \lim _{h \to 0} \frac{[\sec (x+h)-\sec x]}{h}+4 \lim _{h \to 0} \frac{[\cos (x+h)-\cos x]}{h} \\ & =5 \lim _{h \to 0} \frac{1}{h}[\frac{1}{\cos (x+h)}-\frac{1}{\cos x}]+4 \lim _{h \to 0} \frac{1}{h}[\cos (x+h)-\cos x] \\ & =5 \lim _{h \to 0} \frac{1}{h}[\frac{\cos x-\cos (x+h)}{\cos x \cos (x+h)}]+4 \lim _{h \to 0} \frac{1}{h}[\cos x \cos h-\sin x \sin h-\cos x] \\ & =\frac{5}{\cos x} \lim _{h \to 0} \frac{1}{h}[\frac{-2 \sin (\frac{x+x+h}{2}) \sin (\frac{x-x-h}{2})}{\cos (x+h)}]+4 \lim _{h \to 0} \frac{1}{h}[-\cos x(1-\cos h)-\sin x \sin h] \\ & =\frac{5}{\cos x} \cdot \lim _{h \to 0} \frac{1}{h}[\frac{-2 \sin (\frac{2 x+h}{2}) \sin (-\frac{h}{2})}{\cos (x+h)}]+4[-\cos x \lim _{h \to 0} \frac{(1-\cos h)}{h}-\sin x \lim _{h \to 0} \frac{\sin h}{h}] \\ & =\frac{5}{\cos x} \cdot \lim _{h \to 0}[\frac{\sin (\frac{2 x+h}{2}) \cdot \frac{\sin (\frac{h}{2})}{\frac{h}{2}}}{\cos (x+h)}]+4[(-\cos x) \cdot(0)-(\sin x) \cdot 1] \\ & =\frac{5}{\cos x} \cdot[\lim _{h \to 0} \frac{\sin (\frac{2 x+h}{2})}{\cos (x+h)} \cdot \lim _{h \to 0} \frac{\sin (\frac{h}{2})}{\frac{h}{2}}]-4 \sin x \\ & =\frac{5}{\cos x} \cdot \frac{\sin x}{\cos x} \cdot 1-4 \sin x \\ & =5 \sec x \tan x \cdot-4 \sin x \end{aligned} $

(iv) Let $f(x)=cosec x$. Accordingly, from the first principle,

$ \begin{aligned} & f^{\prime}(x)=\lim _{h \to 0} \frac{f(x+h)-f(x)}{h} \\ & f^{\prime}(x)=\lim _{h \to 0} \frac{1}{h}[cosec(x+h)-cosec x] \\ & =\lim _{h \to 0} \frac{1}{h}[\frac{1}{\sin (x+h)}-\frac{1}{\sin x}] \\ & =\lim _{h \to 0} \frac{1}{h}[\frac{\sin x-\sin (x+h)}{\sin (x+h) \sin x}] \\ & =\lim _{h \to 0} \frac{1}{h}[\frac{2 \cos (\frac{x+x+h}{2}) \cdot \sin (\frac{x-x-h}{2})}{\sin (x+h) \sin x}] \\ & =\lim _{h \to 0} \frac{1}{h}[\frac{2 \cos (\frac{2 x+h}{2}) \sin (-\frac{h}{2})}{\sin (x+h) \sin x}] \\ & =\lim _{h \to 0} \frac{-\cos (\frac{2 x+h}{2}) \cdot \frac{\sin (\frac{h}{2})}{(\frac{h}{2})}}{\sin (x+h) \sin x} \\ & =\lim _{h \to 0}(\frac{-\cos (\frac{2 x+h}{2})}{\sin (x+h) \sin x}) \lim _{\frac{h}{2} \to 0} \frac{\sin (\frac{h}{2})}{(\frac{h}{2})} \\ & =(\frac{-\cos x}{\sin x \sin x}) \cdot 1 \\ & =-cosec x \cot x \end{aligned} $

(v) Let $f(x)=3 \cot x+5 cosec x$. Accordingly, from the first principle,

$$ \begin{align*} f^{\prime}(x) & =\lim _{h \to 0} \frac{f(x+h)-f(x)}{h} \\ & =\lim _{h \to 0} \frac{3 \cot (x+h)+5 cosec(x+h)-3 \cot x-5 cosec x}{h} \\ & =3 \lim _{h \to 0} \frac{1}{h}[\cot (x+h)-\cot x]+5 \lim _{h \to 0} \frac{1}{h}[cosec(x+h)-cosec x] \tag{1} \end{align*} $$

Now, $\lim _{h \to 0} \frac{1}{h}[\cot (x+h)-\cot x]$

$ =\lim _{h \to 0} \frac{1}{h}[\frac{\cos (x+h)}{\sin (x+h)}-\frac{\cos x}{\sin x}] $

$$ \begin{align*} & =\lim _{h \to 0} \frac{1}{h}[\frac{\cos (x+h) \sin x-\cos x \sin (x+h)}{\sin x \sin (x+h)}] \\ & =\lim _{h \to 0} \frac{1}{h}[\frac{\sin (x-x-h)}{\sin x \sin (x+h)}] \\ & =\lim _{h \to 0} \frac{1}{h}[\frac{\sin (-h)}{\sin x \sin (x+h)}] \\ & =-(\lim _{h \to 0} \frac{\sin h}{h}) \cdot(\lim _{h \to 0} \frac{1}{\sin x \cdot \sin (x+h)}) \\ & =-1 \cdot \frac{1}{\sin x \cdot \sin (x+0)}=\frac{-1}{\sin ^{2} x}=-cosec^{2} x \tag{2} \end{align*} $$

$$ \begin{aligned} & \lim _{h \to 0} \frac{1}{h}[cosec(x+h)-cosec x] \\ & =\lim _{h \to 0} \frac{1}{h}[\frac{1}{\sin (x+h)}-\frac{1}{\sin x}] \\ & =\lim _{h \to 0} \frac{1}{h}[\frac{\sin x-\sin (x+h)}{\sin (x+h) \sin x}] \\ & =\lim _{h \to 0} \frac{1}{h}[\frac{2 \cos (\frac{x+x+h}{2}) \cdot \sin (\frac{x-x-h}{2})}{\sin (x+h) \sin x}] \\ & =\lim _{h \to 0} \frac{1}{h}[\frac{2 \cos (\frac{2 x+h}{2}) \sin (-\frac{h}{2})}{\sin (x+h) \sin x}] \\ & \frac{-\cos (\frac{2 x+h}{2}) \cdot \frac{\sin (\frac{h}{2})}{(\frac{h}{2})}}{\sin (x+h) \sin x} \\ & =\lim _{h \to 0}(\frac{-\cos (\frac{2 x+h}{2})}{\sin (x+h) \sin x}) \lim _{\frac{h}{2} \to 0} \frac{\sin (\frac{h}{2})}{(\frac{h}{2})} \\ & =(\frac{-\cos x}{\sin x \sin x}) \cdot 1 \\ & =-cosec x \cot x \end{aligned} $$

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

$ f^{\prime}(x)=-3 cosec^{2} x-5 cosec x \cot x $

(vi) Let $f(x)=5 \sin x$ - $6 \cos x+7$. Accordingly, from the first principle,

$$ \begin{aligned} & f^{\prime}(x)=\lim _{h \to 0} \frac{f(x+h)-f(x)}{h} \\ & =\lim _{h \to 0} \frac{1}{h}[5 \sin (x+h)-6 \cos (x+h)+7-5 \sin x+6 \cos x-7] \\ & =\lim _{h \to 0} \frac{1}{h}[5{\sin (x+h)-\sin x}-6{\cos (x+h)-\cos x}] \\ & =5 \lim _{h \to 0} \frac{1}{h}[\sin (x+h)-\sin x]-6 \lim _{h \to 0} \frac{1}{h}[\cos (x+h)-\cos x] \\ & =5 \lim _{h \to 0} \frac{1}{h}[2 \cos (\frac{x+h+x}{2}) \sin (\frac{x+h-x}{2})]-6 \lim _{h \to 0} \frac{\cos x \cos h-\sin x \sin h-\cos x}{h} \\ & =5 \lim _{h \to 0} \frac{1}{h}[2 \cos (\frac{2 x+h}{2}) \sin \frac{h}{2}]-6 \lim _{h \to 0}[\frac{-\cos x(1-\cos h)-\sin x \sin h}{h}] \\ & =5 \lim _{h \to 0}(\cos (\frac{2 x+h}{2}) \frac{\sin \frac{h}{2}}{\frac{h}{2}})-6 \lim _{h \to 0}[\frac{-\cos x(1-\cos h)}{h}-\frac{\sin x \sin h}{h}] \\ & =5[\lim _{h \to 0} \cos (\frac{2 x+h}{2})][\frac{\lim _{h} \to 0}{\frac{\sin }{2}} \frac{h}{2}]-6[(-\cos x)(\lim _{h \to 0} \frac{1-\cos h}{h})-\sin x \lim _{h \to 0}(\frac{\sin h}{h})] \\ & =5 \cos x \cdot 1-6[(-\cos x) \cdot(0)-\sin x \cdot 1] \\ & =5 \cos x+6 \sin x \end{aligned} $$

(vii) Let $f(x)=2 \tan x$ - $7 \sec x$. Accordingly, from the first principle,

$$ \begin{aligned} & f^{\prime}(x)=\lim _{h \to 0} \frac{f(x+h)-f(x)}{h} \\ & =\lim _{h \to 0} \frac{1}{h}[2 \tan (x+h)-7 \sec (x+h)-2 \tan x+7 \sec x] \\ & =\lim _{h \to 0} \frac{1}{h}[2{\tan (x+h)-\tan x}-7{\sec (x+h)-\sec x}] \\ & =2 \lim _{h \to 0} \frac{1}{h}[\tan (x+h)-\tan x]-7 \lim _{h \to 0} \frac{1}{h}[\sec (x+h)-\sec x] \\ & =2 \lim _{h \to 0} \frac{1}{h}[\frac{\sin (x+h)}{\cos (x+h)}-\frac{\sin x}{\cos x}]-7 \lim _{h \to 0} \frac{1}{h}[\frac{1}{\cos (x+h)}-\frac{1}{\cos x}] \\ & =2 \lim _{h \to 0} \frac{1}{h}[\frac{\sin (x+h) \cos x-\sin x \cos (x+h)}{\cos x \cos (x+h)}]-7 \lim _{h \to 0} \frac{1}{h}[\frac{\cos x-\cos (x+h)}{\cos x \cos (x+h)}] \\ & =2 \lim _{h \to 0} \frac{1}{h}[\frac{\sin (x+h-x)}{\cos x \cos (x+h)}]-7 \lim _{h \to 0} \frac{1}{h}[\frac{-2 \sin (\frac{x+x+h}{2}) \sin (\frac{x-x-h}{2})}{\cos x \cos (x+h)}] \\ & =2 \lim _{h \to 0}[(\frac{\sin h}{h}) \frac{1}{\cos x \cos (x+h)}]-7 \lim _{h \to 0} \frac{1}{h}[\frac{-2 \sin (\frac{2 x+h}{2}) \sin (-\frac{h}{2})}{\cos x \cos (x+h)}] \\ & =2(\lim _{h \to 0} \frac{\sin h}{h})(\lim _{h \to 0} \frac{1}{\cos x \cos (x+h)})-7(\lim _{\frac{h}{2} \to 0} \frac{\sin \frac{h}{2}}{\frac{h}{2}})(\lim _{h \to 0} \frac{\sin (\frac{2 x+h}{2})}{\cos x \cos (x+h)}) \\ & =2.1 \cdot \frac{1}{\cos x \cos x}-7.1(\frac{\sin x}{\cos x \cos x}) \\ & =2 \sec ^{2} x-7 \sec x \tan x \end{aligned} $$

Answer :

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

By first principle,

$ \begin{aligned} f^{\prime}(x) & =\lim _{h \to 0} \frac{f(x+h)-f(x)}{h} \\ & =\lim _{h \to 0} \frac{-(x+h)-(-x)}{h} \\ & =\lim _{h \to 0} \frac{-x-h+x}{h} \\ & =\lim _{h \to 0} \frac{-h}{h} \\ & =\lim _{h \to 0}(-1)=-1 \end{aligned} $

(ii) Let

$ f(x)=(-x)^{-1}=\frac{1}{-x}=\frac{-1}{x} \text{.Accordingly, } f(x+h)=\frac{-1}{(x+h)} $

By first principle,

$ \begin{aligned} f^{\prime}(x) & =\lim _{h \to 0} \frac{f(x+h)-f(x)}{h} \\ & =\lim _{h \to 0} \frac{1}{h}[\frac{-1}{x+h}-(\frac{-1}{x})] \\ & =\lim _{h \to 0} \frac{1}{h}[\frac{-1}{x+h}+\frac{1}{x}] \\ & =\lim _{h \to 0} \frac{1}{h}[\frac{-x+(x+h)}{x(x+h)}] \\ & =\lim _{h \to 0} \frac{1}{h}[\frac{-x+x+h}{x(x+h)}] \\ & =\lim _{h \to 0} \frac{1}{h}[\frac{h}{x(x+h)}] \\ & =\lim _{h \to 0} \frac{1}{x(x+h)} \\ & =\frac{1}{x \cdot x}=\frac{1}{x^{2}} \end{aligned} $

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

By first principle,

$ \begin{aligned} f^{\prime}(x) & =\lim _{h \to 0} \frac{f(x+h)-f(x)}{h} \\ & =\lim _{h \to 0} \frac{1}{h}[\sin (x+h+1)-\sin (x+1)] \\ & =\lim _{h \to 0} \frac{1}{h}[2 \cos (\frac{x+h+1+x+1}{2}) \sin (\frac{x+h+1-x-1}{2})] \\ & =\lim _{h \to 0} \frac{1}{h}[2 \cos (\frac{2 x+h+2}{2}) \sin (\frac{h}{2})] \\ & =\lim _{h \to 0}[\cos (\frac{2 x+h+2}{2}) \cdot \frac{\sin (\frac{h}{2})}{(\frac{h}{2})}] \\ & .=\lim _{h \to 0} \cos (\frac{2 x+h+2}{2}) \cdot \lim _{\frac{h}{2} \to 0}^{(\frac{h}{2})} \quad[\text{ As } \frac{\sin }{2}) \Rightarrow \frac{h}{2} \to 0] \\ & =\cos (\frac{2 x+0+2}{2}) \cdot 1 \\ & =\cos (x+1) \\ & f(x)=\cos (x-\frac{\pi}{8}) \cdot \text{ Accordingly, } \end{aligned} $

By first principle,

$ \begin{aligned} f^{\prime}(x) & =\lim _{h \to 0} \frac{f(x+h)-f(x)}{h} \\ & =\lim _{h \to 0} \frac{1}{h}[\cos (x+h-\frac{\pi}{8})-\cos (x-\frac{\pi}{8})] \end{aligned} $

$ \begin{aligned} & =\lim _{h \to 0} \frac{1}{h}[-2 \sin \frac{(x+h-\frac{\pi}{8}+x-\frac{\pi}{8})}{2} \sin (\frac{x+h-\frac{\pi}{8}-x+\frac{\pi}{8}}{2})] \\ & =\lim _{h \to 0} \frac{1}{h}[-2 \sin (\frac{2 x+h-\frac{\pi}{4}}{2}) \sin \frac{h}{2}] \\ & =\lim _{h \to 0}[-\sin (\frac{2 x+h-\frac{\pi}{4}}{2}) \frac{\sin (\frac{h}{2})}{(\frac{h}{2})}] \end{aligned} $

$ \begin{aligned} =\lim _{h \to 0}[-\sin (\frac{2 x+h-\frac{\pi}{4}}{2})] .\lim _{\frac{h}{2} \to 0}\frac{\sin (\frac{h}{2})}{(\frac{h}{2})} \hspace{30px} [\text{As } h \to 0 \Rightarrow \frac{h}{2} \to 0] \end{aligned} $

$ \begin{aligned} & =-\sin (\frac{2 x+0-\frac{\pi}{4}}{2}) .1 \\ & =-\sin (x-\frac{\pi}{8}) \end{aligned} $

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):

Answer :

Let $f(x)=x+a$. Accordingly, $f(x+h)=x+h+a$

By first principle,

$ \begin{aligned} f^{\prime}(x) & =\lim _{h \to 0} \frac{f(x+h)-f(x)}{h} \\ & =\lim _{h \to 0} \frac{x+h+a-x-a}{h} \\ & =\lim _{h \to 0}(\frac{h}{h}) \\ & =\lim _{h \to 0}(1) \\ & =1 \end{aligned} $

Answer :

Let $f(x)=(p x+q)(\frac{r}{x}+s)$

By Leibnitz product rule,

$ \begin{aligned} f^{\prime}(x) & =(p x+q)(\frac{r}{x}+s)^{\prime}+(\frac{r}{x}+s)(p x+q)^{\prime} \\ & =(p x+q)(r x^{-1}+s)^{\prime}+(\frac{r}{x}+s)(p) \\ & =(p x+q)(-r x^{-2})+(\frac{r}{x}+s) p \\ & =(p x+q)(\frac{-r}{x^{2}})+(\frac{r}{x}+s) p \\ & =\frac{-p r}{x}-\frac{q r}{x^{2}}+\frac{p r}{x}+p s \\ & =p s-\frac{q r}{x^{2}} \end{aligned} $

Answer :

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

By Leibnitz product rule,

$ \begin{aligned} f^{\prime}(x) & =(a x+b) \frac{d}{d x}(c x+d)^{2}+(c x+d)^{2} \frac{d}{d x}(a x+b) \\ & =(a x+b) \frac{d}{d x}(c^{2} x^{2}+2 c d x+d^{2})+(c x+d)^{2} \frac{d}{d x}(a x+b) \\ & =(a x+b)[\frac{d}{d x}(c^{2} x^{2})+\frac{d}{d x}(2 c d x)+\frac{d}{d x} d^{2}]+(c x+d)^{2}[\frac{d}{d x} a x+\frac{d}{d x} b] \\ & =(a x+b)(2 c^{2} x+2 c d)+(c x+d^{2}) a \\ & =2 c(a x+b)(c x+d)+a(c x+d)^{2} \end{aligned} $

Answer :

Let $f(x)=\frac{a x+b}{c x+d}$

By quotient rule,

$ \begin{aligned} f^{\prime}(x) & =\frac{(c x+d) \frac{d}{d x}(a x+b)-(a x+b) \frac{d}{d x}(c x+d)}{(c x+d)^{2}} \\ & =\frac{(c x+d)(a)-(a x+b)(c)}{(c x+d)^{2}} \\ & =\frac{a c x+a d-a c x-b c}{(c x+d)^{2}} \\ & =\frac{a d-b c}{(c x+d)^{2}} \end{aligned} $

Answer :

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

By quotient rule,

$ \begin{aligned} f^{\prime}(x) & =\frac{(x-1) \frac{d}{d x}(x+1)-(x+1) \frac{d}{d x}(x-1)}{(x-1)^{2}}, x \neq 0,1 \\ & =\frac{(x-1)(1)-(x+1)(1)}{(x-1)^{2}}, x \neq 0,1 \\ & =\frac{x-1-x-1}{(x-1)^{2}}, x \neq 0,1 \\ & =\frac{-2}{(x-1)^{2}}, x \neq 0,1 \end{aligned} $

Answer :

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

By quotient rule,

$ \begin{aligned} f^{\prime}(x) & =\frac{(a x^{2}+b x+c) \frac{d}{d x}(1)-\frac{d}{d x}(a x^{2}+b x+c)}{(a x^{2}+b x+c)^{2}} \\ & =\frac{(a x^{2}+b x+c)(0)-(2 a x+b)}{(a x^{2}+b x+c)^{2}} \\ & =\frac{-(2 a x+b)}{(a x^{2}+b x+c)^{2}} \end{aligned} $

Answer :

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

By quotient rule,

$ \begin{aligned} f^{\prime}(x) & =\frac{(p x^{2}+q x+r) \frac{d}{d x}(a x+b)-(a x+b) \frac{d}{d x}(p x^{2}+q x+r)}{(p x^{2}+q x+r)^{2}} \\ & =\frac{(p x^{2}+q x+r)(a)-(a x+b)(2 p x+q)}{(p x^{2}+q x+r)^{2}} \\ & =\frac{a p x^{2}+a q x+a r-2 a p x^{2}-a q x-2 b p x-b q}{(p x^{2}+q x+r)^{2}} \\ & =\frac{-a p x^{2}-2 b p x+a r-b q}{(p x^{2}+q x+r)^{2}} \end{aligned} $

Answer :

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

By quotient rule,

$ \begin{aligned} f^{\prime}(x) & =\frac{(a x+b) \frac{d}{d x}(p x^{2}+q x+r)-(p x^{2}+q x+r) \frac{d}{d x}(a x+b)}{(a x+b)^{2}} \\ & =\frac{(a x+b)(2 p x+q)-(p x^{2}+q x+r)(a)}{(a x+b)^{2}} \\ & =\frac{2 a p x^{2}+a q x+2 b p x+b q-a p x^{2}-a q x-a r}{(a x+b)^{2}} \\ & =\frac{a p x^{2}+2 b p x+b q-a r}{(a x+b)^{2}} \end{aligned} $

Answer :

$ \begin{aligned} & \text{ Let } f(x)=\frac{a}{x^{4}}-\frac{b}{x^{2}}+\cos x \\ & f^{\prime}(x)=\frac{d}{d x}(\frac{a}{x^{4}})-\frac{d}{d x}(\frac{b}{x^{2}})+\frac{d}{d x}(\cos x) \\ & =a \frac{d}{d x}(x^{-4})-b \frac{d}{d x}(x^{-2})+\frac{d}{d x}(\cos x) \\ & =a(-4 x^{-5})-b(-2 x^{-3})+(-\sin x) \quad[\frac{d}{d x}(x^{n})=n x^{n-1} \text{ and } \frac{d}{d x}(\cos x)=-\sin x] \\ & =\frac{-4 a}{x^{3}}+\frac{2 b}{x^{3}}-\sin x \end{aligned} $

Answer :

$ \begin{aligned} & \text{ Let } f(x)=4 \sqrt{x}-2 \\ & \begin{aligned} f^{\prime}(x) & =\frac{d}{d x}(4 \sqrt{x}-2)=\frac{d}{d x}(4 \sqrt{x})-\frac{d}{d x}(2) \\ & =4 \frac{d}{d x}(x^{\frac{1}{2}})-0=4(\frac{1}{2} x^{\frac{1}{2}-1}) \\ & =(2 x^{-\frac{1}{2}})=\frac{2}{\sqrt{x}} \end{aligned} \end{aligned} $

Answer :

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,

$ \begin{aligned} f^{\prime}(x) & =\lim _{h \to 0} \frac{f(x+h)-f(x)}{h} \\ & =\lim _{h \to 0} \frac{(a x+a h+b)^{n}-(a x+b)^{n}}{h} \\ & =\lim _{h \to 0} \frac{(a x+b)^{n}(1+\frac{a h}{a x+b})^{n}-(a x+b)^{n}}{h} \\ & =(a x+b)^{n} \lim _{h \to 0} \frac{(1+\frac{a h}{a x+b})^{n}-1}{h} \\ & =(a x+b)^{n} \lim _{h \to 0} \frac{1}{n}[{1+n(\frac{a h}{a x+b})+\frac{n(n-1)}{\lfloor 2}(\frac{a h}{a x+b})^{2}+\ldots}-1] \\ & ..=(a x+b)^{n} \lim _{h \to 0} \frac{1}{h}[\frac{n}{a h})^{n}]+\frac{n(n-1) a^{2} h^{2}}{\lfloor(a x+b)^{2}.}+\ldots(\text{ Terms containing higher degrees of } h)] \\ & =(a x+b)^{n} \lim _{h \to 0}[\frac{n a}{(a x+b)}+\frac{n(n-1) a^{2} h}{\lfloor(a x+b)^{2}.}+\ldots] \\ & =(a x+b)^{n}[\frac{n a}{(a x+b)}+0] \\ & .=n a \frac{(a x+b)^{n}}{(a x+b)}[a x+b)^{n-1}] \end{aligned} $

Answer :

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

By Leibnitz product rule,

$$ \begin{equation*} f^{\prime}(x)=(a x+b)^{n} \frac{d}{d x}(c x+d)^{m}+(c x+d)^{m} \frac{d}{d x}(a x+b)^{n} \tag{1} \end{equation*} $$

Now, let $f_1(x)=(c x+d)^{m}$

$f_1(x+h)=(c x+c h+d)^{m}$

$f_1^{\prime}(x)=\lim _{h \to 0} \frac{f_1(x+h)-f_1^{\prime}(x)}{h}$

$=\lim _{h \to 0} \frac{(c x+c h+d)^{m}-(c x+d)^{m}}{h}$

$=(c x+d)^{m} \lim _{h \to 0} \frac{1}{h}[(1+\frac{c h}{c x+d})^{m+}-1]$

$=(c x+d)^{m} \lim _{h \to 0} \frac{1}{h}[(1+\frac{m c h}{(c x+d)}+\frac{m(m-1)}{2} \frac{(c^{2} h^{2})}{(c x+d)^{2}}+\ldots)-1]$

$=(c x+d)^{m} \lim _{h \to 0} \frac{1}{h}[\frac{m c h}{(c x+d)}+\frac{m(m-1) c^{2} h^{2}}{2(c x+d)^{2}}+\ldots(.$ Terms containing higher degrees of $.h)]$

$=(c x+d)^{m} \lim _{h \to 0}[\frac{m c}{(c x+d)}+\frac{m(m-1) c^{2} h}{2(c x+d)^{2}}+\ldots]$

$=(c x+d)^{m}[\frac{m c}{c x+d}+0]$

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

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

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

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

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

$ \begin{aligned} f^{\prime}(x) & =(a x+b)^{n}{m c(c x+d)^{m-1}}+(c x+d)^{m}{n a(a x+b)^{n-1}} \\ & =(a x+b)^{n-1}(c x+d)^{m-1}[m c(a x+b)+n a(c x+d)] \end{aligned} $

Answer :

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

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

By first principle,

$ \begin{aligned} & f^{\prime}(x)=\lim _{h \to 0} \frac{f(x+h)-f(x)}{h} \\ & =\lim _{h \to 0} \frac{\sin (x+h+a)-\sin (x+a)}{h} \\ & =\lim _{h \to 0} \frac{1}{h}[2 \cos (\frac{x+h+a+x+a}{2}) \sin (\frac{x+h+a-x-a}{2})] \\ & =\lim _{h \to \infty} \frac{1}{h}[2 \cos (\frac{2 x+2 a+h}{2}) \sin (\frac{h}{2})] \\ & =\lim _{h \to 0}[\cos (\frac{2 x+2 a+h}{2}){\frac{\sin (\frac{h}{2})}{(\frac{h}{2})}}] \\ & =\lim _{h \to 0} \cos (\frac{2 x+2 a+h}{2}) \lim _{\frac{h}{2} \to 0}{\frac{\sin (\frac{h}{2})}{(\frac{h}{2})}} \quad[\text{ As } h \to 0 \Rightarrow \frac{h}{2} \to 0] \\ & =\cos (\frac{2 x+2 a}{2}) \times 1 \quad[\lim _{x \to 0} \frac{\sin x}{x}=1] \\ & =\cos (x+a) \end{aligned} $

Answer :

Let $f(x)=cosec x \cot x$

By Leibnitz product rule,

$f^{\prime}(x)=cosec x(\cot x)^{\prime}+\cot x(cosec x)^{\prime}$

Let $f_1(x)=\cot x$. Accordingly, $f_1(x+h)=\cot (x+h)$

By first principle,

$$ \begin{align*} f_1^{\prime}(x) & =\lim _{h \to 0} \frac{f_1(x+h)-f_1(x)}{h} \\ & =\lim _{h \to 0} \frac{\cot (x+h)-\cot x}{h} \\ & =\lim _{h \to 0} \frac{1}{h}(\frac{\cos (x+h)}{\sin (x+h)}-\frac{\cos x}{\sin x}) \\ & =\lim _{h \to 0} \frac{1}{h}[\frac{\sin x \cos (x+h)-\cos x \sin (x+h)}{\sin x \sin (x+h)}] \\ & =\lim _{h \to 0} \frac{1}{h}[\frac{\sin (x-x-h)}{\sin x \sin (x+h)}] \\ & =\frac{1}{\sin _{x} x} \cdot \lim _{h \to 0} \frac{1}{h}[\frac{\sin (-h)}{\sin (x+h)}] \\ & =\frac{-1}{\sin ^{x} x} \cdot(\lim _{h \to 0} \frac{\sin h}{h})(\lim _{h \to 0} \frac{1}{\sin (x+h)}) \\ & =\frac{-1}{\sin ^{x}} \cdot 1 \cdot(\frac{1}{\sin (x+0)}) \\ & =\frac{-1}{\sin ^{2} x} \\ & =-cosec^{2} x \tag{2} \end{align*} $$

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

Now, let $f_2(x)=cosec x$. Accordingly, $f_2(x+h)=cosec(x+h)$

By first principle,

$ \begin{aligned} f_2^{\prime}(x) & =\lim _{h \to 0} \frac{f_2(x+h)-f_2(x)}{h} \\ & =\lim _{h \to 0} \frac{1}{h}[cosec(x+h)-cosec x] \end{aligned} $

$ \begin{aligned} & =\lim _{h \to 0} \frac{1}{h}[\frac{1}{\sin (x+h)}-\frac{1}{\sin x}] \\ & =\lim _{h \to 0} \frac{1}{h}[\frac{\sin x-\sin (x+h)}{\sin x \sin (x+h)}] \end{aligned} $

$$ \begin{align*} & =\frac{1}{\sin x} \cdot \lim _{h \to 0} \frac{1}{h}[\frac{2 \cos (\frac{x+x+h}{2}) \sin (\frac{x-x-h}{2})}{\sin (x+h)}] \\ & =\frac{1}{\sin x} \cdot \lim _{h \to 0} \frac{1}{h}[\frac{.2 \cos (\frac{2 x+h}{2}) \sin (\frac{-h}{2})]}{\sin (x+h)}] \\ & =\frac{1}{\sin x} \cdot \lim _{h \to 0}[\frac{-\sin (\frac{h}{2})}{(\frac{h}{2})} \cdot \frac{\cos (\frac{2 x+h}{2})}{\sin (x+h)}] \\ & =\frac{-1}{\sin x} \cdot \lim _{h \to 0} \frac{\sin (\frac{h}{2})}{(\frac{h}{2})} \cdot \lim _{h \to 0} \frac{\cos (\frac{2 x+h}{2})}{\sin (x+h)} \\ & =\frac{-1}{\sin x} \cdot 1 \cdot \frac{\cos (\frac{2 x+0}{2})}{\sin (x+0)} \\ & =\frac{-1}{\sin x} \cdot \frac{\cos x}{\sin x} \\ & =-\cos ec x \cdot \cot x \tag{3} \end{align*} $$

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

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

$ \begin{aligned} f^{\prime}(x) & =cosec x(-cosec^{2} x)+\cot x(-cosec x \cot x) \\ & =-cosec^{3} x-\cot ^{2} x cosec x \end{aligned} $

Answer :

Let $f(x)=\frac{\cos x}{1+\sin x}$

By quotient rule,

$ \begin{aligned} f^{\prime}(x) & =\frac{(1+\sin x) \frac{d}{d x}(\cos x)-(\cos x) \frac{d}{d x}(1+\sin x)}{(1+\sin x)^{2}} \\ & =\frac{(1+\sin x)(-\sin x)-(\cos x)(\cos x)}{(1+\sin x)^{2}} \\ & =\frac{-\sin x-\sin ^{2} x-\cos ^{2} x}{(1+\sin x)^{2}} \\ & =\frac{-\sin x-(\sin ^{2} x+\cos ^{2} x)}{(1+\sin x)^{2}} \\ & =\frac{-\sin x-1}{(1+\sin x)^{2}} \\ & =\frac{-(1+\sin x)}{(1+\sin x)^{2}} \\ & =\frac{-1}{(1+\sin x)} \end{aligned} $

Answer :

Let $f(x)=\frac{\sin x+\cos x}{\sin x-\cos x}$

By quotient rule,

$ \begin{aligned} f^{\prime}(x) & =\frac{(\sin x-\cos x) \frac{d}{d x}(\sin x+\cos x)-(\sin x+\cos x) \frac{d}{d x}(\sin x-\cos x)}{(\sin x-\cos x)^{2}} \\ & =\frac{(\sin x-\cos x)(\cos x-\sin x)-(\sin x+\cos x)(\cos x+\sin x)}{(\sin x-\cos x)^{2}} \\ & =\frac{-(\sin x-\cos x)^{2}-(\sin x+\cos x)^{2}}{(\sin x-\cos x)^{2}} \\ & =\frac{-[\sin ^{2} x+\cos ^{2} x-2 \sin x \cos x+\sin ^{2} x+\cos ^{2} x+2 \sin x \cos x]}{(\sin x-\cos x)^{2}} \\ & =\frac{-[1+1]}{(\sin x-\cos x)^{2}} \\ & =\frac{-2}{(\sin x-\cos x)^{2}} \end{aligned} $

Answer :

Let $f(x)=\frac{\sec x-1}{\sec x+1}$

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

By quotient rule,

$ \begin{aligned} & f^{\prime}(x)=\frac{(1+\cos x) \frac{d}{d x}(1-\cos x)-(1-\cos x) \frac{d}{d x}(1+\cos x)}{(1+\cos x)^{2}} \\ & =\frac{(1+\cos x)(\sin x)-(1-\cos x)(-\sin x)}{(1+\cos x)^{2}} \\ & =\frac{\sin x+\cos x \sin x+\sin x-\sin x \cos x}{(1+\cos x)^{2}} \\ & =\frac{2 \sin x}{(1+\cos x)^{2}} \\ & =\frac{2 \sin x}{(1+\frac{1}{\sec x})^{2}}=\frac{2 \sin x}{\frac{(\sec x+1)^{2}}{\sec ^{2} x}} \\ & =\frac{2 \sin x \sec ^{2} x}{(\sec x+1)^{2}} \\ & =\frac{\frac{2 \sin x}{\cos x} \sec x}{(\sec x+1)^{2}} \\ & =\frac{2 \sec x \tan x}{(\sec x+1)^{2}} \end{aligned} $

Answer :

Let $y=\sin ^{n} x$.

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

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

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

$$ \begin{align*} \therefore \frac{d y}{d x} & =\frac{d}{d x}(\sin x \sin x) \\ & =(\sin x)^{\prime} \sin x+\sin x(\sin x)^{\prime} \\ & =\cos x \sin x+\sin x \cos x \\ & =2 \sin x \cos x \tag{1} \end{align*} $$

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

$ \therefore \frac{d y}{d x}=\frac{d}{d x}(\sin x \sin ^{2} x) $

$$ \begin{matrix} =(\sin x)^{\prime} \sin ^{2} x+\sin x(\sin ^{2} x)^{\prime} & \text{ [By Leibnitz product rule] } \\ =\cos x \sin ^{2} x+\sin x(2 \sin x \cos x) & \text{ [Using (1)] } \\ =\cos x \sin ^{2} x+2 \sin ^{2} x \cos x & \\ =3 \sin ^{2} x \cos x & \end{matrix} $$

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

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

i.e., $\frac{d}{d x}(\sin ^{k} x)=k \sin ^{(k-1)} x \cos x$

Consider

$ \begin{aligned} \frac{d}{d x}(\sin ^{k+1} x) & =\frac{d}{d x}(\sin x \sin ^{k} x) \\ & =(\sin x)^{\prime} \sin ^{k} x+\sin x(\sin ^{k}. \\ & =\cos x \sin ^{k} x+\sin x(k \sin ^{(k-1}. \\ & =\cos x \sin ^{k} x+k \sin ^{k} x \cos x \\ & =(k+1) \sin ^{k} x \cos x \end{aligned} $

$ \begin{matrix} =(\sin x)^{\prime} \sin ^{k} x+\sin x(\sin ^{k} x)^{\prime} & \text{ [By Leibnitz product rule] } \\ =\cos x \sin ^{k} x+\sin x(k \sin ^{(k-1)} x \cos x) & {[\text{ Using (2)] }} \end{matrix} $

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

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

Answer :

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

By quotient rule,

$ \begin{aligned} f^{\prime}(x) & =\frac{(c+d \cos x) \frac{d}{d x}(a+b \sin x)-(a+b \sin x) \frac{d}{d x}(c+d \cos x)}{(c+d \cos x)^{2}} \\ & =\frac{(c+d \cos x)(b \cos x)-(a+b \sin x)(-d \sin x)}{(c+d \cos x)^{2}} \\ & =\frac{c b \cos x+b d \cos ^{2} x+a d \sin x+b d \sin ^{2} x}{(c+d \cos x)^{2}} \\ & =\frac{b c \cos x+a d \sin x+b d(\cos ^{2} x+\sin ^{2} x)}{(c+d \cos x)^{2}} \\ & =\frac{b c \cos x+a d \sin x+b d}{(c+d \cos x)^{2}} \end{aligned} $

Answer :

Let $f(x)=\frac{\sin (x+a)}{\cos x}$

By quotient rule,

$$ \begin{align*} f^{\prime}(x) & =\frac{\cos x \frac{d}{d x}[\sin (x+a)]-\sin (x+a) \frac{d}{d x} \cos x}{\cos ^{2} x} \\ f^{\prime}(x) & =\frac{\cos x \frac{d}{d x}[\sin (x+a)]-\sin (x+a)(-\sin x)}{\cos ^{2} x} \tag{i} \end{align*} $$

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

By first principle,

$$ \begin{align*} g^{\prime}(x) & =\lim _{h \to 0} \frac{g(x+h)-g(x)}{h} \\ & =\lim _{h \to 0} \frac{1}{h}[\sin (x+h+a)-\sin (x+a)] \\ & =\lim _{h \to 0} \frac{1}{h}[2 \cos (\frac{x+h+a+x+a}{2}) \sin (\frac{x+h+a-x-a}{2})] \\ & =\lim _{h \to 0} \frac{1}{h}[2 \cos (\frac{2 x+2 a+h}{2}) \sin (\frac{h}{2})] \\ & =\lim _{h \to 0}{\operatorname { c o s } ( \frac { 2 x + 2 a + h } { 2 } ) [\frac{\sin (\frac{h}{2})}.(\frac{h}{2})]} \quad[\text{ As } h \to 0 \Rightarrow \frac{h}{2} \to 0].. \\ & =\lim _{h \to 0} \cos (\frac{2 x+2 a+h}{2}) \lim _{\frac{h}{2} \to 0}{\frac{\sin (\frac{h}{2})}{(\frac{h}{2})}} \quad[\lim _{h \to 0} \frac{\sin h}{h}=1] \\ & =(\cos \frac{2 x+2 a}{2}) \times 1 \\ & =\cos (x+a) \tag{ii} \end{align*} \text{ (ii) } \quad $$

From (i) and (ii), we obtain

$ \begin{aligned} f^{\prime}(x) & =\frac{\cos x \cdot \cos (x+a)+\sin x \sin (x+a)}{\cos ^{2} x} \\ & =\frac{\cos (x+a-x)}{\cos ^{2} x} \\ & =\frac{\cos a}{\cos ^{2} x} \end{aligned} $

Answer :

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

By product rule,

$ \begin{aligned} f^{\prime}(x) & =x^{4} \frac{d}{d x}(5 \sin x-3 \cos x)+(5 \sin x-3 \cos x) \frac{d}{d x}(x^{4}) \\ & =x^{4}[5 \frac{d}{d x}(\sin x)-3 \frac{d}{d x}(\cos x)]+(5 \sin x-3 \cos x) \frac{d}{d x}(x^{4}) \\ & =x^{4}[5 \cos x-3(-\sin x)]+(5 \sin x-3 \cos x)(4 x^{3}) \\ & =x^{3}[5 x \cos x+3 x \sin x+20 \sin x-12 \cos x] \end{aligned} $

Answer:

Let $f(x)=(x^{2}+1) \cos x$

By product rule,

$ \begin{aligned} f^{\prime}(x) & =(x^{2}+1) \frac{d}{d x}(\cos x)+\cos x \frac{d}{d x}(x^{2}+1) \\ & =(x^{2}+1)(-\sin x)+\cos x(2 x) \\ & =-x^{2} \sin x-\sin x+2 x \cos x \end{aligned} $

Answer :

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

By product rule,

$ \begin{aligned} f^{\prime}(x) & =(a x^{2}+\sin x) \frac{d}{d x}(p+q \cos x)+(p+q \cos x) \frac{d}{d x}(a x^{2}+\sin x) \\ & =(a x^{2}+\sin x)(-q \sin x)+(p+q \cos x)(2 a x+\cos x) \\ & =-q \sin x(a x^{2}+\sin x)+(p+q \cos x)(2 a x+\cos x) \end{aligned} $

Answer :

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

By product rule,

$$ \begin{align*} & \begin{align*} f^{\prime}(x) & =(x+\cos x) \frac{d}{d x}(x-\tan x)+(x-\tan x) \frac{d}{d x}(x+\cos x) \\ & =(x+\cos x)[\frac{d}{d x}(x)-\frac{d}{d x}(\tan x)]+(x-\tan x)(1-\sin x) \\ & =(x+\cos x)[1-\frac{d}{d x} \tan x]+(x-\tan x)(1-\sin x) \end{align*} \\ & \text{ Let } g(x)=\tan x \text{. Accordingly, } g(x+h)=\tan (x+h) \end{align*} $$

By first principle,

$$ \begin{align*} g^{\prime}(x) & =\lim _{h \to 0} \frac{g(x+h)-g(x)}{h} \\ & =\lim _{h \to 0}(\frac{\tan (x+h)-\tan x}{h}) \\ & =\lim _{h \to 0} \frac{1}{h}[\frac{\sin (x+h)}{\cos (x+h)}-\frac{\sin x}{\cos x}] \\ & =\lim _{h \to 0} \frac{1}{h}[\frac{\sin (x+h) \cos x-\sin x \cos (x+h)}{\cos (x+h) \cos x}] \\ & =\frac{1}{\cos x} \cdot \lim _{h \to 0} \frac{1}{h}[\frac{\sin (x+h-x)}{\cos (x+h)}] \\ & =\frac{1}{\cos x} \cdot \lim _{h \to 0} \frac{1}{h}[\frac{\sin h}{\cos (x+h)}] \\ & =\frac{1}{\cos ^{x}} \cdot(\lim _{h \to 0} \frac{\sin h}{h}) \cdot(\lim _{h \to 0} \frac{1}{\cos (x+h)}) \\ & =\frac{1}{\cos ^{2}} \cdot 1 \cdot \frac{1}{\cos (x+0)} \\ & =\frac{1}{\cos ^{2} x} \\ & =\sec ^{2} x \tag{ii} \end{align*} $$

Therefore, from (i) and (ii), we obtain

$$ \begin{aligned} f^{\prime}(x) & =(x+\cos x)(1-\sec ^{2} x)+(x-\tan x)(1-\sin x) \\ & =(x+\cos x)(-\tan ^{2} x)+(x-\tan x)(1-\sin x) \\ & =-\tan ^{2} x(x+\cos x)+(x-\tan x)(1-\sin x) \end{aligned} $$

Answer :

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

By quotient rule,

$ \begin{aligned} f^{\prime}(x) & =\frac{(3 x+7 \cos x) \frac{d}{d x}(4 x+5 \sin x)-(4 x+5 \sin x) \frac{d}{d x}(3 x+7 \cos x)}{(3 x+7 \cos x)^{2}} \\ & =\frac{(3 x+7 \cos x)[4 \frac{d}{d x}(x)+5 \frac{d}{d x}(\sin x)]-(4 x+5 \sin x)[3 \frac{d}{d x} x+7 \frac{d}{d x} \cos x]}{(3 x+7 \cos x)^{2}} \\ & =\frac{(3 x+7 \cos x)(4+5 \cos x)-(4 x+5 \sin x)(3-7 \sin x)}{(3 x+7 \cos x)^{2}} \\ & =\frac{12 x+15 x \cos x+28 \cos x+35 \cos ^{2} x-12 x+28 x \sin x-15 \sin x+35 \sin ^{2} x}{(3 x+7 \cos x)^{2}} \\ & =\frac{15 x \cos x+28 \cos x+28 x \sin x-15 \sin x+35(\cos ^{2} x+\sin ^{2} x)}{(3 x+7 \cos x)^{2}} \\ & =\frac{35+15 x \cos x+28 \cos x+28 x \sin x-15 \sin x}{(3 x+7 \cos x)^{2}} \end{aligned} $

Answer :

Let $f(x)=\frac{x^{2} \cos (\frac{\pi}{4})}{\sin x}$

By quotient rule,

$ \begin{aligned} f^{\prime}(x) & =\cos \frac{\pi}{4} \cdot[\frac{\sin x \frac{d}{d x}(x^{2})-x^{2} \frac{d}{d x}(\sin x)}{\sin ^{2} x}] \\ & =\cos \frac{\pi}{4} \cdot[\frac{\sin x \cdot 2 x-x^{2} \cos x}{\sin ^{2} x}] \\ & =\frac{x \cos \frac{\pi}{4}[2 \sin x-x \cos x]}{\sin ^{2} x} \end{aligned} $

Answer :

Let $f(x)=\frac{x}{1+\tan x}$

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

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

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

By first principle,

$$ \begin{align*} g^{\prime}(x) & =\lim _{h \to 0} \frac{g(x+h)-g(x)}{h} \\ & =\lim _{h \to 0}[\frac{1+\tan (x+h)-1-\tan x}{h}] \\ & =\lim _{h \to 0} \frac{1}{h}[\frac{\sin (x+h)}{\cos (x+h)}-\frac{\sin x}{\cos x}] \\ & =\lim _{h \to 0} \frac{1}{h}[\frac{\sin (x+h) \cos x-\sin x \cos (x+h)}{\cos (x+h) \cos x}] \\ & =\lim _{h \to 0} \frac{1}{h}[\frac{\sin (x+h-x)}{\cos (x+h) \cos x}] \\ & =\lim _{h \to 0} \frac{1}{h}[\frac{\sin h}{\cos (x+h) \cos x}] \\ & =(\lim _{h \to 0} \frac{\sin h}{h}) \cdot(\lim _{h \to 0} \frac{1}{\cos (x+h) \cos x}) \\ & =1 \times \frac{1}{\cos ^{2} x}=\sec ^{2} x \\ \Rightarrow \frac{d}{d x} & (1+\tan x)=\sec ^{2} x \tag{ii} \end{align*} $$

From (i) and (ii), we obtain

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

Answer :

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

By product rule,

$$ \begin{align*} f^{\prime}(x) & =(x+\sec x) \frac{d}{d x}(x-\tan x)+(x-\tan x) \frac{d}{d x}(x+\sec x) \\ & =(x+\sec x)[\frac{d}{d x}(x)-\frac{d}{d x} \tan x]+(x-\tan x)[\frac{d}{d x}(x)+\frac{d}{d x} \sec x] \\ & =(x+\sec x)[1-\frac{d}{d x} \tan x]+(x-\tan x)[1+\frac{d}{d x} \sec x] \tag{i} \end{align*} $$

Let $f_1(x)=\tan x, f_2(x)=\sec x$

Accordingly, $f_1(x+h)=\tan (x+h)$ and $f_2(x+h)=\sec (x+h)$

$$ \begin{align*} f_1^{\prime}(x) & =\lim _{h \to 0}(\frac{f_1(x+h)-f_1(x)}{h}) \\ & =\lim _{h \to 0}(\frac{\tan (x+h)-\tan x}{h}) \\ & =\lim _{h \to 0}[\frac{\tan (x+h)-\tan x}{h}] \\ & =\lim _{h \to 0} \frac{1}{h}[\frac{\sin (x+h)}{\cos (x+h)}-\frac{\sin x}{\cos x}] \\ & =\lim _{h \to 0} \frac{1}{h}[\frac{\sin (x+h) \cos x-\sin x \cos (x+h)}{\cos (x+h) \cos x}] \\ & =\lim _{h \to 0} \frac{1}{h}[\frac{\sin (x+h-x)}{\cos (x+h) \cos x}] \\ & =\lim _{h \to 0} \frac{1}{h}[\frac{\sin h}{\cos (x+h) \cos x}] \\ & =(\lim _{h \to 0} \frac{\sin h}{h}) \cdot(\lim _{h \to 0} \frac{1}{\cos (x+h) \cos x}) \\ & =1 \times \frac{1}{\cos ^{2} x}=\sec ^{2} x \\ \Rightarrow \frac{d}{d x} & \tan x=\sec ^{2} x \tag{ii} \end{align*} $$

$$ \begin{aligned} & f_2^{\prime}(x)=\lim _{h \to 0}(\frac{f_2(x+h)-f_2(x)}{h}) \\ & =\lim _{h \to 0}(\frac{\sec (x+h)-\sec x}{h}) \\ & =\lim _{h \to 0} \frac{1}{h}[\frac{1}{\cos (x+h)}-\frac{1}{\cos x}] \\ & =\lim _{h \to 0} \frac{1}{h}[\frac{\cos x-\cos (x+h)}{\cos (x+h) \cos x}] \\ & =\frac{1}{\cos x} \cdot \lim _{h \to 0} \frac{1}{h}[\frac{-2 \sin (\frac{x+x+h}{2}) \cdot \sin (\frac{x-x-h}{2})}{\cos (x+h)}] \\ & =\frac{1}{\cos x} \cdot \lim _{h \to 0} \frac{1}{h}[\frac{-2 \sin (\frac{2 x+h}{2}) \cdot \sin (\frac{-h}{2})}{\cos (x+h)}] \\ & =\frac{1}{\cos x} \cdot \lim _{h \to 0}[\frac{\sin (\frac{2 x+h}{2}){\frac{\sin (\frac{h}{2})}{\frac{h}{2}}}}{\cos (x+h)}] \\ & =\sec x \frac{{\lim _{h \to 0} \sin (\frac{2 x+h}{2})}{\lim _{\frac{h}{2} \to 0} \frac{\sin (\frac{h}{2})}{\frac{h}{2}}}}{\lim _{h \to 0} \cos (x+h)} \\ & =\sec x \cdot \frac{\sin x \cdot 1}{\cos x} \\ & \Rightarrow \frac{d}{d x} \sec x=\sec x \tan x \end{aligned} $$

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

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

Answer :

Let $f(x)=\frac{x}{\sin ^{n} x}$

By quotient rule,

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

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

Therefore,

$$ \begin{aligned} f^{\prime}(x) & =\frac{\sin ^{n} x \frac{d}{d x} x-x \frac{d}{d x} \sin ^{n} x}{\sin ^{2 n} x} \\ & =\frac{\sin ^{n} x \cdot 1-x(n \sin ^{n-1} x \cos x)}{\sin ^{2 n} x} \\ & =\frac{\sin ^{n-1} x(\sin x-n x \cos x)}{\sin ^{2 n} x} \\ & =\frac{\sin x-n x \cos x}{\sin ^{n+1} x} \end{aligned} $$