Let $\quad f(x)=\cos \left(x^2+1\right)$ and $f(x+h)=\cos \left\{(x+h)^2+1\right\}$

$\therefore \quad \frac{d}{d x} f(x)=\lim _\limits{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}$

$$\begin{aligned} & =\lim _{h \rightarrow 0} \frac{\cos \left\{(x+h)^2+1\right\}-\cos \left(x^2+1\right)}{h} \\ & =\lim _{h \rightarrow 0} \frac{-2 \sin \left\{\frac{(x+h)^2+1+x^2+1}{2}\right\} \sin \left\{\frac{(x+h)^2+1-x^2-1}{2}\right\}}{h} \\ & {\left[\because \cos C-\cos D=-2 \sin \frac{C+D}{2} \cdot \sin \frac{C-D}{2}\right]} \end{aligned}$$

$$ \begin{aligned} & =\lim _{h \rightarrow 0} \frac{1}{h}\left[-2 \sin \left\{\frac{(x+h)^2+x^2+2}{2}\right\} \sin \left\{\frac{(x+h)^2-x^2}{2}\right\}\right] \\ & =\lim _{h \rightarrow 0} \frac{1}{h}\left[-2 \sin \left\{\frac{(x+h)^2+x^2+2}{2}\right\} \sin \left\{\frac{x^2+h^2+2 x h-x^2}{2}\right\}\right] \\ & =\lim _{h \rightarrow 0} \frac{1}{h}\left[-2 \sin \left\{\frac{(x+h)^2+x^2+2}{2}\right\} \sin \left\{\frac{h^2+2 h x}{2}\right\}\right] \\ & =-2 \lim _{h \rightarrow 0} \sin \left\{\frac{(x+h)^2+x^2+2}{2}\right\} \lim _{h \rightarrow 0}\left\{\frac{\sin h\left(\frac{h+2 x}{2}\right)}{h\left(\frac{h+2 x}{2}\right)}\right\} \times\left(\frac{h+2 x}{2}\right) \\ & =-2 \lim _{h \rightarrow 0} \sin \left\{\frac{(x+h)^2+x^2+2}{2}\right\} \lim _{h \rightarrow 0}\left(\frac{h+2 x}{2}\right)\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right] \\ & =-2 x \sin \left(x^2+1\right) \quad {[\because x \rightarrow 0 \Rightarrow k x \rightarrow 0]}\\ \end{aligned}$$

$$\begin{aligned} &\begin{aligned} \text{Let}\quad & f(x)=\frac{a x+b}{c x+d} \\ & f(x+h)=\frac{a(x+h)+b}{c(x+h)+d} \\ \therefore \quad & \frac{d}{d x} f(x)=\lim _{h \rightarrow 0} \frac{1}{h}[f(x+h)-f(x)] \\ & =\lim _{h \rightarrow 0} \frac{1}{h}\left[\frac{a(x+h)+b}{c(x+h)+d}-\frac{a x+b}{c x+d}\right] \\ & =\lim _{h \rightarrow 0} \frac{1}{h}\left[\frac{a x+b+a h}{c(x+h)+d}-\frac{a x+b}{c x+d}\right] \\ & =\lim _{h \rightarrow 0} \frac{1}{h}\left[\frac{(a x+a h+b)(c x+d)-(a x+b\{c(x+h)+d\}}{\{c(x+h)+d\}(c x+d)}\right] \\ & =\lim _{h \rightarrow 0} \frac{1}{h}\left[\frac{(a x+a h+b)(c x+d)-(a x+b)(c x+c h+d)}{\{c(x+h)+d)\}(c x+d)}\right] \\ & =\lim _{h \rightarrow 0} \frac{1}{h}\left[\begin{array}{l} \\ \frac{a c x^2+a c h x+b c x+a d x+a d h+b d-a c x^2+a c h x+a d x+b c x+b c h+b d}{\{c(x+h)+d\}(c x+d)} \end{array}\right] \end{aligned}\\ &\begin{aligned} & =\lim _{h \rightarrow 0} \frac{1}{h}\left[\frac{a d h-b c h}{\{c(x+h)+d\}(c x+d)}\right] \\ & =\lim _{h \rightarrow 0} \frac{a c-b d}{\{c(x+h)+d\}(c x+d)} \\ & =\frac{a c-b d}{(c x+d)^2} \end{aligned} \end{aligned}$$

$$\begin{aligned} \text{Let}\quad f(x) & =x^{2 / 3} \\ f(x+h) & =(x+h)^{2 / 3} \\ \text{Now,}\quad \frac{d}{d x} f(x) & =\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\ & =\lim _{h \rightarrow 0} \frac{1}{h}\left[(x+h)^{2 / 3}-x^{2 / 3}\right] \\ & =\lim _{h \rightarrow 0} \frac{1}{h}\left[x^{2 / 3}\left(1+\frac{h}{x}\right)^{2 / 3}-x^{2 / 3}\right] \end{aligned}$$

$$\begin{aligned} & =\lim _{h \rightarrow 0} \frac{1}{h}\left[x^{2 / 3}\left(1+\frac{h}{x} \cdot \frac{2}{3}+\frac{2}{3}\left(\frac{2}{3}-1\right) \frac{h^2}{x^2}+\cdots\right)-1\right] \\ & \quad\left[\because(1+x)^n=1+n x+\frac{n(n-1)}{2!} x^2+\cdots\right] \\ & =\lim _{h \rightarrow 0} \frac{1}{h}\left[x^{2 / 3}\left(\frac{2}{3} \cdot \frac{h}{x}-\frac{2}{9} \cdot \frac{h^2}{x^2}+\cdots\right)\right] \\ & =\lim _{h \rightarrow 0} \frac{x^{2 / 3}}{h} \cdot \frac{2}{3} \frac{h}{x}\left(1-\frac{1}{3} \cdot \frac{h}{x}+\cdots\right) \\ & =\frac{2}{3} x^{2 / 3-1}=\frac{2}{3} x^{-1 / 3} \end{aligned}$$

Alternate Method

$$ \begin{aligned} & \text { Let } \quad f(x)=x^{2 / 3} \\ & f(x+h)=(x+h)^{2 / 3} \\ & \therefore \quad \frac{d}{d x} f(x)=\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\ & =\lim _{(h \rightarrow 0}\left[\frac{(x+h) 2 / 3-x^{2 / 3}}{h}\right]=\lim _{(x+h) \rightarrow x}\left[\frac{(x+h) 2 / 3-x^{2 / 3}}{(x+h)-x}\right] \\ & =\frac{2}{3}(x)^{2 / 3-1} \left[\because \lim _{x \rightarrow a} \frac{x^x-a^n}{x-a}=n a^{n-1}\right]\\ & =\frac{2}{3} x^{-1 / 3} \end{aligned}$$

$$\begin{aligned} & \text { Let } & f(x) & =x \cos x \\ & \therefore & f(x+h) & =(x+h) \cos (x+h) \\ & \therefore & \frac{d}{d x} f(x) & =\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \end{aligned}$$

$$\begin{aligned} & =\lim _{h \rightarrow 0} \frac{1}{h}[(x+h) \cos (x+h)-x \cos x] \\ & =\lim _{h \rightarrow 0} \frac{1}{h}[x \cos (x+h)+h \cos (x+h)-x \cos x] \\ & =\lim _{h \rightarrow 0} \frac{1}{h}[x\{\cos (x+h)-\cos x\}+h \cos (x+h)] \\ & =\lim _{h \rightarrow 0} \frac{1}{h}\left[x\left\{-2 \sin \left(\frac{2 x+h}{2}\right) \sin \frac{h}{2}\right\}+h \cos (x+h)\right] \\ & =\lim _{h \rightarrow 0}\left[-2 x \sin \left(x+\frac{h}{2}\right) \frac{\sin \frac{h}{2}}{h}+\cos (x+h)\right] \\ & \qquad\left[\because \cos C-\cos D=-2 \sin \frac{C+D}{2} \cdot \sin \frac{C-D}{2}\right] \\ & =-2 \lim _{h \rightarrow 0} x \sin \left(x+\frac{h}{2}\right) \lim _{h \rightarrow 0} \frac{\sin \frac{h}{2}}{\frac{h}{2}} \cdot \frac{1}{2}+\lim _{h \rightarrow 0} \cos (x+h) \\ & =-2 \cdot \frac{1}{2} x \sin x+\cos x \\ & =\cos x-x \sin x \end{aligned}$$

Given, $\lim _\limits{y \rightarrow 0} \frac{(x+y) \sec (x+y)-x \sec x}{y}$

$$\begin{aligned} & =\lim _{y \rightarrow 0} \frac{\frac{x+y}{\cos (x+y)}-\frac{x}{\cos x}}{y} \\ & =\lim _{y \rightarrow 0} \frac{(x+y) \cos x-x \cos (x+y)}{y \cos x \cos (x+y)} \\ & =\lim _{y \rightarrow 0}\left[\frac{x \cos x+y \cos x-x \cos (x+y)}{y \cos x \cos (x+y)}\right] \\ & =\lim _{y \rightarrow 0}\left[\frac{x \cos x-x \cos (x+y)+y \cos x}{y \cos x \cos (x+y)}\right] \\ & =\lim _{y \rightarrow 0} \frac{x\{\cos x-\cos (x+y)\}+y \cos x}{y \cos x \cos (x+y)} \\ & =\lim _{y \rightarrow 0} \frac{x\left[-2 \sin \left(x+\frac{y}{2}\right) \sin \left(\frac{-y}{2}\right)\right]+y \cos x}{y \cos x \cos (x+y)} \\ & {\left[\because \cos C-\cos D=-2 \sin \frac{C+D}{2} \cdot \sin \frac{C-D}{2}\right]} \\ & =\lim _{y \rightarrow 0}\left[\frac{x\left\{2 \sin \left(x+\frac{y}{2}\right) \sin \frac{y}{2}\right\}+y \cos x}{y \cos x \cos (x+y)}\right] \\ & =\lim _{y \rightarrow 0} \frac{2 x \sin \left(x+\frac{y}{2}\right)}{\cos x \cos (x+y)} \cdot \lim _{y \rightarrow 0} \frac{\sin \frac{y}{2}}{\frac{y}{2}} \cdot \frac{1}{2}+\lim _{y \rightarrow 0} \sec (x+y) \\ & {\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1 \text { and } x \rightarrow 0 \Rightarrow k x \rightarrow 0\right]} \\ & =\lim _{y \rightarrow 0} \frac{2 x \sin \left(x+\frac{y}{2}\right)}{\cos x \cos (x+y)} \cdot \frac{1}{2}+\lim _{y \rightarrow 0} \sec (x+y) \\ & =\frac{2 x \sin x}{\cos x \cos x} \cdot \frac{1}{2}+\sec x \\ & =x \tan x \sec x+\sec x \\ & =\sec x(x \tan x+1) \end{aligned}$$