Let, $y=\sin ^m x \cdot \cos ^n x$

$\therefore \quad \frac{d y}{d x}=\frac{d}{d x}\left[(\sin x)^m \cdot(\cos x)^n\right]$

$$\begin{aligned} & =(\sin x)^m \cdot \frac{d}{d x}(\cos x)^n+(\cos x)^n \cdot \frac{d}{d x}(\sin x)^m \\ & =(\sin x)^m \cdot n(\cos x)^{n-1} \cdot \frac{d}{d x} \cos x+(\cos x)^n m(\sin x)^{m-1} \cdot \frac{d}{d x} \sin x \\ & =(\sin x)^m \cdot n(\cos x)^{n-1}(-\sin x)+(\cos x)^n \cdot m(\sin x)^{m-1} \cos x \\ & =-n \sin ^m x \cdot \cos ^{n-1} x \cdot(\sin x)+m \cos ^n x \cdot \sin ^{m-1} x \cdot \cos x \\ & =-n \cdot \sin ^m x \cdot \sin x \cdot \cos ^n x \cdot \frac{1}{\cos x}+m \cdot \sin ^m x \cdot \frac{1}{\sin x} \cdot \cos ^n x \cdot \cos x \\ & =-n \cdot \sin ^m x \cdot \cos ^n x \cdot \tan x+m \sin ^m x \cdot \cos ^n x \cdot \cot x \\ & =\sin ^m x \cdot \cos ^n x[-n \tan x+m \cot x] \end{aligned}$$

$$\begin{aligned} &\begin{aligned} \text { Let } \quad y & =(x+1)^2(x+2)^3(x+3)^4 \\ \therefore \quad \log y & =\log \left\{(x+1)^2 \cdot(x+2)^3(x+3)^4\right\} \\ & =\log (x+1)^2+\log (x+2)^3+\log (x+3)^4 \end{aligned} \end{aligned}$$

$$\begin{aligned} \text { and } \quad & \frac{d}{d y} \log y \cdot \frac{d y}{d x}=\frac{d}{d x}[2 \log (x+1)]+\frac{d}{d x} {[3 \log (x+2)]+\frac{d}{d x}[4 \log (x+3)] } \\ & \frac{1}{y} \cdot \frac{d y}{d x}=\frac{2}{(x+1)} \cdot \frac{d}{d x}(x+1)+3 \cdot \frac{1}{(x+2)} \cdot \frac{d}{d x}(x+2) \\ &+4 \cdot \frac{1}{(x+3)} \cdot \frac{d}{d x}(x+3) \quad\left[\because \quad \frac{d}{d x}(\log x)=\frac{1}{x}\right] \end{aligned}$$

$$\begin{array}{rlrl} & =\left[\frac{2}{x+1}+\frac{3}{x+2}+\frac{4}{x+3}\right] \\ \therefore & \frac{d y}{d x} =y\left[\frac{2}{(x+1)}+\frac{3}{(x+2)}+\frac{4}{(x+3)}\right] \end{array}$$

$$\begin{aligned} & =(x+1)^2 \cdot(x+2)^3 \cdot(x+3)^4\left[\frac{2}{(x+1)}+\frac{3}{(x+2)}+\frac{4}{(x+3)}\right] \\ & =(x+1)^2 \cdot(x+2)^3 \cdot(x+3)^4 \\ & \quad\left[\frac{2(x+2)(x+3)+3(x+1)(x+3)+4(x+1)(x+2)}{(x+1)(x+2)(x+3)}\right] \\ & =\frac{(x+1)^2(x+2)^3(x+3)^4}{(x+1)(x+2)(x+3)} \\ & \quad\left[2\left(x^2+5 x+6\right)+3\left(x^2+4 x+3\right)+4\left(x^2+3 x+2\right)\right] \\ & =(x+1)(x+2)^2(x+3)^3 \\ & \quad\left[2 x^2+10 x+12+3 x^2+12 x+9+4 x^2+12 x+8\right] \\ & =(x+1)(x+2)^2(x+3)^3\left[9 x^2+34 x+29\right] \end{aligned}$$

$$\text{Let}\quad y=\cos ^{-1}\left(\frac{\sin x+\cos x}{\sqrt{2}}\right)$$

$$\begin{aligned} \therefore \quad \frac{d y}{d x} & =\frac{d}{d x} \cos ^{-1}\left(\frac{\sin x+\cos x}{\sqrt{2}}\right) \\ & =\frac{-1}{\sqrt{1-\left(\frac{\sin x+\cos x}{\sqrt{2}}\right)^2}} \cdot \frac{d}{d x}\left(\frac{\sin x+\cos x}{\sqrt{2}}\right) \end{aligned}$$

$\left[\because \frac{d}{d x}(\cos x)=-\frac{1}{\sqrt{1-x^2}}\right]$

$$\begin{aligned} & =\frac{-1}{\sqrt{4-\frac{\left(\sin ^2 x+\cos ^2 x+2 \sin x \cdot \cos x\right)}{2}}} \cdot \frac{1}{\sqrt{2}}(\cos x-\sin x)\\ & =\frac{-1 \cdot \sqrt{2}}{\sqrt{1-\sin 2 x}} \cdot \frac{1}{\sqrt{2}}(\cos x-\sin x) \\ & {\left[\because 1-\sin 2 x=(\cos x-\sin x)^2=\cos ^2 x+\sin ^2 x-2 \sin x \cos x\right]} \\ & =\frac{-1(\cos x-\sin x)}{(\cos x-\sin x)}=-1 \end{aligned}$$

$$\begin{aligned} &\begin{aligned} \text { Let }\quad y & =\tan ^{-1} \sqrt{\frac{1-\cos x}{1+\cos x}} \\ \therefore\quad \frac{d y}{d x} & =\frac{d}{d x} \tan ^{-1} \sqrt{\frac{1-\cos x}{1+\cos x}} \end{aligned} \end{aligned}$$

$=\frac{1}{1+\sqrt{\left(\frac{1-\cos x}{1+\cos x}\right)^2}} \cdot \frac{d}{d x}\left[\frac{1-\cos x}{1+\cos x}\right]^{1 / 2} \quad\left[\because \frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{1+x^2}\right]$

$$\begin{aligned} & =\frac{1}{1+\frac{1-\cos x}{1+\cos x}} \cdot \frac{1}{2}\left[\frac{1-\cos x}{1+\cos x}\right]^{-1 / 2} \cdot \frac{d}{d x}\left(\frac{1-\cos x}{1+\cos x}\right) \\ & =\frac{1}{\frac{1+\cos x+1-\cos x}{1+\cos x}} \cdot \frac{1}{2}\left[\frac{(1-\cos x)}{(1+\cos x)} \cdot \frac{(1-\cos x)}{(1-\cos x)}\right]^{-1 / 2} \cdot \frac{(1+\cos x) \cdot \sin x+(1-\cos x) \cdot \sin x}{(1+\cos x)^2} \end{aligned}$$

$$\begin{aligned} & =\frac{(1+\cos x)}{2} \cdot \frac{1}{2}\left[\frac{(1-\cos x)^2}{\left(1-\cos ^2 x\right)}\right]^{-1 / 2}\left[\frac{\sin x(1+\cos x+1-\cos x)}{(1+\cos x)^2}\right] \\ & =\frac{(1+\cos x)}{2} \cdot \frac{1}{2}\left[\frac{(1-\cos x)^2}{\left(1-\cos ^2 x\right)}\right]^{-1 / 2}\left[\frac{\sin x(1+\cos x+1-\cos x)}{(1+\cos x)^2}\right] \end{aligned}$$

$$\begin{aligned} & =\frac{(1+\cos x)}{2} \cdot \frac{1}{2}\left[\frac{(1-\cos x)^2}{\sin x}\right]^{-1 / 2} \cdot \frac{2 \sin x}{(1+\cos x)^2} \\ & =\frac{(1+\cos x)}{2} \cdot \frac{1}{2} \cdot \frac{\sin x}{(1-\cos x)} \cdot \frac{2 \sin x}{(1+\cos x)^2} \\ & =\frac{2 \sin ^2 x}{4(1+\cos x)(1-\cos x)}=\frac{1}{2} \cdot \frac{\sin ^2 x}{\left(1-\cos ^2 x\right)} \\ & =\frac{1}{2} \cdot \frac{\sin ^2 x}{\sin ^2 x}=\frac{1}{2} \end{aligned}$$

Alternate Method

Let $$y=\tan ^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right)$$

$$\begin{aligned} & =\tan ^{-1}\left(\sqrt{\frac{1-1+2 \sin ^2 \frac{x}{2}}{1+2 \cos ^2 \frac{x}{2}-1}}\right) \quad\left[\because \cos x=1-2 \sin ^2 \frac{x}{2}=2 \cos ^2 \frac{x}{2}-1\right] \\ & =\tan ^{-1}\left(\tan \frac{x}{2}\right)=\frac{x}{2} \end{aligned}$$

On differentiating w.r.t. x, we get

$\frac{d y}{d x}=\frac{1}{2}$

Let $y=\tan ^{-1}(\sec x+\tan x)$

$$\begin{aligned} \therefore \quad \frac{d y}{d x} & =\frac{d}{d x} \tan ^{-1}(\sec x+\tan x) \\ & =\frac{1}{1+(\sec x+\tan x)^2} \cdot \frac{d}{d x}(\sec x+\tan x) \quad\left[\because \frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{1+x^2}\right] \\ & =\frac{1}{1+\sec ^2 x+\tan ^2 x+2 \sec x \cdot \tan x} \cdot\left[\sec x \cdot \tan x+\sec ^2 x\right] \\ & =\frac{1}{\left(\sec ^2 x+\sec ^2 x+2 \sec x \cdot \tan x\right)} \cdot \sec x \cdot(\sec x+\tan x) \\ & =\frac{1}{2 \sec x(\tan x+\sec x)} \cdot \sec x(\sec x+\tan x)=\frac{1}{2} \end{aligned}$$