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

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

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

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

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

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

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

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

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

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

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

$$\begin{aligned} \text { Let } \quad y & =\tan ^{-1}\left(\frac{a \cos x-b \sin x}{b \cos x+a \sin x}\right) \\ & =\tan ^{-1}\left[\frac{\frac{a \cos x}{b \cos x}-\frac{b \sin x}{b \cos x}}{\frac{b \cos x}{b \cos x}+\frac{a \sin x}{b \cos x}}\right]=\tan ^{-1}\left[\frac{\frac{a}{b}-\tan x}{1+\frac{a}{b} \tan x}\right] \\ & =\tan ^{-1} \frac{a}{b}-\tan ^{-1} \tan x \quad \left[\because \tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)\right] \end{aligned}$$

$$\begin{array}{rlr} & =\tan ^{-1} \frac{a}{b}-x & \\ \therefore\quad \frac{d y}{d x} & =\frac{d}{d x}\left(\tan ^{-1} \frac{a}{b}\right)-\frac{d}{d x}(x) & \\ & =0-1 & {\left[\because \frac{d}{d x}\left(\frac{a}{b}\right)=0\right]} \\ & =-1 & \end{array}$$

Let $$y=\sec ^{-1}\left(\frac{1}{4 x^3-3 x}\right)\quad \text{..... (i)}$$

On putting $x=\cos \theta$ in Eq. (i), we get

$$\begin{aligned} y & =\sec ^{-1} \frac{1}{4 \cos ^3 \theta-3 \cos \theta} \\ & =\sec ^{-1} \frac{1}{\cos 3 \theta} \\ & =\sec ^{-1}(\sec 3 \theta)=3 \theta \\ & =3 \cos ^{-1} x \quad \left[\because \theta=\cos ^{-1} x\right]\\ \therefore\quad \frac{d y}{d x} & =\frac{d}{d x}\left(3 \cos ^{-1} x\right) \\ & =3 \cdot \frac{-1}{\sqrt{1-x^2}} \end{aligned}$$