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

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

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

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

$$\begin{aligned} \text{Let}\quad & y=\tan ^{-1}\left(\frac{3 a^2 x-x^3}{a^3-3 a x^2}\right) \\ \text{Put}\quad & x=\operatorname{atan} \theta \Rightarrow \theta=\tan ^{-1} \frac{x}{a} \end{aligned}$$

$\therefore \quad y=\tan ^{-1}\left[\frac{3 \tan \theta-\tan ^3 \theta}{1-3 \tan ^2 \theta}\right] \quad\left[\because \tan 3 \theta=\frac{3 \tan \theta-\tan ^3 \theta}{1-3 \tan ^2 \theta}\right]$

$$\begin{aligned} &\begin{aligned} & =\tan ^{-1}(\tan 3 \theta)=3 \theta \\ & =3 \tan ^{-1} \frac{x}{a} \end{aligned}\\ &\left[\because \theta=\tan ^{-1} \frac{x}{a}\right] \end{aligned}$$

$$\begin{aligned} \therefore \quad \frac{d y}{d x} & =3 \cdot \frac{d}{d x} \tan ^{-1} \frac{x}{a}=3 \cdot\left[\frac{1}{1+\frac{x^2}{a^2}}\right] \cdot \frac{d}{d x} \cdot\left(\frac{x}{a}\right) \\ & =3 \cdot \frac{a^2}{a^2+x^2} \cdot \frac{1}{a}=\frac{3 a}{a^2+x^2} \end{aligned}$$

Let $$y=\tan ^{-1}\left[\frac{\sqrt{1+x^2}+\sqrt{1-x^2}}{\sqrt{1+x^2}-\sqrt{1-x^2}}\right]$$

Put $$x^2=\cos 2 \theta$$

$$\begin{aligned} &\begin{aligned} \text { Put }\quad x^2 & =\cos 2 \theta \\ \therefore\quad y & =\tan ^{-1}\left(\frac{\sqrt{1+\cos 2 \theta}+\sqrt{1-\cos 2 \theta}}{\sqrt{1+\cos 2 \theta}-\sqrt{1-\cos 2 \theta}}\right) \end{aligned} \end{aligned}$$

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

$$\begin{aligned} & =\tan ^{-1} \tan \left(\frac{\pi}{4}+\theta\right) \quad\left[\because \tan (a+b)=\frac{\tan a+\tan b}{1-\tan a \cdot \tan b}\right] \\ & =\frac{\pi}{4}+\theta=\frac{\pi}{4}+\frac{1}{2} \cos ^{-1} x^2 \quad\left[\because 2 \theta=\cos ^{-1} x^2 \Rightarrow \theta=\frac{1}{2} \cos ^{-1} x^2\right] \\ \therefore \quad \frac{d y}{d x} & =\frac{d}{d x}\left(\frac{\pi}{4}\right)+\frac{d}{d x}\left(\frac{1}{2} \cos ^{-1} x^2\right) \\ & =0+\frac{1}{2} \cdot \frac{-1}{\sqrt{1-x^4}} \cdot \frac{d}{d x} x^2=\frac{1}{2} \cdot \frac{-2 x}{\sqrt{1-x^4}}=\frac{-x}{\sqrt{1-x^4}} \end{aligned}$$

$$\begin{array}{lll} \because & x=t+\frac{1}{t} \text { and } y=t-\frac{1}{t} \\ \therefore & \frac{d x}{d t}=\frac{d}{d t}\left(t+\frac{1}{t}\right) \quad \text { and } \quad \frac{d y}{d t}=\frac{d}{d t}\left(t-\frac{1}{t}\right) \\ \Rightarrow & \frac{d x}{d t}=1+(-1) t^{-2} \text { and } \quad \frac{d y}{d t}=1-(-1) t^{-2} \\ \Rightarrow & \frac{d x}{d t}=1-\frac{1}{t^2} \quad \text { and } \quad \frac{d y}{d t}=1+\frac{1}{t^2} \\ \Rightarrow & \frac{d x}{d t}=\frac{t^2-1}{t^2} \quad \text { and } \frac{d y}{d t}=\frac{t^2+1}{t^2} \\ \therefore & \frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{t^2+1 / t^2}{t^2-1 / t^2}=\frac{t^2+1}{t^2-1} \end{array}$$