$$\begin{aligned} &\begin{aligned} \text { Let }\quad u & =\frac{x}{\sin x} \text { and } v=\sin x \\ \therefore\quad \frac{d u}{d x} & =\frac{\sin x \cdot \frac{d}{d x} x-x \cdot \frac{d}{d x} \sin x}{(\sin x)^2} \\ & =\frac{\sin x-x \cos x}{\sin ^2 x}\quad\text{.... (i)} \end{aligned} \end{aligned}$$

$$\begin{aligned} &\begin{aligned} \text { and }\quad & \frac{d v}{d x}=\frac{d}{d x} \sin x=\cos x \quad\text{..... (ii)}\\ \therefore\quad & \frac{d u}{d v}=\frac{d u / d x}{d v / d x}=\frac{\sin x-x \cos x / \sin ^2 x}{\cos x} \end{aligned} \end{aligned}$$

$$\begin{aligned} & =\frac{\sin x-x \cos x}{\sin ^2 x \cos x}=\frac{\frac{\sin x-x \cos x}{\cos x}}{\frac{\sin ^2 x \cos x}{\cos x}} \quad\text{[dividing by cos x in both numerator and denominator]}\\ & =\frac{\tan x-x}{\sin ^2 x} \end{aligned}$$

Let $$u=\tan ^{-1} \frac{\sqrt{1+x^2}-1}{x} \text { and } v=\tan ^{-1} x$$

$$\begin{array}{ll} \therefore & x=\tan \theta \\ \Rightarrow & u=\tan ^{-1} \frac{\sqrt{1+\tan ^2 \theta}-1}{\tan \theta} \end{array}$$

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

$$ \begin{aligned} & =\tan ^{-1}\left[\tan \frac{\theta}{2}\right] \\ & =\frac{\theta}{2}=\frac{1}{2} \tan ^{-1} x \\ \therefore\quad \frac{d u}{d x} & =\frac{1}{2} \frac{d}{d x} \tan ^{-1} x=\frac{1}{2} \cdot \frac{1}{1+x^2} \quad\text{.... (i)}\\ \text{and}\quad \frac{d v}{d x} & =\frac{d}{d x} \tan ^{-1} x=\frac{1}{1+x^2} \quad\text{.... (ii)}\\ \therefore\quad \frac{d u}{d v} & =\frac{d u / d x}{d v / d x} \\ & =\frac{1 / 2\left(1+x^2\right)}{1 /\left(1+x^2\right)}=\frac{\left(1+x^2\right)}{2\left(1+x^2\right)}=\frac{1}{2} \end{aligned}$$

We have, $\quad \sin (x y)+\frac{x}{y}=x^2-y$

On differentiating both sides w.r.t. $x$, we get $$\frac{d}{d x}(\sin x y)+\frac{d}{d x}\left(\frac{x}{y}\right)=\frac{d}{d x} x^2-\frac{d}{d x} y$$

$$\begin{array}{ll} \Rightarrow & \cos x y \cdot \frac{d}{d x}(x y)+\frac{y \frac{d}{d x} x-x \cdot \frac{d}{d x} y}{y^2}=2 x-\frac{d y}{d x} \\ \Rightarrow & \cos x y \cdot\left[x \cdot \frac{d}{d x} y+y \cdot \frac{d}{d x} \cdot x\right]+\frac{y-x \frac{d y}{d x}}{y^2}=2 x-\frac{d y}{d x} \\ \Rightarrow & x \cos x y \cdot \frac{d y}{d x}+y \cos x y+\frac{y}{y^2}-\frac{x}{y^2} \frac{d y}{d x}=2 x-\frac{d y}{d x} \end{array}$$

$$\begin{array}{ll} \Rightarrow & \frac{d y}{d x}\left[x \cos x y-\frac{x}{y^2}+1\right]=2 x-y \cos x y-\frac{y}{y^2} \\ \therefore & \frac{d y}{d x}=\left[\frac{2 x y-y^2 \cos x y-1}{y}\right]\left[\frac{y^2}{x y^2 \cos x y-x+y^2}\right] \end{array}$$

$=\frac{\left(2 x y-y^2 \cos x y-1\right) y}{\left(x y^2 \cos x y-x+y^2\right)}$

We have, $\sec (x+y)=x y$

On differentiating both sides w.r.t. $x$, we get

$$\frac{d}{d x} \sec (x+y)=\frac{d}{d x}(x y)$$

$$\begin{array}{ll} \Rightarrow & \sec (x+y) \cdot \tan (x+y) \cdot \frac{d}{d x}(x+y)=x \cdot \frac{d}{d x} y+y \cdot \frac{d}{d x} x \\ \Rightarrow & \sec (x+y) \cdot \tan (x+y) \cdot\left(1+\frac{d y}{d x}\right)=x \frac{d y}{d x}+y \end{array}$$

$$\begin{array}{ll} \Rightarrow & \sec (x+y) \tan (x+y)+\sec (x+y) \cdot \tan (x+y) \cdot \frac{d y}{d x}=x \frac{d y}{d x}+y \\ \Rightarrow & \frac{d y}{d x}[\sec (x+y) \cdot \tan (x+y)-x]=y-\sec (x+y) \cdot \tan (x+y) \\ \therefore & \frac{d y}{d x}=\frac{y-\sec (x+y) \cdot \tan (x+y)}{\sec (x+y) \cdot \tan (x+y)-x} \end{array}$$

We have, $\tan ^{-1}\left(x^2+y^2\right)=a$

On differentiating both sides w.r.t. $x$, we get

$$\frac{d}{d x} \tan ^{-1}\left(x^2+y^2\right)=\frac{d}{d x}(a)$$

$$\begin{array}{rr} & \frac{d}{d x} \tan ^{-1}\left(x^2+y^2\right)=\frac{d}{d x}(a) \\ \Rightarrow & \frac{1}{1+\left(x^2+y^2\right)^2} \cdot \frac{d}{d x}\left(x^2+y^2\right)=0 \\ \Rightarrow & 2 x+\frac{d}{d y} y^2 \cdot \frac{d y}{d x}=0 \\ \therefore & 2 y \cdot \frac{d y}{d x}=-2 x \\ \Rightarrow & \frac{d y}{d x}=\frac{-2 x}{2 y}=\frac{-x}{y} \end{array}$$