$$\begin{aligned} &\text { We have, }\\ &\begin{aligned} x & =\mathrm{e}^{x / y} \\ \therefore\quad \frac{d}{d x} x & =\frac{d}{d x} e^{x / y} \end{aligned} \end{aligned}$$

$$\begin{array}{ll} \Rightarrow & 1=e^{x / y} \cdot \frac{d}{d x}(x / y) \\ \Rightarrow & 1=e^{x / y} \cdot\left[\frac{y \cdot 1-x \cdot d y / d x}{y^2}\right] \\ \Rightarrow & y^2=y \cdot e^{x / y}-x \cdot \frac{d y}{d x} \cdot e^{x / y} \end{array}$$

$$\begin{aligned} \Rightarrow \quad x \cdot \frac{d y}{d x} \cdot e^{x / y} & =y e^{x / y}-y^2 \\ \therefore \quad \frac{d y}{d x} & =\frac{y\left(e^{x / y}-y\right)}{x \cdot e^{x / y}} \\ & =\frac{\left(e^{x / y}-y\right)}{e^{x / y} \cdot \frac{x}{y}} \quad \left[\because x=\mathrm{e}^{x / y} \Rightarrow \log x=\frac{x}{y}\right]\\ & =\frac{x-y}{x \cdot \log x}\quad\text{Hence proved.} \end{aligned}$$

$$\begin{aligned} &\begin{aligned} \text { We have, }\quad y^x & =\mathrm{e}^{y-x} \\ \Rightarrow\quad \log y^x & =\log e^{y-x} \\ \Rightarrow\quad x \log y & =y-x \cdot \log _e=(y-x) \quad \left[\because \log _e=1\right]\\ \Rightarrow\quad \log y & =\frac{(y-x)}{x}\quad\text{.... (i)} \end{aligned} \end{aligned}$$

Now, differentiating w.r.t. $x$, we get

$$\frac{d}{d y} \log y \cdot \frac{d y}{d x}=\frac{d}{d x} \frac{(y-x)}{x}$$

$$\begin{array}{ll} \Rightarrow & \frac{1}{y} \cdot \frac{d y}{d x}=\frac{x \cdot \frac{d}{d x}(y-x)-(y-x) \cdot \frac{d}{d x} \cdot x}{x^2} \\ \Rightarrow & \frac{1}{y} \frac{d y}{d x}=\frac{x\left(\frac{d y}{d x}-1\right)-(y-x)}{x^2} \end{array}$$

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

$\therefore \quad \frac{d y}{d x}=\frac{-y^2}{x^2-x y}=\frac{-y^2}{x(x-y)}$

$$\begin{aligned} &\begin{aligned} & =\frac{y^2}{x(y-x)} \cdot \frac{x}{x}=\frac{y^2}{x^2} \cdot \frac{1}{\frac{(y-x)}{x}} \\ & =\frac{(1+\log y)^2}{\log y}\left[\because \log y=\frac{y-x}{x} \log y=\frac{y}{x}-1 \Rightarrow 1+\log y=\frac{y}{x}\right] \end{aligned}\\ &\text { Hence proved. } \end{aligned}$$

We have, $y=(\cos x)^{(\cos x)^{\left.(\cos x)^{-\infty}\right)}}$

$$\begin{array}{ll} \Rightarrow & y=(\cos x)^y \\ \therefore & \log y=\log (\cos x)^y \\ \Rightarrow & \log y=y \log \cos x \end{array}$$

$$\begin{aligned} &\text { On differentiating w.r.t. } x \text {, we get }\\ &\begin{aligned} & \frac{1}{y} \cdot \frac{d y}{d x}=y \cdot \frac{d}{d x} \log \cos x+\log \cos x \cdot \frac{d y}{d x} \\ \Rightarrow \quad & \frac{1}{y} \cdot \frac{d y}{d x}=\frac{y}{\cos x} \cdot \frac{d}{d x} \cos x+\log \cos x \cdot \frac{d y}{d x} \end{aligned} \end{aligned}$$

$$\begin{aligned} \Rightarrow \quad \frac{d y}{d x}\left[\frac{1}{y}-\log \cos x\right] & =\frac{-y \sin x}{\cos x}=-y \tan x \\ \therefore \quad \frac{d y}{d x} & =\frac{-y^2 \tan x}{(1-y \log \cos x)} \\ & =\frac{y^2 \tan x}{y \log \cos x-1}\quad \text{Hence proved.} \end{aligned}$$

$$\begin{aligned} &\text { We have, }\\ &\begin{array}{cc} & x \sin (a+y)+\sin a \cdot \cos (a+y)=0 \\ \Rightarrow & x \sin (a+y)=-\sin a \cdot \cos (a+y) \\ \Rightarrow & x=\frac{-\sin a \cdot \cos (a+y)}{\sin (a+y)} \end{array} \end{aligned}$$

$\Rightarrow \quad x=-\sin a \cdot \cot (a+y)$

$\therefore \quad \frac{d x}{d y}=-\sin a \cdot\left[-\operatorname{cosec}^2(a+y)\right] \cdot \frac{d}{d y}(a+y)$

$$\begin{aligned} &\begin{aligned} & =\sin a \cdot \frac{1}{\sin ^2(a+y)} \cdot 1 \\ & =\frac{\sin ^2(a+y)}{\sin a} \end{aligned}\\ &\text { Hence proved. } \end{aligned}$$

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

$\begin{aligned} & \text { On putting } x=\sin \alpha \text { and } y=\sin \beta \text {, we get } \\ & \qquad \sqrt{1-\sin ^2 \alpha}+\sqrt{1-\sin ^2 \beta}=a(\sin \alpha-\sin \beta)\end{aligned}$

$$\begin{array}{ll} \Rightarrow & \cos \alpha+\cos \beta=a(\sin \alpha-\sin \beta) \\ \Rightarrow & 2 \cos \frac{\alpha+\beta}{2} \cdot \cos \frac{\alpha-\beta}{2}=a\left(2 \cos \frac{\alpha+\beta}{2} \cdot \sin \frac{\alpha-\beta}{2}\right) \end{array}$$

$$\begin{array}{ll} \Rightarrow & \cos \frac{\alpha-\beta}{2}=a \sin \frac{\alpha-\beta}{2} \\ \Rightarrow & \cot \frac{\alpha-\beta}{2}=a \end{array}$$

$$\begin{array}{l} \Rightarrow & \frac{\alpha-\beta}{2} =\cot ^{-1} a \\ \Rightarrow & \alpha-\beta =2 \cot ^{-1} a \\ \Rightarrow & \sin ^{-1} x-\sin ^{-1} y =2 \cot ^{-1} a \quad[\because x=\sin \alpha \text { and } y=\sin \beta] \end{array}$$

$$\begin{aligned} &\text { On differentiating both sides w.r.t. } x \text {, we get }\\ &\begin{array}{ll} & \frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-y^2}} \frac{d y}{d x}=0 \\ \therefore & \frac{d y}{d x}=\frac{\sqrt{1-y^2}}{\sqrt{1-x^2}}=\sqrt{\frac{1-y^2}{1-x^2}}\quad\text{Hence proved.} \end{array} \end{aligned}$$