# Solve this

Question:

If $(\sin x)^{y}=x+y$, prove that $\frac{d y}{d x}=\frac{1-(x+y) y \cot x}{(x+y) \log \sin x-1}$

Solution:

Here

$(\sin x)^{y}=x+y$

Taking log both sides,

$\log (\sin x)^{y}=\log (x+y)$

$y \log (\sin x)=\log (x+y)\left[\right.$ Using $\log a^{b}=b \log a$ ]

Differentiating it with respect to $x$ using the chain rule and product rule,

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

$y \frac{d}{d x} \log (\sin x)+\log \sin x \frac{d y}{d x}=\frac{1}{(x+y)} \frac{d}{d x}(x+y)$

$\frac{y}{\sin x} \frac{d}{d x}(\sin x)+\log \sin x \frac{d y}{d x}=\frac{1}{(x+y)} \frac{d}{d x}(x+y)$

$\frac{y \cos x}{\sin x}+\log \sin x \frac{d y}{d x}=\frac{1}{(x+y)}+\frac{1}{(x+y)} \frac{d y}{d x}$

$\frac{d y}{d x}\left(\log \sin x-\frac{1}{(x+y)}\right)=\frac{1}{(x+y)}-y \cot x$

$\frac{d y}{d x}\left(\frac{(x+y) \log \sin x-1}{x+y}\right)=\frac{1-y(x+y) \cot x}{(x+y)}$

$\frac{d y}{d x}=\frac{1-y(x+y) \cot x}{(x+y)} \times \frac{(x+y)}{(x+y) \log \sin x-1}$

$\frac{d y}{d x}=\frac{1-y(x+y) \cot x}{(x+y) \log \sin x-1}$

Hence Proved.