Question

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

Answer 1

$\mathrm{We}\mathrm{have},{\left(\sin x\right)}^y=x+y$
Taking log on both the sides,
$$\begin{gathered} \log {\left(\sin x\right)}^y=\log \left(x+y\right) \\ \Rightarrow y\log \left(\sin x\right)=\log \left(x+y\right) \end{gathered}$$
Differentiating with respect to x using chain rule,
$$\begin{gathered} \frac{d}{dx}\left\{y\log \left(\sin x\right)\right\}=\frac{d}{dx}\left\{\log \left(x+y\right)\right\} \\ \Rightarrow y\frac{d}{dx}\log \sin x+\log \sin x\frac{dy}{dx}=\frac{1}{x+y}\frac{d}{dx}\left(x+y\right) \\ \Rightarrow \frac{y}{\sin x}\frac{d}{dx}\left(\sin x\right)+\log \sin x\frac{dy}{dx}=\frac{1}{\left(x+y\right)}\left[1+\frac{dy}{dx}\right] \\ \Rightarrow \frac{y\left(\cos x\right)}{\left(\sin x\right)}+\log \sin x\frac{dy}{dx}=\frac{1}{\left(x+y\right)}+\frac{1}{\left(x+y\right)}\frac{dy}{dx} \\ \Rightarrow \frac{dy}{dx}\left(\log \sin x-\frac{1}{x+y}\right)=\frac{1}{\left(x+y\right)}-y\cot x \\ \Rightarrow \frac{dy}{dx}\left\{\frac{\left(x+y\right)\log \sin x-1}{\left(x+y\right)}\right\}=\frac{1-y\left(x+y\right)\cot x}{x+y} \\ \Rightarrow \frac{dy}{dx}=\left\{\frac{1-y\left(x+y\right)\cot x}{\left(x+y\right)\log \sin x-1}\right\} \end{gathered}$$