Question

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

Answer 1

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