Question

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

Answer 1

$\mathrm{We}\mathrm{have},{\left(\cos x\right)}^y={\left(\tan y\right)}^x$
Taking log on both sides,
$$\begin{gathered} \log {\left(\cos x\right)}^y=\log {\left(\tan y\right)}^x \\ \Rightarrow y\log \cos x=x\log \tan y \end{gathered}$$
Differentiating it with respect to x using chain,
$$\begin{gathered} \frac{d}{dx}\left(y\log \cos x\right)=\frac{d}{dx}\left(x\log \tan y\right) \\ \Rightarrow y\frac{d}{dx}\left(\log \cos x\right)+\log \cos x\frac{dy}{dx}=x\frac{d}{dx}\left(\log \tan y\right)+\log \tan y\frac{d}{dx}\left(x\right) \\ \Rightarrow y\frac{1}{\cos x}\frac{d}{dx}\left(\cos x\right)+\log \cos x\frac{dy}{dx}=x\frac{1}{\tan y}\frac{d}{dx}\left(\tan y\right)+\log \tan y \\ \Rightarrow \frac{y}{\cos x}\left(-\sin x\right)+\log \cos x\frac{dy}{dx}=\left\{\frac{x}{\tan y}\left({\sec }^{2}y\right)\right\}\frac{dy}{dx}+\log \tan y \\ \Rightarrow -y\tan x+\log \cos x\frac{dy}{dx}=\sec y\mathrm{cosec}y\times x\frac{dy}{dx}+\log \tan y \\ \Rightarrow \frac{dy}{dx}\left[\log \cos x-x\sec y\mathrm{cosec}y\right]=\log \tan y+y\tan x \\ \Rightarrow \frac{dy}{dx}=\left[\frac{\log \tan y+y\tan x}{\log \cos x-x\sec y\mathrm{cosec}y}\right] \end{gathered}$$