If $(\cos x)^y=(\sin y)^x$ then $\frac{dy}{dx}$ is:

$\frac{\log_e \sin y+y \tan x}{\log_e \cos x - x \cot y}$

The correct answer is Option (3) → $\frac{\log_e \sin y+y \tan x}{\log_e \cos x - x \cot y}$

$(\cos x)^y=(\sin y)^x$

$\log((\cos x)^y)=\log((\sin y)^x)$

$y\log\cos x=x\log\sin y$

$\frac{d}{dx}[y\log(\cos x)]=\frac{d}{dx}[x\log(\sin y)]$

$-y\tan x+(\log(\cos x))\frac{dy}{dx}=x\cot y\frac{dy}{dx}+\log(\sin y)$

$⇒\frac{dy}{dx}=\frac{y\tan x+\log(\sin y)}{\log(\cos x)-x\cot y}$