Question

Find $\frac{dy}{dx}$ of $(\cos x{)}^y=(\cos y{)}^x$

Answer 1

Given, $(\cos x{)}^y=(\cos y{)}^x$
Taking logarithm on both sides, we obtain
$y\log \cos x=x\log \cos y$
Differentiating both sides, we obtain
$\log \cos x.\frac{dy}{dx}+y.\frac{d}{dx}\left(\log \cos x\right)=\log \cos y.\frac{d}{dx}\left(x\right)+x.\frac{d}{dx}\left(\log \cos y\right)$
$\Rightarrow \log \cos x\frac{dy}{dx}+y.\frac{1}{\cos x}.\frac{d}{dx}\left(\cos x\right)=\log \cos y.1+x.\frac{1}{\cos y}.\frac{d}{dx}\left(\cos y\right)$
$\Rightarrow \log \cos x\frac{dy}{dx}+\frac{y}{\cos x}.(-\sin x)=\log \cos y+\frac{x}{\cos y}(-\sin y).\frac{dy}{dx}$
$\Rightarrow \log \cos x\frac{dy}{dx}-y\tan x=\log \cos y-x\tan y\frac{dy}{dx}$
$\Rightarrow (\log \cos x+x\tan y)\frac{dy}{dx}=y\tan x+\log \cos y$
$\therefore \frac{dy}{dx}=\frac{y\tan x+\log \cos y}{x\tan y+\log \cos x}$