Question
If $(\cos x)^y= (\sin y)^x$, find $\frac{\text{dy}}{\text{dx}}.$

Answer

$(\cos x)^y = (\sin y)^x $
$\Rightarrow y \log \cos x = x \log \sin y $
$\therefore \log (\cos x)_{^{. }} \frac{\text{dy}}{\text{dx}} - y ^{_{. }}\tan x = \log \sin y + x \cot y ^{_{. }} \frac{\text{dy}}{\text{dx}} $
$\therefore \frac{\text{dy}}{\text{dx}} (\log \cos x - x \cot y) = \log \sin y + y \tan x $
$\therefore \frac{\text{dy}}{\text{dx}}=\frac{\log\sin\text{y + y \tan x}}{\text{\log \cos x - x}\cdot\text{\cot y}}.$

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