Find $\frac{dy}{dx}$, if $(\cos x)^y = (\cos y)^x$.

$\frac{\ln(\cos y) + y \tan x}{\ln(\cos x) + x \tan y}$

The correct answer is Option (2) → $\frac{\ln(\cos y) + y \tan x}{\ln(\cos x) + x \tan y}$ ##

Given, $(\cos x)^y = (\cos y)^x$

Taking log on both sides, we get

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

Differentiating both sides w.r.t. $x$, we get

$\frac{dy}{dx} \cdot \log(\cos x) + y \cdot \left( \frac{-\sin x}{\cos x} \right) = 1 \cdot \log(\cos y) + x \cdot \left( \frac{-\sin y}{\cos y} \right) \frac{dy}{dx}$

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

$\frac{dy}{dx} \left[ \log(\cos x) + \frac{x \sin y}{\cos y} \right] = \log(\cos y) + \frac{y \sin x}{\cos x}$

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

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