Find $\frac{d y}{d x}$ in the following exercise $(\cos x)^{y}=(\cos y)^{x}$

Solution:

Provided: $(\cos x)^{y}=(\cos y)^{x}$

$\log (\cos x)^{y}=\log (\cos y)^{x}$

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

$\frac{d}{d x}(y \log \cos x)=\frac{d}{d x}(x \log \cos y)$

$y \frac{d}{d x} \log \cos x+\log \cos x \frac{d y}{d x}=x \frac{d}{d x} \log \cos y+\log \cos y \frac{d}{d x} x$

$y \frac{1}{\cos x} \frac{d}{d x} \cos x+\log \cos x \frac{d y}{d x}=x \frac{1}{\cos y} \frac{d}{d x} \cos y+\log \cos y$

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

$-y \tan x+\log \cos x \frac{d y}{d x}=-x \tan y \cdot \frac{d y}{d x}+\log \cos y$

$x \tan y \frac{d y}{d x}+\log \cos x \cdot \frac{d y}{d x}=y \tan x+\log \cos y$

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

$\frac{d y}{d x}=\frac{y \tan x+\log \cos y}{x \tan y+\log \cos x}$