Continuity and Differentiability — Maths STD 12 Science — Question

On taking log both sides, we get
log(cos x)^y = log(cos y)^x
$\Rightarrow$ y log(cos x) = x log(cos y)
On differentiating both sides w.r.t x, we get
$y \cdot \frac { d } { d x } \log ( \cos x ) + \log \cos x \cdot \frac { d } { d x } ( y )$
$= x \frac { d } { d x } \log \left( \cos y) + \log ( \cos y ) \frac { d } { d x } ( x )\right.$ [by using product rule of derivative]
$\Rightarrow \quad y \cdot \frac { 1 } { \cos x } \frac { d } { d x } ( \cos x ) + \log ( \cos x ) \frac { d y } { d x }$$= x \cdot \frac { 1 } { \cos y } \frac { d } { d x }$(cos y) + log cos y.1
$\Rightarrow y \cdot \frac { 1 } { \cos x } ( - \sin x ) + \log ( \cos x ) \cdot \frac { d y } { d x }$$= x \cdot \frac { 1 } { \cos y } $(-sin y)$\frac{dy}{dx}$ + log cos y.1
$\Rightarrow$ - y tanx + log(cos x)$\frac { d y } { d x }$ =-x tan y$\frac { d y } { d x }$+ log(cos y)
$\Rightarrow$[ x tan y + log (cos x)]$​​\frac{dy}{dx}$= log(cos y) + y tan x
$\therefore \quad \frac { d y } { d x } = \frac { \log ( \cos y ) + y \tan x } { x \tan y + \log ( \cos x ) }$