Find d y d x of the function ( x )^ y =( y )^ x . - Vidyadip

Question

Find $\frac{d y}{d x}$ of the function $(\cos x )^{ y }=(\cos y )^{ x }$.

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Answer

We have, $(\cos x)^y=(\cos y)^x$
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 $\text{w.r.t x},$ we get
$y \cdot \frac{d}{d x} \log (\cos x)+\log \cos x \cdot \frac{d}{d x}(y)$
$=x \frac{d}{d x} \log (\cos y)+\log (\cos y) \frac{d}{d x}(x)\ [$by using product rule of derivative$]$
$\Rightarrow y \cdot \frac{1}{\cos x} \frac{d}{d x}(\cos x)+\log (\cos x) \frac{d y}{d x}$
$=x \cdot \frac{1}{\cos y} \frac{d}{d x}(\cos y )+\log \cos y \cdot 1$
$\Rightarrow y \cdot \frac{1}{\cos x}(-\sin x)+\log (\cos x) \cdot \frac{d y}{d x}$
$=x \cdot \frac{1}{\cos y}(-\sin y ) \frac{d y}{d x}+\log \cos y \cdot 1$
$\Rightarrow- y \tan x +\log (\cos x ) \frac{d y}{d x}$
$=- x \tan y \frac{d y}{d x}+\log (\cos y )$
$\Rightarrow[ x \tan y +\log (\cos x )] \frac{d y}{d x}$
$=\log (\cos y )+ y \tan x $
$\therefore \frac{d y}{d x}=\frac{\log (\cos y)+y \tan x}{x \tan y+\log (\cos x)}$

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