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Q.
If $y^{\cos x}=x^{\sin y}$, then $\frac{d y}{d x}=$
TS EAMCET 2019
Solution:
It is given that
$y^{\cos x}=x^{\sin y}$
On taking logrithm both sides, we get
$\cos\,x \,\log \,y=\sin\, y \,\log \,x$
On differentiating both side w.r.t. $x$, we get
$\frac{1}{y}(\cos x) \frac{d y}{d x}-(\sin x) \log y $
$=\frac{1}{x} \sin y+(\log x)(\cos y) \frac{d}{d x} $
$\Rightarrow \frac{d y}{d x}\left(\frac{1}{y} \cos x-(\log x) \cos y\right)=\frac{1}{x} \sin y+(\sin x) \log y $
$\Rightarrow \frac{d y}{d x}=\frac{y(\sin y+x \sin x \log y)}{x(\cos x-y \log x \cos y)}$