Solve this


If $(\cos x)^{y}=(\cos y)^{x}$ find $\frac{d y}{d x}$.


Here, $(\cos x)^{y}=(\cos y)^{x}$

Taking log on both sides,

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

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

Differentiating it with respect to $x$ using the chain rule and product rule,

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

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

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

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

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

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

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