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Question:

If $(\cos x)^{y}=(\tan y)^{x}$, prove that $\frac{d y}{d x}=\frac{\log \tan y+y \tan x}{\log \cos x-x \sec y \operatorname{cosec} y}$

Solution:

Here,

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

Taking log on both sides,

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

$y \log (\cos x)=x \log (\tan y)\left[U \sin g \log a^{b}=b \log a\right]$

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

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

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

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

\begin{aligned}\left(\frac{y}{\cos x}(-\sin x)\right.&\left.+\log \cos x \frac{d y}{d x}\right) \\=&\left(\frac{x}{\tan y}\left(\sec ^{2} x\right)\right) \frac{d y}{d x} \\ &+\log \tan y \\ &-y \tan x+\log \cos x \frac{d y}{d x}=\left(\sec y \operatorname{cosec} y \times x \frac{d y}{d x}+\log \tan y\right) \end{aligned}

$\frac{d y}{d x}[\log \cos x-x \sec y \operatorname{cosec} y]=\log \tan y+y \tan x$

$\frac{d y}{d x}=\frac{\log \tan y+y \tan x}{\log \cos x-x \sec y \operatorname{cosec} y}$