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Answer: $\frac{(\cos x)^{\sin x}\{\cos x \cdot \log \cos x-\sin x \tan x\}}{(\sin x)^{\cos x}\{-\sin x \log \sin x+\cos x \cot x\}}$

Hint: $\text { Let } u=(\cos x)^{\operatorname{sin} x}, v=(\sin x)^{\cos x}$

Given: $(\cos x)^{\sin x} \text { w.r.t }(\sin x)^{\cos x}$

Explanation:  Apply log on both sides

\begin{aligned} &\log u=\log (\cos x)^{\sin x} \\\\ &\log u=\sin x \log (\cos x) \end{aligned}

Differentiate both side w.r.t. $x$

$\frac{1}{u} \frac{d u}{d x}=\sin x\left[\frac{1}{\cos x}(-\sin x)\right]+\log (\cos x) \cos x$

$=-\sin x \tan x+\cos x \log (\cos x)$

\begin{aligned} &\frac{d u}{d x}=(\cos x)^{\sin x}\{-\sin x \tan x+\cos x \cdot \log \cos x\} \\\\ &v=(\sin x)^{\cos x} \end{aligned}

Apply log on both sides

\begin{aligned} &\log v=\log (\sin x)^{\cos x} \\\\ &\log v=\cos x \cdot \log (\sin x) \end{aligned}

Differentiate both side w.r.t. $x$

$\frac{1}{v} \frac{d v}{d x}=\cos x\left[-\frac{1}{\sin x}(\cos x)\right]+\log (\sin x)(-\sin x)$

$=\cos x \cot x-\sin x \log \sin x$

\begin{aligned} &\frac{d v}{d x}=(\sin x)^{\cos x}\{-\sin x \log \sin x+\cos x \cot x\} \\\\ &\frac{d u}{d v}=\frac{\frac{d u}{d x}}{\frac{d v}{d x}} \end{aligned}

$=\frac{(\cos x)^{\sin x}\{\cos x \cdot \log \cos x-\sin x \tan x\}}{(\sin x)^{\cos x}\{-\sin x \log \sin x+\cos x \cot x\}}$

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