(sin x)cos x

Let $y=(\sin x)^{\cos x}$

Taking log on both sides,

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

$\Rightarrow \log y=\cos x \cdot \log (\sin x)$ $\left[\because \log x^{y}=y \log x\right]$

Differentiating both sides w.r.t. $x$,

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

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

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

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

$\frac{d y}{d x}=y[\cot x \cdot \cos x-\sin x \cdot \log (\sin x)]$

Thus, $\frac{d y}{d x}=(\sin x)^{\cos x}\left[\frac{\cos ^{2} x}{\sin x}-\sin x \cdot \log (\sin x)\right]$