Answer
$$ y^{\prime}=(\sin x)^{x}[\ln (\sin x)+x \cot x]$$
Work Step by Step
Given $$ y=(\sin x)^{x} $$
So, we have
\begin{aligned}
&y=(\sin x)^{x} \\
&\Rightarrow \ln y=\ln (\sin x)^{x}\\
& \ \ \ \ \ \ \ \ \ \ \ \ \ =x \ln (\sin x) \\
&\text{ differentiate with respect to } \ x, \\
&\Rightarrow \frac{y^{\prime}}{y}=\ln (\sin x)+x\left(\frac{\cos x}{\sin x}\right) \\
&\Rightarrow y^{\prime}=y[\ln (\sin x)+x \cot x]\\
&\Rightarrow y^{\prime}=(\sin x)^{x}[\ln (\sin x)+x \cot x]
& \ \ \ \ \ \ \ \ \ \ \ \
\end{aligned}
