Answer
$$ y^{\prime} =x^{x}(1+\ln x) \cos x^{x}$$
Work Step by Step
Given $$ y=x^{\sin x} $$
So, we have
\begin{aligned}
&y=\sin x^{x} \\
&\Rightarrow y^{\prime}=\cos x^{x} \frac{d}{d x}\left(x^{x}\right)\\
& \text{let} \ \ u=x^{x} \Rightarrow \ln u=\ln x^{x}\\
&\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =x \ln x\\
&\text{ differentiate with respect to } \ x,\\
& \Rightarrow \frac{u^{\prime}}{u}=x \cdot \frac{1}{x}+1 \cdot \ln x=1+\ln x
\\
&\Rightarrow u^{\prime}=x^{x}(1+\ln x) \\
& \text{so, we get}\\
&\Rightarrow y^{\prime}=\cos x^{x} \cdot x^{x}(1+\ln x)\\
&\ \ \ \ \ \ \ \ \ \ =x^{x} (1+\ln x)\cos x^{x}
\end{aligned}
