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