Solution:

Let’s take $y=(\sin x)^{x}+\sin ^{-1} \sqrt{x}=u+v$

In which $u=(\sin x)^{x} \text { and } v=\sin ^{-1} \sqrt{x}$

$\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}$………(1)

So now $u=(\sin x)^{x}$

$\log u=\log (\sin x)^{x}=x \log (\sin x)$

$\frac{d}{d x} \log u=\frac{d}{d x}[x \log (\sin x)]$

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

$\frac{1}{u} \frac{d u}{d x}=x \frac{1}{\sin x} \frac{d}{d x} \sin x+\log (\sin x) .1$

$\frac{1}{u} \frac{d u}{d x}=x \frac{1}{\sin x} \cos x+\log (\sin )=x \cot x+\log \sin x$

$\frac{d u}{d x}=u[x \cot x+\log \sin x] \\\frac{d u}{d x}=(\sin x)^{x}[x \cot x+\log \sin x] \\\dots \dots (2)$

Now again $v=\sin ^{-1} \sqrt{x}$

$\begin{array}{l}
\log v=\log \sin ^{-1} \sqrt{x} \\
\frac{d v}{d x}=\frac{1}{\sqrt{1-(\sqrt{x})^{2}}} \frac{d}{d x} \sqrt{x}\left[\because \frac{d}{d x} \sin ^{-1} f(x)=\frac{1}{\sqrt{1-(f(x))^{2}}} \frac{d}{d x} f(x)\right]
\end{array}$

$\frac{d v}{d x}=\frac{1}{\sqrt{1-x}} \cdot \frac{1}{2 \sqrt{x}}$

$=\frac{1}{2 \sqrt{x} \sqrt{1-x}}$

$=\frac{1}{2 \sqrt{x-x^{2}}} .$………..(3)

Putting the values from equations (2) and (3) in equation (1),

$\frac{d y}{d x}=(\sin x)^{x}[x \cot x+\log \sin x]+\frac{1}{2 \sqrt{x-x^{2}}}$