Differentiate the function with respect to $x$: $(\sin x)^{x} + \sin^{-1} \sqrt{x}$.

Let $y = (\sin x)^{x} + \sin^{-1} \sqrt{x}$.
Let $u = (\sin x)^{x}$ and $v = \sin^{-1} \sqrt{x}$.
Then $y = u + v$,so $\frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx}$ ............$(1)$
For $u = (\sin x)^{x}$:
Taking logarithm on both sides,$\log u = x \log(\sin x)$.
Differentiating with respect to $x$:
$\frac{1}{u} \frac{du}{dx} = \frac{d}{dx}(x) \cdot \log(\sin x) + x \cdot \frac{d}{dx}(\log(\sin x))$
$\frac{1}{u} \frac{du}{dx} = 1 \cdot \log(\sin x) + x \cdot \frac{1}{\sin x} \cdot \cos x$
$\frac{du}{dx} = (\sin x)^{x} [\log(\sin x) + x \cot x]$ ............$(2)$
For $v = \sin^{-1} \sqrt{x}$:
Differentiating with respect to $x$ using the chain rule:
$\frac{dv}{dx} = \frac{1}{\sqrt{1 - (\sqrt{x})^2}} \cdot \frac{d}{dx}(\sqrt{x})$
$\frac{dv}{dx} = \frac{1}{\sqrt{1 - x}} \cdot \frac{1}{2\sqrt{x}} = \frac{1}{2\sqrt{x - x^2}}$ ............$(3)$
Substituting $(2)$ and $(3)$ into $(1)$:
$\frac{dy}{dx} = (\sin x)^{x} (x \cot x + \log(\sin x)) + \frac{1}{2\sqrt{x - x^2}}$.