Let $ (\sin x)^ x = u \; \; \implies log u = x \; log \; \sin x$
on differentiating
$\frac{1}{u} . \frac{dy}{dx} = x . \frac{1}{\sin} \cos x + log\; \sin x . 1$
$\implies \frac{du}{dx} = (\sin x) ^x [x \cot x + log \; \sin x]$
$Let \; v = \sin ^ {-1} \sqrt x$ on differentiating,
$\frac{dv}{dx} = \frac{1}{\sqrt {1-x}} . \frac{1}{2\sqrt x}$
$\frac{dv}{dx} = \frac{1}{2\sqrt {x(1-x)}}$
$\therefore \frac{dy}{dx} = (\sin x)^x [ x \cot x + log \; \sin x ] +\frac{1}{2 \sqrt{x(1-x)}}$