Differentiate the function with respect to $x$: $\sin^{-1}(x\sqrt{x})$,where $0 \le x \le 1$.

Let $y = \sin^{-1}(x\sqrt{x})$.
Using the chain rule,we have:
$\frac{dy}{dx} = \frac{d}{dx} \sin^{-1}(x\sqrt{x})$
$= \frac{1}{\sqrt{1 - (x\sqrt{x})^2}} \cdot \frac{d}{dx}(x \cdot x^{1/2})$
$= \frac{1}{\sqrt{1 - x^3}} \cdot \frac{d}{dx}(x^{3/2})$
$= \frac{1}{\sqrt{1 - x^3}} \cdot \left(\frac{3}{2} x^{1/2}\right)$
$= \frac{3\sqrt{x}}{2\sqrt{1 - x^3}}$
$= \frac{3}{2} \sqrt{\frac{x}{1 - x^3}}$