Derivative of arcsin composed with a square root

Write $f(x) = \arcsin(u)$ where $u = \sqrt{x} = x^{1/2}$. By the chain rule, $f'(x) = \dfrac{1}{\sqrt{1-u^2}} \cdot \dfrac{du}{dx}$. Since $u^2 = x$, this becomes $f'(x) = \dfrac{1}{\sqrt{1-x}} \cdot \dfrac{du}{dx}$.

Now compute $\dfrac{du}{dx} = \dfrac{d}{dx}\left[x^{1/2}\right] = \dfrac{1}{2}x^{-1/2} = \dfrac{1}{2\sqrt{x}}$. Multiplying, $f'(x) = \dfrac{1}{\sqrt{1-x}} \cdot \dfrac{1}{2\sqrt{x}} = \dfrac{1}{2\sqrt{x}\sqrt{1-x}}$, which can also be written as $\dfrac{1}{2\sqrt{x(1-x)}}$.

The outer function was arcsine. In part a, f of x was written as arcsine of u, where u equals the square root of x, so the outer function whose derivative rule was applied first is arcsine, giving one over the square root of one minus u squared before multiplying by du dx.