Product rule with arcsine

Using the product rule, $h'(x) = \dfrac{d}{dx}[x] \cdot \arcsin(x) + x \cdot \dfrac{d}{dx}[\arcsin(x)] = \arcsin(x) + \dfrac{x}{\sqrt{1-x^2}}$.

Now substitute $x = \frac{1}{2}$: $\arcsin\left(\frac{1}{2}\right) = \frac{\pi}{6}$, and $\sqrt{1 - \frac{1}{4}} = \frac{\sqrt{3}}{2}$, so the second term is $\dfrac{1/2}{\sqrt{3}/2} = \dfrac{1}{\sqrt{3}}$. Thus $h'\left(\frac{1}{2}\right) = \dfrac{\pi}{6} + \dfrac{1}{\sqrt{3}}$.

The answer is product rule. Since h of x was defined as x times arcsine of x, a product of two functions, the derivative required applying the product rule, not the quotient rule. The quotient rule only applies when one function is divided by another.