Chain rule with a square root composite function

Rewrite $h(x) = (4x^3-2x)^{1/2}$. By the chain rule, $h'(x) = \frac{1}{2}(4x^3-2x)^{-1/2} \cdot (12x^2 - 2)$, since the derivative of the outer power function is $\frac{1}{2}u^{-1/2}$ and the derivative of the inner function $u = 4x^3-2x$ is $12x^2-2$.

This simplifies to $h'(x) = \dfrac{12x^2-2}{2\sqrt{4x^3-2x}}$. Now evaluate at $x=2$: the inner function gives $4(2)^3 - 2(2) = 32-4=28$, and $12(2)^2-2 = 48-2=46$. So $h'(2) = \dfrac{46}{2\sqrt{28}} = \dfrac{23}{\sqrt{28}}$.