Answer
$$ g'(x)= \frac{1}{12}(\frac{4}{x+1})^{2/3} .$$
Work Step by Step
Since $f(x)=4x^3-1$, then $f'(x)=12x^2$. Hence by Theorem 2 and assuming that $g(x)=f^{-1}(x)$, we have
$$g(x)= (\frac{x+1}{4})^{1/3}, \quad g'(x)=\frac{1}{f'(g(x))}=\frac{1}{12((x+1)/4))^{2/3} }=\frac{1}{12}(\frac{4}{x+1})^{2/3} .$$
