Second derivative of an inverse function

Question

By the inverse function theorem, we know that $G'(x)=1/F'(G(x))$, where $G=F^{-1}$. I want to obtain $G''(x)$, but I don't know how to get the derivative of $F'(G(x))$. Any hints?

Answer 1

Just apply the chain rule: $$ \frac{d}{dx} (F'\circ G)(x) = F''(G(x))G'(x), $$ and insert the value for $G'$ that you have obtained before.

Answer 2

\begin{align} \frac{d}{dx}\left(\frac{1}{f'(g(x))}\right)&=-\frac{1}{\big(f'(g(x))\big)^2} \cdot \underbrace{\frac{d}{dx}\Big(f'(g(x))\Big)}_{\text{Chain Rule}}\\ &=-\frac{1}{\big(f'(g(x))\big)^2} \cdot f''(g(x))\underbrace{\frac{d}{dx}\Big(g(x)\Big)}_{\text{Chain Rule}}\\ &=-\frac{1}{\big(f'(g(x))\big)^2} \cdot f''(g(x)) \cdot g'(x)\\ \end{align}