What's are the derivatives of $f^{-1}g(x)$

Question

What are the first and second derivatives of $f^{-1}g(x)$?

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My attempt: I know that the derivative $f^{-1}{'}(x)=\frac{1}{f'(x)}$ and $f^{-1}{''}(x)=- \frac{f''(x)}{f'(x)^3}$. The first derivative of $f(g(x))$ is $f'(g(x))*g'(x)$. Hence, the first derivative of $f^{-1}g(x)$ is

$$\frac{g'(x)}{f'(g(x))}.$$

The second derivative of $f(g(x))$ is $$g'(x)^2*f''(g(x))+g''(x)f'(g(x)).$$

Hence, the second derivative of $f'(g(x))*g'(x)$ is given by:

$$\frac{g'(x)^2*f''(g(x))}{f'(g(x))^3}+\frac{g''(x)}{f'(g(x))}.$$

Answer 1

Your computation of the derivative of the inverse is wrong:

$$ (f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))} $$

You have to evaluate it in the right point.

Answer 2

Consider $h(x) =f^{-1}(g(x)) $ then $$h'(x) =g'(x)(f^{-1})'(g(x)) =g'(x)\frac{1}{f'f^{-1}(g(x))}$$ $$h''(x) =g''(x)\frac{1}{f'f^{-1}(g(x))} +g'(x)\left(\frac{1}{f'f^{-1}(g(x))}\right)'\\=g''(x)\frac{1}{f'f^{-1}(g(x))} -g'(x)\frac{\left(f'(f^{-1}(g(x)))\right)'}{\left(f'f^{-1}(g(x))\right)^2}$$

and $$\left(f'(f^{-1}(g(x)))\right)'=f''(f^{-1}(g(x)))\left(f^{-1}(g(x))\right)' \\= f''(f^{-1}(g(x)))g'(x)\left(f^{-1}\right)'(g(x))\\= \frac{g'(x)f''(f^{-1}(g(x)))}{f'f^{-1}(g(x))}$$

Thus, $$h''(x) =\frac{g''(x)}{f'f^{-1}(g(x))} -g'(x)^2\frac{f''(f^{-1}(g(x)))}{\left(f'f^{-1}(g(x))\right)^3}$$