Finding $g'(x)$ if $f'(x)$ is given and $g$ is the inverse of $f$.

Question

If $g$ is the inverse of $f$ and $f'(x)=\dfrac{1}{1+x^3}$. Find $g'(x)$.

How to proceed with the question? I thought of integrating $f'(x)$ and then finding the inverse of $f$, i.e., $g$ and then differentiating it. But the answer is given as a function of $g(x)$ itself so I’m stuck there.

Answer 1

Hint: Note that $f(g(x))=x.$ Differentiate and use the Chain Rule.

Answer 2

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

Answer 3

If $f(x) = {1 \over 1+x^3}$ and $y=f(x)$, then $x= \sqrt[3]{{1 \over y}-1}$.

$g'(x) = -{1 \over 3 x^2 (\sqrt[3]{{1 \over y}-1})^2 } = -{1 \over 3 x^2 g^2(x) }$.