Ordinary Differential Equations – Second Derivative of the Inverse Function

Question

I know that the derivative of the inverse function of $f(x)$ is $g'(y) = \frac{1}{f'(x)}$
But how to derive the formula for the second derivative of g(y) knowing that $\left[\frac{1}{f(x)}\right]' = -\frac{f'(x)}{(f(x))^2}$ ?

I just started studying this chapter, so please try to be as simple as possible 😉
Thank you.

Answer 1

It is $$g(f(x)) = x \Longrightarrow g'(f(x))\cdot f'(x) = 1 \Longrightarrow g''(f(x))\cdot f'(x)^2+g'(f(x))\cdot f''(x) = 0 \Longrightarrow g''(y) = -\frac{f''(x)}{f'(x)^3}$$