# second derivative of the inverse function

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.

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It is $$g(f(x)) = x \Longrightarrow g'(f(x))\cdot f'(x) = 1 \Longrightarrow g''(f(x))\cdot f'(x)^2+g'(f(x))\cdot f''(x) = 0 \Longrightarrow g''(y) = -\frac{f''(x)}{f'(x)^3}$$
Note that $y = f(x)$ and $x = g(y)$ in the formula $g'(y) = \frac{1}{f'(x)}$. Hence you can now write $$g'(y) = \frac{1}{f'(g(y))} \, .$$ Now use the chain rule and the known formula for $g'(y)$.