# 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.

-

## 2 Answers

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

-
thank you! it was really that simple! Just a single question: if f(x) = g(x), can I also precisely say that f'(x) = g'(x)? Because you used this property at the second implication... –  Umar Jamil Dec 2 '12 at 16:30
Yes .. if you have two identical functions they have the same derivation. I used this fact twice (for the first implication and for the second implication also) –  user127.0.0.1 Dec 2 '12 at 18:49

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 $g'(y)$.

-