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 2


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

  • $\begingroup$ 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... $\endgroup$ Dec 2, 2012 at 16:30
  • 1
    $\begingroup$ 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) $\endgroup$ Dec 2, 2012 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 $g'(y)$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .