# Thread: Derivative of Inverse Functions

1. ## Derivative of Inverse Functions

I am studying for the Calculus Clep exam and I am having trouble understanding how to find the derivative of the inverse of a function.

Here is the problem I am having trouble with and what I have done with it:

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Let $\displaystyle f(x)=x^3+x$. If $\displaystyle h$ is the inverse function of $\displaystyle f$, then $\displaystyle h'(2)=$ (A) $\displaystyle \frac{1}{13}$ (B) $\displaystyle \frac{1}{4}$ (C) $\displaystyle 1$ (D) $\displaystyle 4$ (E) $\displaystyle 13$

I believe $\displaystyle [f^{-1}(x)]'=\frac{1}{f'(x)}$. If so, then $\displaystyle h'(x)=[f^{-1}(x)]'=\frac{1}{f'(x)}=\frac{1}{3x^2+1}$ and $\displaystyle h'(2)=\frac{1}{3(2)^2+1}=\frac{1}{13}$ but this is not the answer. Am I mistaking an $\displaystyle x$ for a $\displaystyle y$ or something like that? Also, if that is not the correct general equation to find the derivative of the inverse of a function, what is it?

Thank you!
Raul

2. Originally Posted by rualin
I believe $\displaystyle [f^{-1}(x)]'=\frac{1}{f'(x)}$.
If $\displaystyle f(x)=y$ then $\displaystyle \displaystyle \left(f^{-1}\right)'(y)=\frac{1}{f'(x)}$.
We have $\displaystyle f^{-1}(2)=x\Leftrightarrow f(x)=2\Leftrightarrow x^3+x-2=0\Rightarrow x=1$.
So $\displaystyle \displaystyle \left(f^{-1}\right)'(2)=\frac{1}{f'(1)}=\frac{1}{4}$

3. Perfect! That's exactly what I wanted to know.