Derivative of Inverse Functions

**rualin** · July 2nd 2007, 12:18 PM

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

I believe $[f^{-1}(x)]'=\frac{1}{f'(x)}$ . If so, then $h'(x)=[f^{-1}(x)]'=\frac{1}{f'(x)}=\frac{1}{3x^2+1}$ and $h'(2)=\frac{1}{3(2)^2+1}=\frac{1}{13}$ but this is not the answer. Am I mistaking an $x$ for a $y$ 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

**red_dog** · July 2nd 2007, 12:40 PM

Originally Posted by rualin

I believe $[f^{-1}(x)]'=\frac{1}{f'(x)}$ .

If $f(x)=y$ then $\displaystyle \left(f^{-1}\right)'(y)=\frac{1}{f'(x)}$ .
We have $f^{-1}(2)=x\Leftrightarrow f(x)=2\Leftrightarrow x^3+x-2=0\Rightarrow x=1$ .
So $\displaystyle \left(f^{-1}\right)'(2)=\frac{1}{f'(1)}=\frac{1}{4}$

**rualin** · July 2nd 2007, 12:49 PM

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

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