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Math Help - Derivative of Inverse Functions

  1. #1
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    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:

    -------------

    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
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  2. #2
    MHF Contributor red_dog's Avatar
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    Quote Originally Posted by rualin View Post
    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}
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  3. #3
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    Perfect! That's exactly what I wanted to know.
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