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Thread: Help with finding the derivative of inverse functions

  1. #1
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    Help with finding the derivative of inverse functions

    Hey I need some help with the following problems:

    Find (f^-1)'(a):

    1. f(x)= 2x^3 + 3x^2 + 7x + 4, a=4

    2. f(x)= x^3 + 3sinx + 2cosx, a=2
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  2. #2
    Senior Member JaneBennet's Avatar
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    If $\displaystyle \mathrm{f}^{-1}(a)=b$, then $\displaystyle a=\mathrm{f}(b)$. Find b from the given a.

    Now use this rule: If $\displaystyle y=\mathrm{f}(x)$, $\displaystyle (\mathrm{f}^{-1})'(y)=\frac{1}{\mathrm{f}'(x)}$. Hence $\displaystyle (\mathrm{f}^{-1})'(a)=\frac{1}{\mathrm{f}'(b)}$.

    For example, for (1): $\displaystyle \mathrm{f}(x)=4\$ $\displaystyle \Rightarrow\$ $\displaystyle 2x^3 + 3x^2 + 7x + 4=4$ $\displaystyle \Rightarrow\$ $\displaystyle x(2x^2+3x+7)=0$ $\displaystyle \Rightarrow\$ $\displaystyle x=0$ (the quadratic has no real roots). So $\displaystyle (\mathrm{f}^{-1})'(4)=\frac{1}{\mathrm{f}'(0)}=\frac{1}{7}$

    Similarly for (2). $\displaystyle \mathrm{f}(x)=2\$ $\displaystyle \Rightarrow\$ $\displaystyle x^3 + 3\sin{x} + 2\cos{x}=2$. By inspection, $\displaystyle x=0$ is a solution. Hence $\displaystyle (\mathrm{f}^{-1})'(2)=\frac{1}{\mathrm{f}'(0)}$.
    Last edited by JaneBennet; Jan 23rd 2008 at 06:06 PM.
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