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Math Help - 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 \mathrm{f}^{-1}(a)=b, then a=\mathrm{f}(b). Find b from the given a.

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

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

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