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Math Help - Inverses

  1. #1
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    Inverses

    If f is a one-to-one, twice differentiable function with inverse function g, show that:

    \displaystyle g''(x) = - \frac{f''(g(x))}{[f'(g(x))]^3}

    OK, here's where I am:

    f^{-1}(x) = g(x)

    \displaystyle g'(x) = \frac {1}{f'(g(x))}

    \displaystyle g''x = \frac{f'(g(x))\cdot(0) - (1)(f''(g(x))(g'(x))}{[f'(g(x))]^2}

    \displaystyle g''(x) = \frac{-f''(g(x))(g'(x))}{[f'(g(x))]^2}

    So I'm close to the answer, but I don't see how to take it from here. Have I made any mistakes thus far or am I just missing the next step? (or both).

    Thanks.
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  2. #2
    Behold, the power of SARDINES!
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    You answer is equivalent look up one line and what does

    g'(x)=
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  3. #3
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    Ack! Right in front of my eyes. Thanks.
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