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Thread: integral

  1. #1
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    integral

    $\displaystyle f(x) = \int_{2}^{x} \ln(t) dt$

    find $\displaystyle (f^{-1})'(0)$
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  2. #2
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    Use the fact that,
    $\displaystyle [f^{-1}(c)]'=\frac{1}{f'(f^{-1}(c))}$
    In this case,
    $\displaystyle c=0$.

    $\displaystyle [f^{-1}(0)]'=\frac{1}{f'(f^{-1}(0))}$
    Now,
    $\displaystyle f^{-1}(0)=2$ because $\displaystyle \int_2^2 \ln t dt =0$.
    And,
    $\displaystyle f'(x)=\ln x$.
    Thus,
    $\displaystyle [f^{-1}(0)]'=\frac{1}{\ln 2}$
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