integral

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Feb 4th 2007, 07:17 PM
viet

integral

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

find $(f^{-1})'(0)$
Feb 4th 2007, 07:38 PM
ThePerfectHacker

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

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

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