$\displaystyle f(x) = \int_{2}^{x} \ln(t) dt$
find $\displaystyle (f^{-1})'(0)$
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,
$\displaystyle [f^{-1}(0)]'=\frac{1}{\ln 2}$