Since the function is differentiable in $(0,x)$, by Lagrange's theorem, we can say that there exists a number $\epsilon$ in $(0,x)$ such that
\[\frac{\log(1+x) - \log(1+0)}{x-0} = f'(\epsilon)\]
where $f'(\epsilon)$ is the derivative of $f(x)$ at $\epsilon$.
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