TITLE: Let $f$ be a function such that $f'(x)=\frac{1}{x}$ and $f(1) = 0$ , show that $f(xy) = f(x) + f(y)$
QUESTION [8 upvotes]: Let $f:\mathbb{R^+}\rightarrow \mathbb{R} $ a differentiable function such that $f'(x) = \frac{1}{x}$ and $f(1)=0$ . Show that $f(xy) = f(x) + f(y)$ for all $x,y \in \mathbb{R^+}$
It seems to be the logarithm function but I am unable to prove this statement without the fundamental theorem of calculus, is there any other way to prove this fact?
Thanks for your help

REPLY [5 votes]: Consider $G:\Bbb{R}^+\to\Bbb{R}$ such that $G(x,y)=f(xy)-f(x)-f(y)$.
One has
$${\partial{G(x,y)}\over\partial{x}}={y\over xy}-{1\over x}=0$$
$${\partial{G(x,y)}\over\partial{y}}={x\over xy}-{1\over y}=0$$
Therefore $G$ is constant. $G(x,y)=G(1,1)=0$