$g'(x) = \frac{1}{x}$ for all $x > 0$ and $g(1) = 0$. Prove that $g(xy) = g(x) + g(y)$ for all $x, y > 0$. I'm trying to write a proof for this, but I don't know how to begin. Any help appreciated.
Suppose $g$ is a function  such that $g'(x) = \frac{1}{x}$ for all $x > 0$ and $g(1) = 0$. Prove that $g(xy) = g(x) + g(y)$ for all $x, y > 0$.  
Let $h(x) = g(x,y)$
 A: $$g(xy)=\int_1^{xy} \frac{dt}{t}=\int_1^x\frac{dt}{t}+\underbrace{\int_x^{xy}\frac{dt}{t}}_{t\,\mapsto\, xt}=\int_1^x\frac{dt}{t}+\int_1^{y}\frac{dt}{t}=g(x)+g(y)$$
A: If you know what the antiderivative is, this is easy. If you don't, note that $$\frac{d}{dx}g(xy) = g'(xy)y = \frac{y}{xy} = \frac{1}{x} = g'(x) = \frac{d}{dx}g(x),$$ so $g(xy) = g(x) + c$. If $x = 1$, we have $g(y) = g(1) + c = 0 + c$, so then $g(xy) = g(x) + g(y)$.

Added later: In the above calculation, $y$ is treated as a constant so that $g(xy)$ is a function of $x$ alone. Alternatively, if you know something about partial derivatives, we can let $f(x, y) = g(xy)$. Then, noting that $g$ is a function of one variable, $$\frac{\partial}{\partial x} f(x, y) = \frac{\partial g}{\partial x}(xy)\frac{\partial}{\partial x}(xy) = g'(xy)y = \frac{1}{xy}y = \frac{1}{x} = g'(x)$$ so $f(x, y) = g(xy) = g(x) + h(y)$ where $h$ is some arbitrary differentiable function yet to be determined. Then $f(1, y) = g(y) = g(1) + h(y) = 0 + h(y)$, so $h(y) = g(y)$. Therefore $f(x, y) = g(xy) = g(x) + g(y)$.
A: Hint (relying on the chain-rule): Consider $h(t)=\frac{e^{g(t)}}{t}$. What could be said about its gradient?
