Suppose f is a function such that f'(x) = 1/x for all x > 0. Prove that if f(1) = 0, then f(xy) = f(x) + f(y) for all x,y > 0 Suppose f is a function such that $f'(x) = \frac{1}{x}$ for all $x > 0$. Prove that if $f(1) = 0, then f(xy) = f(x) + f(y)$ for all $x,y > 0$
As stated above, how do i go about doing this question? 
Hi all, i would like to apologize if the question is missing context or other details. 
This question was part of my calculus tutorial.
I was on the topic of application of differentiation when this question was given to me.
Thus far i have learnt about intermediate value theorem, mean value theorem, rolle's theroem, Fermat's thereom.
As of this point in my learning, i have not touched upon integration. The only reason why i accepted the answer with integration in it was because i understood it.
Thus i believe i am suppose to use one the few theorems that i've stated to solve this question. But i have no idea where to start.
At first i differentiated the equation $f(xy)=f(x)+f(y)$ with respect to x
i believed the answer to be $f'(xy)\frac{dy}{dx}=f'(x)+f'(y)\frac{dy}{dx}$
which would be $\frac{1}{xy}\frac{1}{x}=\frac{1}{x}+\frac{1}{y}\frac{1}{x}$
At this point a few questions came to my mind 
1) am i suppose to differentiate the eqn in the first place
2) what has $f(1)=0$ got to do with anything
3) what theorem am i suppose to use to solve this question
This was my thought process throughout the question and i hope someone would point out my mistakes. Thank you
 A: Let $\phi(x) = f(xy)-f(x)-f(y)$. Note that $\phi(1) = 0$, and $\phi'(x) = 0$. Hence $\phi = 0$.
A: Use integration. You can either use the fundamental theorem of calculus directly as shown by @Simon s, or else do the basic integration, and solve for the constant of integration later. I will do the latter. Notice that I use the variable $u$ initially. This is to prevent confusion with the second statement you give: $f(xy)=f(x)+f(y)$.
$$f'(u)=\frac{1}{u} \rightarrow \int \rightarrow f(u)=\ln(|u|)+C$$
$$f(1)=\ln(1)+C=0 \leftrightarrow 0+C=0 \rightarrow C=0$$
Therefore:
$$f(u)=\ln(|u|)$$
Since we have constrained: $x,y>0$, we can drop the absolute value signs at this point. At this point, we have the basic logarithm rule:
$$\ln(u*v)=\ln(u)+\ln(v)$$
Therefore, we say:
$$f(xy)=\ln(xy)=\ln(x)+\ln(y)$$
A: Fix $y\in \mathbb{R}_{+}$ and let $g(x)=f(xy)-f(x)$ for $x \in \mathbb{R}_+$, then $g'(x)={y \over xy}-{1 \over x}=0$, so that $g(x)$ is constant. Let $x=1$, then by hypothesis, $g(1)=f(y)-f(1)=f(y)$, and the result immediately follows for any $x$.
