let $f : \mathbb{R} \setminus \{0\} \to \mathbb{R}$ be differentiable on $\mathbb{R} \setminus \{0\}$ 
let $f : \mathbb{R} \setminus \{0\} \to \mathbb{R}$ be differentiable on $\mathbb{R} \setminus \{0\}$
  And :
$$f(xy)=f(x)+f(y)$$
  And:
  $$f'(1)=1$$
  prove that : $$f(x)=\int_1^{|x|} \dfrac{dt}{t}$$

Suppose:Not aware of $e^x , \ln x $ .
I can not. please help me !
 A: As stated in a comment, $f$ can actually be any scalar multiple of what you claimed it is. First, note that $f(1)=f(1\cdot 1) = 2f(1)$, and so $f(1)=0$. Further,
$$f'(x)=\frac{\partial }{\partial x} (f(x)+f(y))=\frac{\partial}{\partial x}f(xy) = yf'(xy)$$
Letting $x=1$ gives us $f'(1)=yf'(y)$, i.e.
$$f'(y) = \frac{f'(1)}{y}$$
An application of the FTOC tells us that, for $y>0$
$$f(y) = \int_1^y\frac{f'(1)}{t}\,dt = \int_1^{\vert y\vert}\frac{f'(1)}{t}\,dt$$
Now note that $0=f(1)=f(-1)+f(-1)\implies f(-1)=0$. Then the FTOC tells us that, for $y<0$, 
$$f(y) = \int_{-1}^y \frac{f'(1)}{t}\, dt=-\int_{-y}^{1}\frac{f'(1)}{t}\,dt = \int_{1}^{\vert y \vert}\frac{f'(1)}{t}\, dt $$
The second equality holds because the integrand is odd (or alternatively, using the substitution $t\mapsto -t$).
A: Let $x > 0$ .  Now Define $g(t)=f(e^t)$ for all $t \in R$. We come up $$g(t+s)=g(t)+g(s)$$  Now by differentiating both side with respect to $t$ we have $$g'(t+s)=g'(t)$$
for all $s \in R$, this means $g'(t)=g'(0)=f'(1)=1$ (contant) for all $t \in R$.  Therefore $$g(t)=t+g(0) = t$$  now by changing variable $x=e^t$ we have $$f(x)= \ln x = \int_1^{x} \dfrac{dt}{t}$$ 
