Finding derivative of $f(x)$ where $f(xy) = f(x) + f(y)$ - without change of variable Let $f(x)$ be a function $(0,\infty) \to R$ and for every $x,y$ in the domain we have: $$f(xy) = f(x) + f(y)$$
It is like logarithm but we don't know the exact form of the function. we know it is differentiable at x=1. Now we want to show it is differentiable at its domain and its derivative is $f'(x) = \frac{1}{x} f'(1)$.
Solution:
I can find that $f(1) = 0$ and $f(x/y) = f(x) - f(y)$ So we have:
$f'(1) = \lim_{h\to 0} \frac{f(1+h) - f(1)}{h} = \lim_{h\to 0}\frac{f(1+h)}{h}  $
so
$f'(x) = \lim_{h\to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h\to 0} \frac{f(\frac{x+h}{x})}{h}  = \lim_{h\to 0} \frac{f(1 + h/x)}{h}$.
I know I can solve it with a simple change of variables $h \to 0 ~~~\rightarrow~~~~ h/x \to 0$, But I want to know is there any way to solve this without changing variable? Maybe using definition of limit?
 A: Set $y=1$ to verify that $f(1)=0$. Now when $h\to 0$
\begin{align}
\frac{f(x+h)-f(x)}{h}&=\frac{f((1+h/x)x)-f(x)}{h}=\frac{f(1+h/x)+f(x)-f(x)}{h}=\\
&=\frac{f(1+h/x)}{h/x}\cdot \frac{1}{x}=\frac{f(1+h/x)-f(1)}{h/x}\cdot\frac{1}{x}\to\ ?
\end{align}
A: You pointed out that $f'(1) = \displaystyle \lim_{h \to 0} \frac{f(1+h)}{h}$ exists using the fact that $f$ is differentiable at $x=1$.
A simple substitution shows that $\displaystyle \lim_{h \to 0} \frac{f(1+h/x)}{h/x} = f'(1)$ for all $x > 0$.

The above statement follows from a basic $\epsilon$ argument. Let $\epsilon > 0$ be given. There exists $\delta > 0$ with the property that
$$0 < |h| < \delta \implies \left| \frac{f(1+h)}{h} - f'(1) \right| < \epsilon.$$
Thus
$$0 < |h| < x \delta \implies 0 < \left| \frac hx \right| <  \delta \implies \left| \frac{f(1+h/x)}{h/x} - f'(1) \right| < \epsilon$$ so that $$\frac{f(1+h/x)}{h/x} \to f'(1).$$

Now let $x > 0$ be arbitrary. Then $$f(x+h) - f(x) = f(x(1 + h/x)) - f(x) = f(1 + h/x)$$ so that $$\frac{f(x+h) - f(x)}{h} = \frac{f(1 + h/x)}{h} = \frac 1x \frac{f(1 + h/x)}{h/x}$$
and consequently
$$\lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \frac 1x \lim_{h \to 0} \frac{f(1 + h/x)}{h/x} = \frac 1x \cdot f'(1).$$
