# Rewriting of functional equation $f(xy)=f(x)+f(y)$

Given the following equation:

$f(xy)=f(x)+f(y)$

and the fact that $y=x^{-1}$,

I've to find how this could became:

$f'(x) = f'(1)/x$,

where it is said that $f'(x)$ is the total derivative of $f$

I really miss some steps..someone could help?

First if we put $x = y = 1$ we see that $f(1) = 0$ and further putting $y = 1/x$ (with $x \neq 0$) we get $f(1/x) = -f(x)$ and therefore $$f(x/y) = f(x) - f(y)$$ From the wording of the question it appears that $f'(1)$ exists. Now it is easy to see that \begin{align} f'(x) &= \lim_{h \to 0}\frac{f(x + h) - f(x)}{h}\notag\\ &= \lim_{h \to 0}\dfrac{f\left(\dfrac{x + h}{x}\right)}{h}\notag\\ &= \lim_{h \to 0}\dfrac{f\left(1 + \dfrac{h}{x}\right)}{h}\notag\\ &= \lim_{h \to 0}\frac{1}{x}\cdot\dfrac{f\left(1 + \dfrac{h}{x}\right)}{\dfrac{h}{x}}\notag\\ &= \frac{1}{x}\lim_{t \to 0}\frac{f(1 + t)}{t}\text{ (putting }t = h/x)\notag\\ &= \frac{1}{x}\lim_{t \to 0}\frac{f(1 + t) - f(1)}{t}\text{ (putting }t = h/x)\notag\\ &= \frac{f'(1)}{x}\notag\\ \end{align} In the above I have assumed that $x \neq 0$ (to justify the expression $((x + h)/x)$. Also we have used the existence of $f'(1)$ to write the limit $$\lim_{t \to 0}\frac{f(1 + t) - f(1)}{t}$$ as $f'(1)$.