Suppose $f:(0,\infty) \to \mathbb{R}$ satisfies $f(x/y) = f(x) - f(y)$ for all $x,y>0$ and that $f(1) = 0$.
Question: Show that $f$ is differentiable on $(0,\infty)$ if and only if $f$ is differentiable at $1$.
My attempt on this problem:
Let $a \in (0,\infty)$. Then $f'(x) = \lim_{h\to 0} \frac{f(a+h) - f(a)}{h} = \lim_{h\to 0} \frac{f(\frac{a+h}{a})}{h} = \lim_{h\to 0} \frac{f(1 +\frac{h}{a})}{h}.$
All I know is that the top will go to $f(1) = 0$ and the bottom will also go to $0$ but I don't know how to show that this is differentiable on $(0,\infty)$. Any help would be appreciated.
If $f$ is differentiable on $(0,\infty)$,then $f$ is differentiable at $1$.
For the other direction, suppose $f$ is differentiable at $1$,i.e.$$\lim_{h\to 0} \frac{f(1+h) - f(1)}{h}=\lim_{h\to 0} \frac{f(1+h)-0}{h}$$ exists. Now let $a \in (0,\infty)$. Then \begin{align} f'(a) &= \lim_{h\to 0} \frac{f(a+h) - f(a)}{h}\\ & = \lim_{h\to 0} \frac{f(\frac{a+h}{a})}{h} \\ &= \lim_{h\to 0} \frac{f(1 +\frac{h}{a})}{h} \end{align} Let $u=\frac{h}{a}$. Then we have \begin{align} f'(a) &=\lim_{u\to 0} \frac{f(1 +u)}{au}\\ &=\frac{f'(1)}{a} \end{align} Since $a\neq 0$, $f'(a)$ exists.