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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 $f'(x) = lim_{hto 0} frac{f(a+h) - f(a)}{h} = lim_{hto 0} frac{f(frac{a+h}{a})}{h} = lim_{hto 0} frac{f(1 +frac{h}{a})}{h}.$

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.$$lim_{hto 0} frac{f(1+h) - f(1)}{h}=lim_{hto 0} frac{f(1+h)-0}{h}$$ exists. Now let $a in (0,infty)$. Then
begin{align}
f'(a) &= lim_{hto 0} frac{f(a+h) - f(a)}{h}\
& = lim_{hto 0} frac{f(frac{a+h}{a})}{h} \
&= lim_{hto 0} frac{f(1 +frac{h}{a})}{h}
end{align}
Let $u=frac{h}{a}$. Then we have
begin{align}
f'(a) &=lim_{uto 0} frac{f(1 +u)}{au}\
&=frac{f'(1)}{a}
end{align}
Since $aneq 0$, $f'(a)$ exists.