A function $f:(0, \infty) \to \Bbb{R} $ satisfies the condition $f(xy) =f(x) +f(y) $ for all $x>0,y>0$. If $f$ is differentiable at $1$, prove that $f$ is differentiable at every $c \in(0,\infty)$ and $f'(c) = \frac{1}{c}f'(1)$.
My thinking:
$f(1)=f(1\cdot 1) =f(1) + f(1)$ which implies that $f(1)=0$.
Since, $f$ is differentiable at $1$, we have $\displaystyle \lim_{x\to 1} \frac{f(x) - f(1)}{x-1} =f'(1)$ which implies that $\displaystyle \lim_{x\to 1} \frac{f(x)}{x-1} = f'(1)$ since $f(1)=0$.
Now, I'm stucked. If I'm on the correct path then please help me. Otherwise if there is any independent approach needed then please tell me. Thanks in advance.
Notice that $$f \Big(\frac xy \Big) + f(y) = f(x)$$ for all $x$ and $y$ in $(0,\infty)$. So, if $c \in (0,\infty)$ observe that $$\lim_{x \to c} \frac{f(x)-f(c)}{x-c} = \frac{1}{c} \lim_{x \to c} \frac{f\Big(\dfrac xc \Big)}{\dfrac xc - 1} = \frac{1}{c} \lim_{t \to 1} \frac{f(t)}{t - 1}.$$