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Let $f:(0,\infty )\rightarrow \mathbb{R}$ satisfy $f(xy)=f(x)+f(y)$ and let f be differentiable at x=1. Prove f is differentiable over it's entire domain with derivative $f'(x)=\frac{f'(1)}{x}$

Using the definition of a derivative, $$f'(1)=\lim_{h\to0}(\frac{f(1+h)-f(1)}{h})=\lim_{h\to0}(\frac{f(1+\frac{h}{1})}{h})=\lim_{h\to0}(\frac{f(1+h)}{h})$$

Now similarly, $$f'(x)=\lim_{h\to0}(\frac{f(x+h)-f(x)}{h})=\lim_{h\to0}(\frac{f(1+\frac{h}{x})}{h})$$

This is where I get really stuck, I've tried manipulating this expression as much as I can and can't seem to get any further. Searching online I found this old post: Functional equation $f(xy)=f(x)+f(y)$ and differentiability, and managed to use it to manipulate my expression to the form:

$$f'(x)=\frac{1}{x}\lim_{h\to0}(\frac{f(1+\frac{h}{x})}{\frac{h}{x}})$$

By using the limit laws and dividing through by x. From this point I just can't seem to proceed. I can't see how $f'(1)=\lim_{h\to0}(\frac{f(1+h)}{h}) $ could equal $\lim_{h\to0}(\frac{f(1+\frac{h}{x})}{\frac{h}{x}})$ Any help would be much appreciated.

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CoffeeCrow
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    You just need to put $h/x = t$ so that $t \to 0$. This will mean that $$\lim_{h \to 0}\dfrac{f\left(1 + \dfrac{h}{x}\right)}{\dfrac{h}{x}} = \lim_{t \to 0}\frac{f(1 + t)}{t}$$ Also note that your question must have differentiable at $x = 1$ instead of $x = 0$ (because $x = 0$ is not in the domain $(0, \infty)$). – Paramanand Singh May 10 '16 at 10:28
  • Sorry, x=0 was a typo. I've fixed it with an edit. Sorry about the duplicate too, I didn't find that post while searching. This helped alot though, so thank you. – CoffeeCrow May 10 '16 at 10:32
  • No need to feel sorry for typos and neither for duplicates! The community takes care of both. Enjoy asking and answering on MSE. – Paramanand Singh May 10 '16 at 13:30

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