Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I want to prove the following claim:

If $f:(0,\infty)\to\mathbb{R}$ satisfying $f(xy)=f(x)+f(y)$, and if $f$ differentiable on $x_0=1$, then $f$ differentiable for all $x_0>0$.

Thank you.

share|cite|improve this question
up vote 5 down vote accepted

Let $y=1+h/x$. Then $$f'(x)=\lim\limits_{h\to 0}\frac{f(x+h)-f(x)}{h}=\lim\limits_{h\to 0}\frac{f(xy)-f(x)}{h}=\lim\limits_{h\to 0}\frac{f(y)}{h}=\frac{1}{x}\lim\limits_{h\to 0}\frac{f(1+h/x)}{h/x}=\frac{f'(1)}{x}.$$

share|cite|improve this answer

Set $g(x) = \exp(f(x))$ and then go here or here. ${}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}$

share|cite|improve this answer

It's a bit hazy, but I' do as follows: 1). Prove that $f(1+x)/x\to f'(1)$ when $x\to 0$ 2). Fix $x>0$, observe that $\frac{f(x+xh)-f(x)}{xh}=\frac{f(1+h)}{xh}$; rewriting it as
$$\frac{f(x+\delta)-f(x)}{\delta}=\frac{1}{x}\frac{f(1+\delta/x)}{(\delta/x)}$$ and conclude by remarking that $$\frac{f(1+\delta/x)}{(\delta/x)} \xrightarrow[\delta\to 0^+]{}f'(1)$$ and thus $$\frac{f(x+\delta)-f(x)}{\delta}=\frac{f'(1)}{x}$$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.