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A function $f$ is defined in $R$, and $f'(0)$ exist.
Let $f(x+y)=f(x)f(y)$ then prove that $f'$ exists for all $x$ in $R$.

I think I have to use two fact:
$f'(0)$ exists
How to combine these two things to prove that statement?

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Possible duplicate:… – akkkk Dec 16 '12 at 12:40
Related:… – akkkk Dec 16 '12 at 12:41
Related:… – akkkk Dec 16 '12 at 12:41

We have $$f(0)=f(0+0)=f(0)f(0)=f^2(0)\Rightarrow f(0)=0\text{ or }f(0)=1$$ If $f(0)=1$ by definition $$f^{\prime}(0)=\lim_{h\to 0}\frac{f(h)-1}{h}$$ and so $$\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0}\frac{f(x)f(h)-f(x)}{h}=f(x)\lim_{h\to 0}\frac{f(h)-1}{h}=f(x)f^{\prime}(0)$$ Thus $f$ is differentiable in $\mathbb{R}$

If $f(0)=0$, $f(x)=f(x+0)=0\ \forall x\in \mathbb{R}$ and again $f$ is differentiable.

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what's missing is just proving that $f(0)=1$. – tst Dec 16 '12 at 12:23
@tst I am editing it. – Nameless Dec 16 '12 at 12:24
You mean $f'(x)$ in the second line? – landolf Dec 16 '12 at 12:25
It's a bit easier, $f(x)=f(x+0)=f(x)f(0)$. – tst Dec 16 '12 at 12:26

$f$ is continuos at $0$ and hence shhow that $f$ is continuos at whole $\mathbb{R}$ and hence is of the form $e^{ax}$

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You make quite a stronger statement than what was asked for, but it's technically correct. – akkkk Dec 16 '12 at 15:53

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