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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 find this problem in here, and wonder the proof stated below has any problems?

Since $f'(0)=\lim_{h \rightarrow 0} \frac{f(0)(f(h)-1)}{h}=f(0)\lim_{h \rightarrow 0} \frac{f(h)-1}{h}$
it means $\lim_{h \rightarrow 0} \frac{f(h)-1}{h}$ exists.
$f'(x)==\lim_{h \rightarrow 0} \frac{f(x)(f(h)-1)}{h}=f(x)\lim_{h \rightarrow 0} \frac{f(h)-1}{h}$
thus $f'(x)$ exists if $f(x)$ exists.
But we can differentiate a function if it has a function value $f(x)$, thus $f'$ exists for all $x$ in $R$.
I know it's not a perfect one,
just want to know it can be another way to solve that problem or not.

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$f(0)=0$ case:
If $f(0)=0$, because $f(x+y)=f(x)f(y)$,
$f(x)=f(x+0)=f(x)f(0)=0$ thus $f'(x)=f'(0)$. Is that right?

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    You're right. That's my misspelling, I'll fix it.2012-12-21
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    When you say "Let $f(x+y)=f(x)f(y)$" do you mean that the equality holds for all $x,y$? (You are not "letting" anything, as $f$ was already given. One would "suppose" or "assume" the equality.)2012-12-21
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    Your solution is fine, there is one minor mistake though: When you conclude " It means0$ \lim_{h \rightarrow 0} \frac{f(h)-1}{h}$" : what if $f(0)=0$? You should threat that case separately....2012-12-21
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    You forgot one case in your sketch. If $f(0)=0$, the limit you wrote does not exist. (And another comment appeared just before mine indicating the same issue.) On the other hand, this case is very easy to deal with.2012-12-21
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    Then, if I add $f(0)=0$ case, this proof can be a solution of that problem?2012-12-21
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    Definitely. One can polish the presentation a little bit, but the key idea is there, and the approach is efficient.2012-12-21
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    Just to be clear, the first part of your proof shows that $f'(x) = f(x) f'(0)$, hence $f$ is differentiable everywhere. For the second part, since $f(x) = f(x+0)=f(x)f(0)$, you have $f(x) = 0$ everywhere. It follows that $f'(x) = 0$.2012-12-21

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