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This question already has an answer here:

Let $f$: $\mathbb R$ $\to$ $\mathbb R$ be a function such that $f(x+y)$ = $f(x)f(y)$ for all $x,y$ $\in$ $\mathbb R$. Suppose that $f'(0)$ exists. Prove that $f$ is a differentiable function.

This is what I've tried: Using the definition of differentiability and taking arbitrary $x_0$ $\in$ $\mathbb R$.

$\lim_{h\to 0}$ ${f(x_0 + h)-f(x_0)\over h}$ $=$ $\cdots$ $=$ $f(x_0)$$\lim_{h\to 0}$ ${f(h) - 1\over h}$.

Then since $x_0$ arbitrary, using $f(x_0+0) = f(x_0) = f(x_0)f(0)$ for $y = 0$, can I finish the proof?

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marked as duplicate by Martin Sleziak, Community Dec 1 '15 at 13:23

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Yes, since $f(0) = f(0 +0) = f(0)^2$, it follows that either $f(0) = 1$ or $f(0) = 0$. If $f(0) = 0$, then again use the defining property to conclude that $f$ is identically zero. Hence if $f\neq 0$, then $f(0) = 1$, and the limit you have at the end of your equalities is $$ \lim_{h\to 0} \frac{f(h) -1}{h} = f'(0) $$ Hence the function is differentiable at $x_0$ with $f'(x_0) = f(x_0)f'(0)$.

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By assumption $$ \lim_{h \to 0}\frac{f(h) - f(0)}{h} = f(0)\lim_{h \to 0}\frac{f(h) - 1}{h} $$ exists, implying $$ l := \lim_{h \to 0}\frac{f(h)-1}{h} $$ exists; hence, if $x \in \Bbb{R}$, then $$ \frac{f(x+h) - f(x)}{h} = \frac{f(x)[f(h) - 1]}{h} \to f(x)l $$ as $h \to 0$.

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