TITLE: If $f(x+y)=f(x)\cdot f(y)$, then $f(x)=e^{ax}$
QUESTION [5 upvotes]: I'm trying to prove this question:

Let $f:\mathbb R\to \mathbb R$ a function such that $f(x+y)=f(x)\cdot
 f(y)$, $f(0)=1$ and $f'(0)=a$. Show that $f(x)=e^{ax}$, for every
  $x\in \mathbb R$.

First of all I'm trying to prove that this function is positive, but even that it's difficult to me, I know that this function is never zero, because if there is $y \in \mathbb R$ such that $f(y)=0$, then $f(x-y+y)=f(x-y)\cdot f(y)=0$, then $f(x)=0$ for every $x\in \mathbb R$.
I need help to prove the positivity and hints to follow from that point.
Thanks a lot in advance.

REPLY [6 votes]: Pardon me if my solution seems like a cop out
They've already said the derivative at $0$ is defined, so $\frac{f(h) - 1}{h} \rightarrow a$.
Now $\frac{f(x+h)-f(x)}{h} = f(x) \frac{f(h) - 1}{h}$ which as $h \rightarrow 0$ is $a f(x)$.  So $f'(x) = a f(x)$ and $f(0) = 1$.
This has solution $f(x) = e^{a x}$.