Function that satisfy the properties of the exponential function Let $E:\mathbb{R} \to \mathbb{R}$ be an infinitely continuously differentiable function and $E$ is not zero function 
such that $$E(u+v)=E(u)E(v).$$
Show that $E(x)=e^{ax}$ for some $a\in \mathbb{R}$.
My partial answer:
The function $x\mapsto e^{ax}$ satisfies the properties immediately.
For $y=0$, we have $E(x)=E(x)E(0)$. Thus, $E(0)=1$. Let $y\in \mathbb{R}$ is fixed, then 
$$E'(x+y)=E(y)E'(x)$$.
I don't know how to continue the answer. 
Could you give me some hint? 
Thanks.
 A: Using the definition of the derivative,
$$E'(x) = \lim_{h\to 0}\frac{E(x+h)-E(x)}{h} = \ldots 
=\left(\lim_{h\to 0}\frac{E(h)-1}{h} \right)E(x).  $$
By the functional equation, $E(0)=E(0)^2$, so $E(0)=0$ or $1$.  If $E(0)=0$, the functional equation gives $E(x)=0$ for all $x$, and you don't want the zero solution.  So $E(0)=1$.  Therefore 
the last limit exists and is equal to $E'(0)$.   So the derivative of $E$ is proportional to $E$.  $E$ has the form $E(x) = ce^{ax}$ for some constants $c$ and $a$. Again, since $E(0)=1$, $c=1$.
This is how I would do it.  Your work looks valid, but I'm not sure how to progress beyond the last equation you wrote.
I just noticed that this is pretty similar to Daniel Fischer's suggestion, which I think he made before this answer.  If Daniel makes his suggestion an answer, it would not bother me if you accepted it instead of mine.
A: Following up on Dan's suggestion we have (d/dx)E(x+y) = E'(x)E(y).  Setting x  = 0 we get E'(y) = E'(0)E(y).  The unique solution of this differential equation is E(y) = $Ce^{ay}$ where a = E'(0).  But if E(0) = 1, C = 1.
The case of E $\equiv$ 1 comes when a = 0; E $\equiv$  0 is from C = 0.
