Is this proof about functions correct? Did I do this proof correctly? You'll see why I think it is a little shaky in a minute:
Problem: Let E be a function satisfying $E(0)=E'(0)=1$. Prove that if $E(a+b)=E(a)E(b)$ for all $a,b$, then $E$ is differentiable and $E'(x)=E(x)$ for all x.
Let $E(x)=y$ and $E'(x)=dy/dx$


*

*$E(x+a)=E(x)*E(a)\ldots$

*$E(x+a)-E(x)=E(x)(E(a)-1)\ldots$

*$\lim_{a\to0} E(x+a)-E(x)=\lim_{a\to 0}E(x)(E(a)-1)\ldots$

*$dy=E(x)\cdot \lim_{a\to 0}(E(a)-1)\ldots$

*Since $E(0)=1$, then $\lim_{a\to0}(E(a)-1)=dy$. Since $E'(0)=1$, $dy$ is approximately equal to $dx\ldots$

*$dy=E(x)dx$

*$dy/dx=E(x)$

*$E'(x)=E(x)$


My math teacher skimmed over it and said it is ok, but I still have a feeling it is wrong because of the step which converts $(E(a)-1)$ to $dx$. The problem I see here is that instead of $dy=E(x)dx$ I could have just said $dy=E(x)dy$, so $E(x)=1$ which is clearly wrong. Any commentary is appreciated. Thanks!

I don't like all that $dx$ and $dy$ they are confusing me here (maybe it is just the formatting), just use the definition of differentiable:
$$\lim_{h\to 0} \frac{E(x+h)-E(x)}{h}$$ 
exists.  With $E(a+b)=E(a)\cdot E(b)$ you get
$$\lim_{h\to 0 } \frac{E(x+h)-E(x)}{h}=E(x) \lim_{h\to 0 } \frac{E(h)-1}{h} = E(x) \cdot 
\lim_{h\to 0} \frac{E(h)-E(0)}{h}= E(x) \cdot E'(0)= E(x)$$
