Function with multiplicative property and $f(0) = f'(0) = 1$ In my differential equations book, the question is posed to find a function $f$ such that $f(0) = 1$, $f'(0) = 1$, and $f(x+y) = f(x)f(y)$ for $x,y \in \mathbb{R}$. The book does nothing more than indicate that $f(x) = e^{x}$, but I'm struggling to see how to move forward to the solution... so far all I have been able to conject is that $f'(x) = f(x) \ \forall \ x$, but I'm having a hard time proving this and am not sure how to move forward. 
Any ideas or hints?
 A: We have
$$\frac{f(x+h)-f(x)}{h}=\frac{f(x)f(h)-f(x)}{h}=f(x)\cdot\frac{f(h)-1}{h} $$
and as $f(0)=1$, $\lim_{h\to0}\frac{f(h)-1}{h} =f'(0)=1$., so that indeed $f'(x)=f(x)$ for all $x$. In particular, the derivative of $\ln f(x)$ is $\frac {f'(x)}{f(x)}=1$, which means that $f\ln f(x)=x$, or $f(x)=e^x$.
To be precise, we should first show that $f(x)>0$ for all $x$ before applying $\ln$: If $f(x)<0$ then by the IVT, $f(x)=0$ for some $y$ between $0$ and $x$. If $f(y)=0$ then $f(0)=f(-y+y)=f(-y)f(y)=0$, contradiction. 
A: Taking the derivative of the functional equation with respect to $x$, we get that
$$f'(x+y) = f(y)f'(x)$$
Similarly, taking the derivative with respect to $y$ yields
$$f'(x+y) = f(x)f'(y)$$
so we can combine these two to get
$$f(y)f'(x) = f(x)f'(y)$$
Note that, if $f(x)=0$ for some $x$, we would have that $f$ is identically $0$, contradicting that $f(0)=1$. Thus, we can divide by $f(x)f(y)$ to get that
$$\frac{f'(x)}{f(x)} = \frac{f'(y)}{f(y)}$$
which means that
$$\frac{f'(x)}{f(x)} = c$$
for all real $x$. Given that $f(0) = f'(0) = 1$, we get that this constant must be $1$, so we have that
$$f'(x) = f(x)$$
and we can proceed from there.
Generally, when you have a functional equation with a function that you know is differentiable, try taking the derivative and see what happens. 
