Problem (from Rudin): Suppose $f(x)f(y) = f(x+y)$ for all real $x$ and $y$.
Assuming that $f$ is differentiable and not zero, prove that $f(x) = e^{cx}$ where $c$ is a constant.
I've shown that $f > 0$, and that $f(0) = 1$. I need help showing that $f'(x) = f(x)$. When I apply the product rule, I get $$f'(x) = f'(x+0) = (f(x)f(0))' = f'(x)f(0) + f(x)f'(0),$$ but I don't know where to go from there.
Fix $x$. Since \begin{align} f(x+y)=f(x)f(y) \end{align} then differentiating with respect to $y$ yields \begin{align} f'(x+y)=f(x)f'(y). \end{align} Now set $y =0$ and you get \begin{align} f'(x)=f'(0)f(x). \end{align}