Given a function $f$ satisfying the following two conditions for all $x$ and $y$:
(a) $f(x+y)=f(x)\cdot f(y)$,
(b) $f(x)=1+xg(x)$, where $\displaystyle \lim_{x\rightarrow 0}g(x)=1$.
Prove that $f'(x)=f(x)$.
The only thing I know is that $f'(x)=f(x)$ is true for $x=\{0,1\}$ , but how do we know that it's true for all $x$?