PROBLEM:

Suppose that a function $\displaystyle f$ satisfies the following conditions for all real values $\displaystyle x$ and $\displaystyle y$:

i. $\displaystyle f(x+y)=f(x)*f(y)$

ii. $\displaystyle f(x)=1+xg(x)$, where $\displaystyle \lim_{x\to\0}g(x)=1$

Show that the derivative $\displaystyle f'(x)$ exists at every value of $\displaystyle x$ and that $\displaystyle f'(x)=f(x)$.

I honestly don't have much of a clue here. Instead, I've tried to ask myself questions to get some idea going but to no avail. First, I tried to make sense of what the conditions imply and discovered that because $\displaystyle x$ and $\displaystyle y$ belong in the same set of real numbers without restriction, then everything said about $\displaystyle x$ is also true about $\displaystyle y$.

ATTEMPT:

Therefore, $\displaystyle f(y)=1+yg(y)$, where $\displaystyle \lim_{y\to\0}g(y)=1$

I thought, maybe, if I could rewrite condition (i) with a substitution in terms of (ii), I might get somewhere.

Any thoughts?