Proving that $f(x) \le\ g(x)$, for all $x \in\ [c,b)$ Let $f$ and $g$ be functions which are differentiable on $(a,b)$ with 
$$f(c) = g(c)$$
for some $c\in (a,b)$.
If
$$ f'(x) \le\ g'(x),\qquad \forall \;x \in [c,b)
$$
prove that
$$
f(x) \le g(x),\qquad \forall \;x \in [c,b)
$$ 
We are not allowed to use Riemann Integrals. However since $$f'(x) \le g'(x)$$ 
We can state that g - f is increasing. This is where I'm lost.

You have that $g' - f' \ge 0$ on $(c,b)$ so that $g-f$ is nondecreasing. Thus $x \in (c,b)$ will satisfy $$g(x) - f(x) \ge g(c) - f(c) = 0$$ so that $f(x) \le g(x)$.
