Strengthening a result on the growth of continuously differentiable functions. 
Given two continuously differentiable functions $f,\
 g:\mathbb{R}\rightarrow \mathbb{R}$ such that $f(x_0) = g(x_0)$ and
   $f'(x_0) < g'(x_0)$, there exists $\epsilon > 0$ such that $f(x) <
 g(x)$ for $x \in (x_0, x_0 + \epsilon)$.

The above result is not too difficult to prove, but I was wondering if the condition that the functions be continuously differentiable is absolutely necessary. I have not taken much analysis, but I'm wondering if the fact that the derivative exists at $x_0$ would be enough to prove this result without needing the derivative to be continuous. The reason I ask this is because it seems that the existence of a derivative function already implies it must satisfy some quite stringent requirements (i.e. Darboux's Theorem), perhaps these are enough?
 A: You are right, it would be enough to assume that $f$ and $g$ are differentiable at $x_0$.
By considering a translate of the difference $g - f$ it suffices to prove that if $h$ is differentiable at $0$ and $h(0) = 0$ and $h'(0) > 0$, then there is $\delta > 0$ with the property that $h(x) > 0$ for all $x \in (0, \delta)$.  
But this is clear: put the positive number $\epsilon = h'(0)/2$ into the "$\epsilon-\delta$" definition of the statement $\lim_{x \to 0} \frac{h(x) - h(0)}{x - 0} = h'(0)$ to learn that there is $\delta > 0$ with the property that for all $x \in (0, \delta)$ one has
$$
\left|\frac{h(x)}{x} - h'(0)\right| < \frac{h'(0)}{2}.
$$
This implies, in particular, that for all $x \in (0, \delta)$ one has $-\frac{h'(0)}{2} < \frac{h(x)}{x} - h'(0)$ and hence that $h(x) > \frac{x}{2} h'(0) > 0$ whenever $x$ is in $(0, \delta)$.
A: If $f$ and $g$ are differentiable at $x_0$, there exist two functions $\omega_f$ and $\omega_g$ such that $\omega_f$ and $\omega_g$ are continuous at $x_0$ and $$f(x)=f(x_0)+\omega_f(x)(x-x_0), \quad g(x)=g(x_0)+\omega_g (x)(x-x_0)$$ for all $x$ (in some neighborhood of $x_0$). From the assumptions you get $\omega_f(x_0) < \omega_g(x_0)$ and there fore $f(x)-g(x)=\left(\omega_f(x)-\omega_g(x)\right)(x-x_0)$. You can immediately conclude that $f(x)<g(x)$ in a right neighborhood of $x_0$.
