Show that $f '(x_0) =g'(x_0)$. Assume that $f$ and $g$ are differentiable on interval $(a,b)$ and $f(x) \le g(x)$ for all $x \in (a,b)$.
There exists a point $x_0\in (a,b)$ such that $f(x_0) =g(x_0)$.
Show that $f '(x_0) =g'(x_0)$.
I am guessing we create a function $h(x) = f(x)-g(x)$ and try to come up with the conclusion using the definition  of differentiability at a point. I am not sure how. 
Some useful facts: A function $f:(a,b)\to \mathbb R$ is differentiable at a point $x_0$ in $(a,b)$ if
$$\lim_{h\to 0}\frac{f(x_0 + h)-f(x_0)}{h}=\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}.$$

Let $h(x)=g(x)-f(x)$. Then $h$ is differentiable on $(a,b)$, with $h(x)\geq 0$ for all $x$ in $(a,b)$, and in particular, $h(x_0)=0$ for some $x_0$ in $(a,b)$. We wish to show that $h'(x_0)=0$. Suppose not.
If $h'(x_0)=m<0$, then $$m=\lim_{x\to x_0^+}\frac{h(x)-h(x_0)}{x-x_0}=\lim_{x\to x_0^+}\frac{h(x)}{x-x_0},$$ so since $x-x_0$ is always positive as $x$ approaches $x_0$ from the right, we'll eventually need for $h(x)$ to be negative for all $x>x_0$ sufficiently close to $x_0$. This is impossible, though, since we know that $h(x)\geq 0$ for $x_0<x<b$. Similarly, if $h'(x_0)>0,$ we can approach from the left and derive a contradiction. Hence, we must have $h'(x_0)=0$, as desired.
Put another way, for $x$ sufficiently close to $x_0$, we can always approximate $h(x)$ by $$h(x_0)+h'(x_0)\cdot(x-x_0)=h'(x_0)\cdot(x-x_0)$$ as closely as we like. If we had $h'(x_0)\neq 0$, then there would be some $x$ near $x_0$ in the interval $(a,b)$ such that $h(x)<0$, which we assumed not to be the case.
