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 $(x_n^+)$ be a sequence converging to $x_0$ from above and $(x_n^-)$ be a sequence converging to $x_0$ from below.
Then we have $$\frac{f(x_n^+)-f(x_0)}{x_n^+-x_0}\leq\frac{g(x_n^+)-g(x_0)}{x_n^+-x_0}$$
Since $f$ and $g$ are differentiable the sequences converge to $f'(x_0)$ and $g'(x_0)$, respectively. We conclude $f'(x_0)\leq g'(x_0)$.
Similarly, we can use $(x_n^-)$ to show $f'(x_0)\geq g'(x_0)$.
