Prove that g is differentiable Question: Suppose f, g, and h are defined on (a,b) and $a < x_0 < b$. Assume f and h are differentiable at $x_0$, $f(x_0) = h(x_0)$, and $f(x) \le g(x) \le h(x)$ for all x in (a,b). Prove that g is differentiable at $x_0$ and $f'(x_0) = g'(x_0) = h'(x_0)$.
I know the case where $g(x) = f(x)$ or $h(x)$ makes this problem very simple so we don't need to look at that one.
Other than that, all I really know is, based on the definition, that 
$lim_x \to x_0$$(f(x)-f(x_0))/(x-x_0)$ and the same for $g(x)$ but I'm not really sure where to go from there.
Thanks for the help!

First prove that $f'(x_0) = h'(x_0)$.  This is easy (and has been stated by another answer).  We have the following inequality: $f(x) \leq h(x)$ which leads us to $f(x) - h(x) \leq 0$.  The maximum value is clearly $0$ and must occur when $f(x_0) = h(x_0)$ and therefore $f'(x_0) - h'(x_0) = 0$ (it's possible that there are also local minimums of $f(x) - h(x)$, but we know that $x_0$ is a local max by the original inequality and thus must satisfy the equation $f'(x_0) - h'(x_0) = 0$), therefore $f'(x_0) = h'(x_0)$ where $x_0$ is such that $f(x_0) = h(x_0)$.
So now we have that $f'(x_0) = h'(x_0) = L$.
Assuming that $g'(x_0)$ exists (I realize this needs to be proved), there is a simple proof by contradiction that shows that $g'(x_0) = f'(x_0) = h'(x_0) = L$.
Assume that $g'(x_0) = L + \Delta L$, such that $\Delta L > 0$.  In the linear approximation, this means that for small $\Delta x > 0$, we would have:
$$
g(x_0 + \Delta x) \approx g(x_0) + (L + \Delta L)\Delta x \\
f(x_0 + \Delta x) \approx f(x_0) + L\Delta x \\
h(x_0 + \Delta x) \approx h(x_0) + L\Delta x \\
$$
Since we are given that $g(x_0) = f(x_0) = h(x_0)$, we would find that $g(x_0 + \Delta x) \approx h(x_0 + \Delta x) + \Delta L \Delta x \geq h(x_0 + \Delta x)$ which is a contradiction to our given, which is that $g(x) \leq h(x)$, therefore it's not true that $\Delta L > 0$ and you can prove the same for $\Delta L < 0$ using $f(x)$, therefore it must be the case that $\Delta L = 0$ and thus $g'(x_0) = L = f'(x_0) = h'(x_0)$.
Granted, this is missing the component of actually proving that $g'(x_0)$ exists, but it may be helpful.  I think this sets up the $\epsilon$-$\delta$ proof that is necessary (but I might be wrong).
