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!

We have: $f(x_0) \leq g(x_o) \leq h(x_0)$, and $f(x_0) = h(x_0)$, this implies $f(x_0) = g(x_0) = h(x_0)$, you can continue. From this we have: $f'(x_0^{+}) \leq g'(x_0^{+}) \leq h'(x_0^{+})$, and also $f'(x_0^{-}) \geq g'(x_0^{-}) \geq h'(x_0^{-})$. But $f'(x_0^{+}) = f'(x_0^{-}) = f'(x_0) = h'(x_0) = h'(x_0^{+}) = h'(x_0^{-}) \to g'(x_0^{+}) = g'(x_0^{-}) = g'(x_0) = f'(x_0) = h'(x_0)$. 
