If $f$ and $h$ are differentiable in $a$ and $h'(a)=f'(a)$. Then $g$ is differentiable and $g'(a)=f'(a).$ Let $f,g,h:X\to\mathbb{R}$ such that $f(x)\leq g(x)\leq h(x)$ for all $x\in X$. If $f$ and $h$ are differentiable in $a$ and $h(a)=f(a)$. Then $g$ is differentiable and $g'(a)=f'(a).$
How can I can prove that $g$ is differentiable in $a$.? Thanks
 A: This is not even close to being true -- for example a counterexample could be
$$ f(x) = -4 \qquad g(x)=\arctan(x) \qquad h(x)=4 $$
(where $g'(a)$ exists for all $a$ but never equals $f'(a)$) or
$$ f(x) = 0 \qquad g(x)=\begin{cases}1 & x\in\mathbb Q \\ 0 & x\notin\mathbb Q \end{cases} \qquad h(x)=1 $$
(where $g'(a)$ doesn't exist anywhere).

For the edited question where the assumption is $f(a)=h(a)$ rather than $f'(a)=h'(a)$:
Without loss of generality we can assume that $f(x)$ is constant, $f(x)=c$ everywhere -- otherwise just subtract $f(x)$ from all three functions.
Then since $f(a)=h(a)=c$ we must also have $g(a)=c0$.
Furthermore neither $g$ nor $h$ can be less than the constant value of $f$, so
$$ \frac{h(x)-h(a)}{x-a} $$
is either zero or has the same sign as $x-a$ -- so if it has a limit for $x\to a$ (which it must becaue $h$ is differentiable), this limit must be $0$.
Now since $g$ is between $f$ and $h$, $$\frac{g(x)-g(a)}{x-a}$$
is always between $0=\frac{f(x)-c)}{x-a}$ and $\frac{h(x)-c}{x-a}$ which both converge to $0$, so by the squeeze theorem it too converges to $0$.
A: Without loss of generality, let $a=0$ and $f(0) = g(0) = h(0) = 0$
Then
$$\frac{f(\epsilon)}{\epsilon} \leq \frac{g(\epsilon)}{\epsilon} \leq \frac{h(\epsilon)}{\epsilon} $$
Now, take the limit when $\epsilon \to 0$, and as $f$ and $h$ are differentiable at $0$, you have that $\lim_{\epsilon \to 0} \frac{g(\epsilon)}{\epsilon}$ exists and is equal to $f'(0)$. So we have the result
