Why can we only talk about derivatives on an open interval? For instance, in my calculus class, all theorems are in the following form:
For example, Rolle's theorem:
If $f(x)$ is continuous on $[a,b]$, differentiable on $(a,b)$ ... (etc)
My question is, when presented with a closed interval, why must we talk about its derivatives on an open interval? Why doesn't Rolle's theorem say `differentiable on $[a,b]$' instead of $(a,b)$
 A: As @PeterFrank says in the comments, one can talk about so-called one-sided differentiability in the end points, e.g.
$$
f'(a)=\lim_{h \downarrow 0} \frac{f(a+h)-f(a)}{h}. 
$$
But the point is that if you only assume that $f$ has to be differentiable on $(a,b)$, the theorem is better/stronger. 
For example, we can apply Rolle to $x\mapsto \sqrt{x}$ on $[0,1]$, although this map is not one-sidedly differentiable in $0$. 
If you would assume that $f$ has to be differentiable on $[0,1]$, you could not apply Rolle in this case.  
A: For this reason: (I talk about Mean Value Theorem here)
If we consider differentiability on the closed interval $\mathbf[a, b\mathbf]$, we are then losing the generality of our theorem; because we're asking for a more strict condition to be satisfied.
Meanwhile, considering differentiability on $\mathbf(a, b\mathbf)$ makes the theorem more general and stronger without affecting the validity of our theorem.
For example, take a look at the graph of the semicircle with radius $1$
$$
f(x) = \sqrt{1 - x^2}
$$
where $f(x)$ is continuous on its domain $\mathbf[-1,1\mathbf]$, but it happens to be differentiable only on $\mathbf(-1, 1\mathbf)$
And my point is that, with the stricter closed interval condition; this function is not qualified, even though there really is a point $\mathbf{c}$ in $\mathbf(a, b\mathbf)$ where the slope of the tangent there is equal to that of the secant cutting through the endpoints $\mathbf{a}$ & $\mathbf{b}$.
And as far as I know, and as far as I am convinced, there doesn't exist a function where you'd need such a strict condition for the theorem to remain true.
So, we conclude that there is not any need for considering differentiability on a closed interval instead of an open one; since that'd cause a loss of generality for no acceptable reason.
And as always, Allah S.W.T. knows best. I hope this helps insha'Allah.
