TITLE: Differentiable only on interior points
QUESTION [5 upvotes]: I have been pondering this question for quite a while and I would really like to get some closure on it. Whenever I have read about differentiation of functions defined on an interval, they almost always require differentiability only being valid for inner points i.e. on the interval (a,b). 
Why? Can somebody thoroughly and sensibly delineate why this is the case? 
One explanation I have come up with is say we attempted to define the derivative at b. In that case the left-hand and right-hand limit (the difference quotient which defines the derivative) must coincide. Points to the left of b certainly belong to the domain, however points the right of b do not belong to the domain. Therefore we can not really say anything about them. Since we cannot examine any right-hand limit, the notion of a derivative is not sensible at the boundary point.
However I am not sure the explanation entirely suffices; I am curious as to what other reasons there are to impose existence of derivatives only on interior points.

REPLY [3 votes]: It's just so that differentiability at a point is conserved in case you decide to expand the domain of your function. 
In general, for a function $f:A\rightarrow{}B$, where the domain $A$ is a subset of some topological space $X$ and $B$ is a subset of some other topological space $Y$, a limit can exist for any limit point of the domain $A$. The definition being that some $L\in{}Y$ is the limit as $x$ approaches $x_0$ of $f(x)$, i.e.
$$ \lim_{x\rightarrow{}x_0}f(x)=L, $$
if and only if for all neighborhoods $V$ of $L$ there exists a neighborhood $U$ of $x_0$ such that $f(U\cap{}A-\{x_0\})\subseteq{}V\cap{}B$. Notice that the limit would converge everywhere if $x_0$ was not a limit point of the domain $A$ (and also, irrelevant to this answer but also interesting, $L$ has to be in the closure of $B$).
So again, limits for a function only make sense for limit points of the function's domain. Then a function
$$F(x)=\frac{f(x)-f(x_0)}{x-x_0},$$
defined in terms of the function $f$ above, would have a limit as $x$ approaches $x_0$ if $x_0$ was a limit point of $A-\{x_0\}$, the domain of $F$. This means that, as far as limits are concerned, the derivative could exist for any limit point of $A-\{x_0\}$.
But what changes when we restrict the limit points of $A-\{x_0\}$ to the points in the interior of $A$? Formally, the definition of
$$ \lim_{x\rightarrow{}x_0}F(x)=M $$
is that for all neighborhoods $V$ of $M$ there exists a neighborhood $U$ of $x_0$ such that $F(U\cap{}(A-\{x_0\})-\{x_0\})=F(U\cap{}A-\{x_0\})\subseteq{}V\cap{}Im(F)$. 
But by making $x_0$ a point in the interior of $A$ we are requiring that there exists a neighborhood $W$ of $x_0$ such that $W\subseteq{}A$. And since $W\cap{}U$ is also a neighborhood of $x_0$ and $(W\cap{}U)\cap{}A=W\cap{}U$, expanding $A$, the domain of $f$, does not change whether the limit exists or not, which makes differentiability at a point immune to expansions of the domain of the original function.