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Assume $f$ has a domain of $[a,b]$. Is it possible that $f$ is differentiable on the closed interval $[a,b]$, or must the maximal domain for $f'$ be $(a,b)$?

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  • $\begingroup$ It is common to say that a function $f : [a, b] \to\mathbb R$ is differetiable at $a$ (or $b$) if $f$ can be extended to a function $g : (c, d)\to \mathbb R$ for some $(c, d)$ containing $[a, b]$, and $g$ is differentiable at $a$ (or $b$) $\endgroup$ – user99914 May 23 '15 at 9:50
  • $\begingroup$ Depends exactly on what you mean. At first differentiability is only defined for interior points. However, one is then typically introduced to the concept of lateral derivative and with that the definition of differentiability is extended to the end points. $\endgroup$ – Git Gud May 23 '15 at 9:51
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It depends on the definition of differentiability you have. Some textbooks only define it for interior points. But there is also a more general definition (see this answer for references):

A function $f: A \rightarrow \mathbb R$ is differentiable for an accumulation point $a \in A$ of $A$ with derivative $f^\prime(a)$, iff for each $x_n \rightarrow a$ with $x_n\in A\setminus\{a\}$ you have $f^\prime(a) = \lim_{n\rightarrow\infty} \frac{f(x_n)-f(a)}{x_n-a}$.

So the answer is yes: You can define the derivative in a way, such that $f'$ is also defined for the end points of a closed interval. Note that for some theorem like the mean value theorem you only need continuity at the end points of the interval.

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    $\begingroup$ @StephenKulla: I'm very interested in the text you've seen this definition in. I've never seen it before. Do you remember where you've seen it? $\endgroup$ – SOULed_Outt Aug 9 '17 at 5:35
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    $\begingroup$ @SOULed_Outt See math.stackexchange.com/a/162172/32951 $\endgroup$ – Stephan Kulla Aug 10 '17 at 7:21

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