Let $f: A \subseteq \mathbb{R} \to \mathbb{R}$ and $a\in A.$ We say that, when $$\lim_{x\to a} \frac{f(x)-f(a)}{x-a}$$ exists, then the function $f$ differentiable at $a \in A$. So, does the point $a \in A$ is a limit point of $A$ or is interior point of $A$?
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When you are in $\Bbb R$ you can take $a$ to be any point in $A$ (notice $f(a)$ appears in the definition) – William M. Apr 18 '17 at 17:19
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2The definition just requires $a \in A$. That's it. – A. B. Marnie Apr 18 '17 at 17:21
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@WillM. let $f(x) =x, D_f =[1,2]$ function$ f $ differentiable at $1$or $2$? – Almot1960 Apr 18 '17 at 17:27
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Are you sure that there are no assumptions on the domain $A$? – Siminore Apr 18 '17 at 17:49
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Actually you need that $a$ is in the closure of $A\setminus{a}$. Otherwise the limit is not well defined. – user251257 Apr 18 '17 at 18:19
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If the point $a $ is not in the interior of $A $, we speak about differentiability on the left $...,a] $ or on the right $[a,.. $.
hamam_Abdallah
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let $f(x)=x , D_f =\mathbb{R} - {\frac{1}{n} : n\in \mathbb{Z}} $ function $f$ differentiable $x=0$ – Almot1960 Apr 18 '17 at 17:34
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In a typical calculus textbook you will note that many authors would put it as a requirement that $A$ is an open set (that is, every point of $A$ is an interior point of $A$). This is because $a$ being interior to $A$ allows us to take both left and right limits at $a$. So you may take it as something out of which an author tries not to confuse the reader and at the same time probably not to get ahead of himself, which is just like an elementary-school teacher usually "lets" his students think that division only applies to positive integers.
But there is no further "technical" reason to stop at requiring $A$ open. If $A := [1, 2[$, a half-open interval, for instance, then we still can define the derivative of $f$ at $1$ by taking merely the right limit (there is no left limit to take in this case) of the difference quotient about $f$ at $1$.
Yes
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