What is the difference between the terms "undefined" and "does not exist", especially in the context of differential calculus?

Most calculus materials state, for example, that $\frac{d}{dx}{|x|}$ does not exist at $x = 0$. Why don't we say that the derivative is undefined at $x = 0$?


2 Answers 2


In the particular example you gave: The derivative is defined as $\lim_{h\to 0} \frac{f(x+h) - f(x)}{h}$ and, as it is with limits, this limit may or may not exist. In the case $f(x) = |x|$ and $x=0$ the limit just does not exist and hence, this is the right wording. On the other hand there are possible definitions of a derivative of $f(x) = |x|$ at zero (e.g. using convex analysis one may define it to be the whole interval $[-1,1]$) and hence, it seems appropriate to say that the derivative is undefined.

In general "does not exists" and "is undefined" are very different things at a practical level. The former says that there is a definition for something which does not lead to a mathematical object in a specific case. The latter says that there is just no definition for a specific case. Of course, one can interchange both formulation some times (as in you example, at least in my opinion).

  • $\begingroup$ I've changed the example to $\frac{d}{dx}(\frac{1}{x})$. $\endgroup$
    – bwDraco
    Feb 6, 2013 at 18:48
  • 2
    $\begingroup$ Why?$\mbox{}\mbox{}$ $\endgroup$
    – Dirk
    Feb 6, 2013 at 19:17
  • $\begingroup$ Alright, I've rolled it back. $\endgroup$
    – bwDraco
    Feb 6, 2013 at 20:31
  • $\begingroup$ Indeed, editing the question such that right answers get wrong is not always a good idea. If you ask for the function $f(x) = 1/x$ and whether the derivative exists or is undefined at $x=0$, then the answer would be (in my understanding) that the derivative is undefined (because its definition involves the term $f(0)$ which is undefined). $\endgroup$
    – Dirk
    Feb 7, 2013 at 8:16
  • $\begingroup$ By your saying that "there is a definition for something which does not lead to a mathematical object in a specific case", do you mean that the result is defined, but is something like $\infty$? $\endgroup$
    – bwDraco
    Feb 8, 2013 at 21:20

There is a little subtlety not addressed so far:

Given a function $f:\ A\to\Bbb R$ on an open set $A\subset\Bbb R$ and a point $x\in A$, you can ask whether $f$ is differentiable at $x$. The function $f$ is differentiable at $x$ (or: has a derivative at $x$) if the limit $$\lim_{h\to0}{f(x+h)-f(x)\over h}\tag{1}$$ exists and is finite. This limit is called the derivative of $f$ at $x$ and is denoted by $f'(x)$.

The set $A'$ of all $x\in A$ where the limit $(1)$ exists, is the domain of a new function $f'$ associated to $f$. This new function is called the derivative of $f$.

When $x\in A'$ then we say that $f'$ is defined at $x$.

  • 5
    $\begingroup$ Continuing with your example, say $x\notin A'$. Then would we say that $f'(x)$ does not exist or is undefined or both? $\endgroup$
    – user547493
    Oct 24, 2018 at 4:48
  • $\begingroup$ (bump) I would also appreciate an answer to the above comment! $\endgroup$
    – S.C.
    Sep 8, 2021 at 8:56

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .