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My roommates and I have an argument you guys can help to settle (peace is at stake, don't let us down!) In undergrad calculus courses, one usually explains what it means for a function to be differentiable at a point x, and then differentiable in a domain. Then the focus is entirely on this latter notion. My question is:

Has the notion of differentiability at a point any interest?

That is, I'm looking for a theorem which is valid for a function regular at some point, but which needs significantly less regularity in a neighborhood of this point, or a good reason for which such a theorem doesn't exist.

Of course, this question is very flexible, and any insight is welcome.

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up vote 5 down vote accepted

Differentiability at a point is useful when you want to formulate a theorem like the fundamental theorem of calculus in the Lebesgue setting. For instance a continuous nondecreasing $f:[a,b]\to\mathbb{R}$ that satisfies some nice properties (maps sets of measure 0 to sets of measure 0 and is absolutely continuous) will be differentiable almost everywhere and satisfy the familiar $f(b) - f(a) = \int_a^b f'(x)d\mu(x)$ where $\mu$ is the Lebesgue measure.

Here of course $f'(x)$ really means any function which is equal to $f'(x)$ whenever $f'$ exists.

Often functions will be differentiable almost everywhere or, you'll only need to consider functions differentiable almost everywhere, and in this case differentiability at a point is inherent in such a definition.

Of course, sets of measure $0$ are small in some sense but such sets can still be dense so you can still easily have functions whose points of nondifferentiability are lurking in any interval about a point, although to construct such weird functions isn't trivial.

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Though somewhat trivial, there is the Local Extremum Theorem: If $f$ is differentiable at $c$ and has a local extremum at $c$, then $f'(c) = 0$.

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There are functions that are only differentiable at one point, here is an example:

Consider the function $d:\mathbb{R}\to\mathbb{R}$ defined by $d(x)=0$ if $x$ is rational and $d(x)=1$ if $x$ is irrational. This function is not differentiable anywhere, since it is not continuous, however, it is surely bounded. Now look at $f(x)=x^2\cdot d(x)$.

Note that for a rational $x$ we have $f(x)=0$ and while for irrational $x$ we have $f(x)=x^2$.

Next since there are rational numbers arbitrary close to any irrational $x$ our function is not continuous at irrationals. The same argument holds for rational $x$'s whenever $x\ne0$.

At $x=0$ we have, for $h$ rational and $\ne0$, $$\frac{f(x+h)-f(x)}{h}= \frac{f(0+h)-f(0)}{h}=\frac{f(h)-f(0)}{h}=\frac{0}{h}=0$$ while for irrational $h$ we get $$\frac{f(x+h)-f(x)}{h} = \frac{f(h)-f(0)}{h}=\frac{h^2}{h}=h$$ which tends to $0$ as $h\to0$, in particular $f$ is differentiable at $0$ and $f'(0)=0$.

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This evokes very old memories. It strikes me that even the Dirac derivative may not exist. It is easy to see that the Dirac on each rational number would be -1, but then it needs to be +1 infinitely close to rational numbers but not at rational numbers. Hmmm - no Dirac for this function! Or is it +1 for every non rational number? I would not be able to begin proving or disproving this. – asoundmove Mar 1 '11 at 21:09

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