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Consider the following function (I think it has a name, but I don't remember it): $$ f(x) = \cases{-1 & $x < 0$ \\ 0 & $x = 0$ \\ 1 & $x > 0$} $$

$f'(x)$ is zero everywhere except at $x=0$, where $f$ is not continuous. But suppose we ignore the right half of the real line and define $f(0)$ to be $-1$. Then $f$ has a left derivative at $x=0$, and it is zero. We can do the same thing from the right, so in a way it could make a little bit of sense to say that $f'(0) =0$.

Of course, I understand that going by the definition $f$ isn't differentiable at $x=0$. But one could imagine an alternative definition of derivative for discontinous functions, in which one calculates lateral derivatives by redefining the function to be continuous, and then we see if the lateral derivatives match. This doesn't always work; for example it's hard to meaningfully assign a derivative to $x \mapsto |x|$ at $x=0$.

Are there other functions with this property? Does it have a name?

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    Your $f$ is generally called $\operatorname{sgn}$ or the sign function.2012-09-24
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    You might be interested in [Dini derivatives](http://en.wikipedia.org/wiki/Dini_derivative) and [removable singularities](http://en.wikipedia.org/wiki/Removable_singularity), which is what your derivative function has.2012-09-24
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    Although this does not apply to your particular function, there is a notion of differentiability that does not imply continuity: [approximate differentiability](http://www.encyclopediaofmath.org/index.php/Approximate_derivative). (It does imply approximate continuity, though, and your function is not approximately continuous at zero.)2012-09-24
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    @RobinSaunders Four years later, I don't really recall what I was trying to get at with the comment. Going to remove it.2016-08-15
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    Yes, sorry, I only noticed the date after responding! This question came up as related to a recent one, and I didn't think to check the date.2016-08-15

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