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This is a quick question, and perhaps a quick answer.

What keeps us from defining the derivative, formally, as:

$$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x-h)}{2h}$$?

Graphically this would mean that we could calculate the tangent to a graph by approaching the desired point from both sides at equal rates, rather than from one side.

As a side effect, functions could be differentiated at removable discontinuities, as well as "sharp ends". For example, the derivative of $|x|$ at $0$ would unambiguously be $0$. Or, the derivative of $\frac{x^2-3x+2}{x-1}$ at $1$ would be defined unambiguously as $-1$.

I didn't ever think about this until today. And since I thought about this, I can't let this go. And it bothers me: why do we define derivative as a tangent approaching from one side, while the point of interest is fixed? It seems more natural to approach from both sides at once, now that I played with this idea in my head.

So, why isn't the derivative defined like this? As a bonus question, if this definition was widely used, what would be the effects on other aspects of calculus? Or would there be none, besides the absorption of local discontinuities?

KKZiomek
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    This seems to have been asked here: https://math.stackexchange.com/q/170948 The main answer basically states your observations as cons (whereas you present them as pros). – Maximal Ideal Jun 29 '19 at 04:46
  • I tried to search for a similar question, but apparently I didn't search enough. Thank you for the links, I'll look into them @SpiralRain – KKZiomek Jun 29 '19 at 04:47
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    Quick answer: $\sqrt {|x|}$ is differentiable at $0$ according to this definition but not according to the usual definition. – Kavi Rama Murthy Jun 29 '19 at 04:50
  • @KKZiomek Not sure if this should be an answer, but one important thing in real analysis is that we can think of a sort-of "smoothness hierarchy." A $C^{0}$-function is continuous, then a $C^{1}$-function is differentiable with a continuous derivative, then a $C^{2}$-function is twice-differentiable with a continuous second-deriv, and so on. With the usual definition, we have $C^{0}\supseteq C^{1}\supseteq C^{2}\supseteq\cdots$, but in the symmetric deriv, there is no relationship (because, like you said, a discontinuous function may still be differentiable in your model). – Maximal Ideal Jun 29 '19 at 05:14
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    Since this is about definitions, it ultimately comes down to convention and utility. The strongest argument against "symmetric derivatives" I'd say is the fact that the mean value theorem wouldn't work with it. – Maximal Ideal Jun 29 '19 at 05:16
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    "the derivative of |x| at 0 would unambiguously be 0" Why would that be a good thing? If the two limits are unequal then the function has two different and inconsistent behaviors. And if that's the case, we want to know about it. Why don't want to average the difference and ignore it. If the derivative of $|x|$ is unambiguously $0$ then the slope of the tangent line is unambiguously $0$. But if there is no one tangent line and no determinable slope, then we want the derivative to be ambiguous. Truth is more important than ease and if the truth is ambiguous then we don't want clarity. – fleablood Jun 29 '19 at 05:51
  • Another difference is that $f$ might be discontinuous at $x$. E.g. if $f(0)=1$ but $f(h)=0$ when $h\ne 0$ then according to the alternate def'n we would have $f'(0)=0. $ So the alternate applies to a broader class of functions but yields less info about $f$. It is a valid concept but should be given a different name than "derivative". – DanielWainfleet Jun 29 '19 at 06:42

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