The symmetric derivative is always equal to the regular derivative when it exists, and still isn't defined for jump discontinuities. From what I can tell the only differences are that a symmetric derivative will give the 'expected slope' for removable discontinuities, and the average slope at cusps. These seem like extremely reasonable quantities to work with (especially the former), so I'm wondering why the 'typical' derivative isn't taken to be this one. What advantage is there to taking $\lim\limits_{h\to0}\frac{f(x+h)-f(x)} h$ as the main quantity of interest instead? Why would we want to use the one that's defined less often?
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J. M. ain't a mathematician
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asked Jul 15, 2012 at 1:37
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10$\begingroup$ Mean Value Theorem: The basis of almost all theory of the derivative. Oops! doesn't work for symmetric derivative! $\endgroup$– GEdgarJul 15, 2012 at 2:04
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2$\begingroup$ @GEdgar oh god, my engineering education is showing. Embarrassing. We barely looked at the MVT. I'm trading in my Stewart for Spivak, so hopefully the real depth of this will be clearer soon. $\endgroup$– Robert MastragostinoJul 15, 2012 at 2:16
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1$\begingroup$ If you do want to explore symmetric derivatives further, you might want to check out Thomson's Symmetric properties of real functions. $\endgroup$– Jonas MeyerJul 15, 2012 at 5:21
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$\begingroup$ A related question. $\endgroup$– J. M. ain't a mathematicianJul 15, 2012 at 11:05
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2$\begingroup$ @J.M.: an interesting difference in emphasis. THIS question asks why not use the s.d. all the time, the other question asks whether s.d. is ever useful. $\endgroup$– GEdgarJul 15, 2012 at 13:01
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2 Answers
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The symmetric derivative being defined at more places isn't a good thing.
In my mind, the main point of differentiation is to locally approximate a function by a linear function. That is, the heart of saying that the derivative $f'(a)$ exists at a point $a$ is the statement that
$$f(x) = f(a) + f'(a) (x - a) + o(|x - a|)$$
as $x \to a$, and if I were the King of Calculus this is how the derivative would actually be defined. (Among other things, this definition generalizes smoothly to higher dimensions.) Removable discontinuities are a non-issue as they should just be removed, but at a cusp we do not have this property for any possible value of $f'(a)$, so we shouldn't be talking about derivatives at such points at all. (We can talk about left or right derivatives, but this is something different.)
The symmetric derivative at $a$ is not a natural definition. It has the utterly strange property that any weirdness in a neighborhood of $a$ is ignored if it happens to be canceled by equivalent weirdness after reflecting around $a$. Let me give an example. Consider the function $f(x) = 1_{\mathbb{Q}}(x)$ which is equal to $1$ if $x$ is rational and $0$ otherwise. The symmetric derivative of $f$ at any rational point exists and is equal to $0$! Is there any reasonable sense in which $f$ is differentiable at a rational point?
The ordinary derivative, on the other hand, is sensitive to weirdness around $a$ because it compares all of that weirdness to $f(a)$.
answered Jul 15, 2012 at 1:44
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$\begingroup$ @Qiochu Yuan This is indeed the correct definition. $\endgroup$ Jul 15, 2012 at 1:49
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14$\begingroup$ the "King of Calculus"... I'll have to remember that phrase for the next time I teach such a course. $\endgroup$– KCdJul 15, 2012 at 5:33
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6$\begingroup$ Not worth a full answer but: all this hints at the fact that a proper two-sided definition of the derivative of $f$ at $x$ should be the limit of $(f(x+r)-f(x-s))/(r+s)$ when $r\gt0$ and $s\gt0$ both go to $0$. Fortunately (or unfortunately, maybe), this seems to be strictly equivalent to the usual definition (provided removable discontinuities are... well, removed). $\endgroup$– DidJul 15, 2012 at 13:01
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1$\begingroup$ @Did Not sure what you want. :-) Say you are teaching Calculus I to people who don't want to be there but must and that you want to "talk to them". At least in the case of a gap, the lack of continuity does not involve looking beyond $a$. Then they do see that any kind of "smoothness" requires looking at a neighborhood of $a$ but, as I tried to explain, not why they should still look at $a$ which, in their eyes, makes the definition of the ordinary derivative "weird". Hence my interest in the asymmetric derivative you presented and my question as to whether you had presented it to students. $\endgroup$ Jan 6, 2018 at 5:32
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Following my comment on the Mean Value Theorem. Since MVT fails, anything we prove from MVT is likely to fail as well. For example:
Find the minimum of the (symetrically) differentiable function
$f(x) = x+2|x|$ on the interval $[-1,1]$.
Usual solution: find where the derivative is zero. Answer: nowhere!
Since $f'(x) = -1$ on $[-1,0)$, $f'(0)=1$,
and $f'(x)=3$ on $(0,1]$.
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1$\begingroup$ An interesting note is that there is a generalization of the MVT for the symmetric difference quotient, which asserts that difference quotients are bounded by the symmetric difference quotient but not necessarily equal to the symmetric difference quotient as with the usual MVT. $\endgroup$ Jun 17, 2020 at 22:09
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1$\begingroup$ I saw today that there is a whole book on this. MVT: A Most Valuable Theorem by C. Smorynski $\endgroup$– GEdgarJan 23, 2021 at 12:39