Why use the derivative and not the symmetric derivative?

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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Mean Value Theorem: The basis of almost all theory of the derivative. Oops! doesn't work for symmetric derivative! – GEdgar Jul 15 '12 at 2:04
@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. – Robert Mastragostino Jul 15 '12 at 2:16
If you do want to explore symmetric derivatives further, you might want to check out Thomson's Symmetric properties of real functions. – Jonas Meyer Jul 15 '12 at 5:21
A related question. – J. M. Jul 15 '12 at 11:05
@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. – GEdgar Jul 15 '12 at 13:01

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 $f(a)$.

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@Qiochu Yuan This is indeed the correct definition. – ncmathsadist Jul 15 '12 at 1:49
@QiochuYuan Linearity makes it exceedingly obvious in retrospect. Always a sign of a fantastic answer. – Robert Mastragostino Jul 15 '12 at 2:10
the "King of Calculus"... I'll have to remember that phrase for the next time I teach such a course. – KCd Jul 15 '12 at 5:33
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). – Did Jul 15 '12 at 13:01
What happens if we remove the restriction that r > 0 and s > 0 (and just ask that they not sum to zero)? What extra properties must a function differentiable at x satisfy for the two-variable difference quotient limit (in r and s) to exist? [E.g., I believe it suffices for the function to be continuously differentiable. How tight a restriction is this?] – Sridhar Ramesh Aug 30 '12 at 4:33

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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