# Symmetric derivative

In mathematics, the symmetric derivative is an operation generalizing the ordinary derivative.

It is defined as:[1][2] ${\displaystyle \lim _{h\to 0}{\frac {f(x+h)-f(x-h)}{2h}}.}$

The expression under the limit is sometimes called the symmetric difference quotient.[3][4] A function is said to be symmetrically differentiable at a point x if its symmetric derivative exists at that point.

If a function is differentiable (in the usual sense) at a point, then it is also symmetrically differentiable, but the converse is not true. A well-known counterexample is the absolute value function f(x) = |x|, which is not differentiable at x = 0, but is symmetrically differentiable here with symmetric derivative 0. For differentiable functions, the symmetric difference quotient does provide a better numerical approximation of the derivative than the usual difference quotient.[3]

The symmetric derivative at a given point equals the arithmetic mean of the left and right derivatives at that point, if the latter two both exist.[1][2]: 6

Neither Rolle's theorem nor the mean-value theorem hold for the symmetric derivative; some similar but weaker statements have been proved.

## Examples

### The absolute value function

For the absolute value function ${\displaystyle f(x)=|x|}$, using the notation ${\displaystyle f_{s}(x)}$ for the symmetric derivative, we have at ${\displaystyle x=0}$ that {\displaystyle {\begin{aligned}f_{s}(0)&=\lim _{h\to 0}{\frac {f(0+h)-f(0-h)}{2h}}=\lim _{h\to 0}{\frac {f(h)-f(-h)}{2h}}\\&=\lim _{h\to 0}{\frac {|h|-|{-h}|}{2h}}\\&=\lim _{h\to 0}{\frac {|h|-|h|}{2h}}\\&=\lim _{h\to 0}{\frac {0}{2h}}=0.\\\end{aligned}}}

Hence the symmetric derivative of the absolute value function exists at ${\displaystyle x=0}$ and is equal to zero, even though its ordinary derivative does not exist at that point (due to a "sharp" turn in the curve at ${\displaystyle x=0}$).

Note that in this example both the left and right derivatives at 0 exist, but they are unequal (one is −1, while the other is +1); their average is 0, as expected.

### The function x−2

For the function ${\displaystyle f(x)=1/x^{2}}$, at ${\displaystyle x=0}$ we have {\displaystyle {\begin{aligned}f_{s}(0)&=\lim _{h\to 0}{\frac {f(0+h)-f(0-h)}{2h}}=\lim _{h\to 0}{\frac {f(h)-f(-h)}{2h}}\\[1ex]&=\lim _{h\to 0}{\frac {1/h^{2}-1/(-h)^{2}}{2h}}=\lim _{h\to 0}{\frac {1/h^{2}-1/h^{2}}{2h}}=\lim _{h\to 0}{\frac {0}{2h}}=0.\end{aligned}}}

Again, for this function the symmetric derivative exists at ${\displaystyle x=0}$, while its ordinary derivative does not exist at ${\displaystyle x=0}$ due to discontinuity in the curve there. Furthermore, neither the left nor the right derivative is finite at 0, i.e. this is an essential discontinuity.

### The Dirichlet function

The Dirichlet function, defined as: ${\displaystyle f(x)={\begin{cases}1,&{\text{if }}x{\text{ is rational}}\\0,&{\text{if }}x{\text{ is irrational}}\end{cases}}}$ has a symmetric derivative at every ${\displaystyle x\in \mathbb {Q} }$, but is not symmetrically differentiable at any ${\displaystyle x\in \mathbb {R} \setminus \mathbb {Q} }$; i.e. the symmetric derivative exists at rational numbers but not at irrational numbers.

## Quasi-mean-value theorem

The symmetric derivative does not obey the usual mean-value theorem (of Lagrange). As a counterexample, the symmetric derivative of f(x) = |x| has the image {−1, 0, 1}, but secants for f can have a wider range of slopes; for instance, on the interval [−1, 2], the mean-value theorem would mandate that there exist a point where the (symmetric) derivative takes the value ${\displaystyle {\frac {|2|-|-1|}{2-(-1)}}={\frac {1}{3}}}$.[5]

A theorem somewhat analogous to Rolle's theorem but for the symmetric derivative was established in 1967 by C. E. Aull, who named it quasi-Rolle theorem. If f is continuous on the closed interval [a, b] and symmetrically differentiable on the open interval (a, b), and f(a) = f(b) = 0, then there exist two points x, y in (a, b) such that fs(x) ≥ 0, and fs(y) ≤ 0. A lemma also established by Aull as a stepping stone to this theorem states that if f is continuous on the closed interval [a, b] and symmetrically differentiable on the open interval (a, b), and additionally f(b) > f(a), then there exist a point z in (a, b) where the symmetric derivative is non-negative, or with the notation used above, fs(z) ≥ 0. Analogously, if f(b) < f(a), then there exists a point z in (a, b) where fs(z) ≤ 0.[5]

The quasi-mean-value theorem for a symmetrically differentiable function states that if f is continuous on the closed interval [a, b] and symmetrically differentiable on the open interval (a, b), then there exist x, y in (a, b) such that[5][2]: 7

${\displaystyle f_{s}(x)\leq {\frac {f(b)-f(a)}{b-a}}\leq f_{s}(y).}$

As an application, the quasi-mean-value theorem for f(x) = |x| on an interval containing 0 predicts that the slope of any secant of f is between −1 and 1.

If the symmetric derivative of f has the Darboux property, then the (form of the) regular mean-value theorem (of Lagrange) holds, i.e. there exists z in (a, b) such that[5] ${\displaystyle f_{s}(z)={\frac {f(b)-f(a)}{b-a}}.}$

As a consequence, if a function is continuous and its symmetric derivative is also continuous (thus has the Darboux property), then the function is differentiable in the usual sense.[5]

## Generalizations

The notion generalizes to higher-order symmetric derivatives and also to n-dimensional Euclidean spaces.

### The second symmetric derivative

The second symmetric derivative is defined as[6][2]: 1  ${\displaystyle \lim _{h\to 0}{\frac {f(x+h)-2f(x)+f(x-h)}{h^{2}}}.}$

If the (usual) second derivative exists, then the second symmetric derivative exists and is equal to it.[6] The second symmetric derivative may exist, however, even when the (ordinary) second derivative does not. As example, consider the sign function ${\displaystyle \operatorname {sgn}(x)}$, which is defined by ${\displaystyle \operatorname {sgn}(x)={\begin{cases}-1&{\text{if }}x<0,\\0&{\text{if }}x=0,\\1&{\text{if }}x>0.\end{cases}}}$

The sign function is not continuous at zero, and therefore the second derivative for ${\displaystyle x=0}$ does not exist. But the second symmetric derivative exists for ${\displaystyle x=0}$: ${\displaystyle \lim _{h\to 0}{\frac {\operatorname {sgn}(0+h)-2\operatorname {sgn}(0)+\operatorname {sgn}(0-h)}{h^{2}}}=\lim _{h\to 0}{\frac {\operatorname {sgn}(h)-2\cdot 0+(-\operatorname {sgn}(h))}{h^{2}}}=\lim _{h\to 0}{\frac {0}{h^{2}}}=0.}$