Symmetric derivative

The absolute value function of the absolute value function. Note the sharp turn at , leading to non-differentiability of the curve at . The function hence possesses no ordinary derivative at . The symmetric derivative, however, exists for the function at . For the absolute value function f(x) = |x|, using the notation f_s(x) for the symmetric derivative, we have at x = 0 that \begin{align} 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{2h} \\ &= \lim_{h \to 0}\frac{2h} \\ &= \lim_{h \to 0}\frac{0}{2h} = 0. \\ \end{align} Hence the symmetric derivative of the absolute value function exists at 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 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 f(x) = 1/x^2, at x = 0 we have \begin{align} 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{align} Again, for this function the symmetric derivative exists at x = 0, while its ordinary derivative does not exist at 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: 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 x \in \Q, but is not symmetrically differentiable at any x \in \R \setminus \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 has the image {{math|{−1, 0, 1}}}, but secants for f can have a wider range of slopes; for instance, on the interval , the mean-value theorem would mandate that there exist a point where the (symmetric) derivative takes the value {{nowrap|\frac{2 - (-1)} = \frac{1}{3}.}} The quasi-mean-value theorem for a symmetrically differentiable function states that if is continuous on the closed interval and symmetrically differentiable on the open interval , then there exist , in such that f_s(x) \leq \frac{f(b) - f(a)}{b - a} \leq f_s(y). As an application, the quasi-mean-value theorem for on an interval containing 0 predicts that the slope of any secant of is between −1 and 1. If the symmetric derivative of has the Darboux property, then the (form of the) regular mean-value theorem (of Lagrange) holds, i.e. there exists in such that 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. == 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 The second symmetric derivative may exist, however, even when the (ordinary) second derivative does not. As example, consider the sign function \sgn(x), which is defined by \sgn(x) = \begin{cases} -1 & \text{if } x 0. \end{cases} The sign function is not continuous at zero, and therefore the second derivative for x = 0 does not exist. But the second symmetric derivative exists for x = 0: \lim_{h \to 0} \frac{\sgn(0 + h) - 2\sgn(0) + \sgn(0 - h)}{h^2} = \lim_{h \to 0} \frac{\sgn(h) - 2\cdot 0 + (-\sgn(h))}{h^2} = \lim_{h \to 0} \frac{0}{h^2} = 0. == See also ==