Symmetric derivative

Remarks

If $f$ is differentiable at $x$, then $\frac{f(x + h) - f(x - h)}{2h}$ is the average of two difference quotients with increments $h$ and $-h$. Both tend to $f'(x)$, so $f_s(x) = f'(x)$. The symmetric derivative does not use the value $f(x)$. For example, the indicator function of $\set{0}$ has symmetric derivative $0$ at $0$, although it is not continuous there.

Example: $f(x) = \abs{x}$ at $x = 0$. For $h \ne 0$, the quotient is $\frac{\abs{h} - \abs{-h}}{2h} = 0$, so $f_s(0) = 0$, while $f'(0)$ does not exist.

Non-example: $f(x) = \sqrt[3]{x}$ at $x = 0$. The quotient is $\frac{h^{1/3} + h^{1/3}}{2h} = h^{-2/3}$, which tends to $+\infty$, so the limit is not a real number.