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  "body": "\n## Definition\n\nLet $I \\subseteq \\R$ be an open interval, let $f \\colon I \\to \\R$ be a [[function|0XHDTE5YMZ]], and let $x \\in I$.\nThe function $f$ is *symmetrically differentiable* at $x$ if the [[limit|0TXTSEMHR5]]\n$$f_s(x) = \\lim_{h \\to 0} \\frac{f(x + h) - f(x - h)}{2h}$$\nexists and is a real number.\nThe number $f_s(x)$ is the *symmetric derivative* of $f$ at $x$.\n\n## Remarks\n\nIf $f$ is differentiable at $x$, then $\\frac{f(x + h) - f(x - h)}{2h}$ is the average of two [[difference quotients|AZBDB5F4M4]] with increments $h$ and $-h$.\nBoth tend to $f'(x)$, so $f_s(x) = f'(x)$.\nThe symmetric derivative does not use the value $f(x)$.\nFor example, the indicator function of $\\set{0}$ has symmetric derivative $0$ at $0$, although it is not continuous there.\n\nExample: $f(x) = \\abs{x}$ at $x = 0$.\nFor $h \\ne 0$, the quotient is $\\frac{\\abs{h} - \\abs{-h}}{2h} = 0$, so $f_s(0) = 0$, while $f'(0)$ does not exist.\n\nNon-example: $f(x) = \\sqrt[3]{x}$ at $x = 0$.\nThe 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.\n",
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