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  "body": "\n## Definition\n\nLet $q$ be a real number with $q \\ne 0$ and $q \\ne 1$.\nLet $A \\subseteq \\R$ be a set with $qx \\in A$ for every $x \\in A$, and let $f \\colon A \\to \\R$.\nFor $x \\in A$ with $x \\ne 0$, the *$q$-derivative* of $f$ at $x$ is\n$$D_q f(x) = \\frac{f(qx) - f(x)}{(q - 1) x}.$$\nIf $0 \\in A$, one sets $D_q f(0) = f'(0)$ when the ordinary [[derivative|0V1EJCT8G6]] $f'(0)$ exists.\n\n## Remarks\n\nIt is a [[difference quotient|AZBDB5F4M4]] of $f$ between $x$ and $qx$.\nExample: for an integer $n \\ge 1$ and $f(x) = x^n$ on $\\R$, one gets $D_q x^n = \\frac{q^n - 1}{q - 1} x^{n-1} = [n]_q \\, x^{n-1}$, where $[n]_q = 1 + q + \\cdots + q^{n-1}$ is the $q$-analog of $n$.\nAs $q \\to 1$, $[n]_q \\to n$, and more generally $D_q f(x) \\to f'(x)$ at each $x \\ne 0$ where $f$ is differentiable.\nNon-example: with $q = 2$ and $f$ defined only on $A = [1, 2]$, the point $qx$ lies outside $A$ for $x \u003e 1$, so $D_2 f(x)$ is not defined there.\nThe operator is linear and satisfies the product rule $D_q (fg)(x) = f(qx) \\, D_q g(x) + g(x) \\, D_q f(x)$.\n",
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