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  "body": "\n## Proof\n\n1. Fix a real number $a$. For $x \u003e 0$, the [[real power|0TV07AQT62]] is $x^a = e^{a \\ln x}$.\n2. Let $u(x) = a \\ln x$ on the open interval $(0, \\infty)$. By the [[derivative of the logarithm|07X1QA6BRW]] and the [[constant multiple rule|0Z9JB71BS0]], $u$ is differentiable with $u'(x) = \\frac{a}{x}$.\n3. By [[$\\exp' = \\exp$|0EX9GFJ0B3]], $\\exp$ is differentiable on $\\mathbb{R}$ with derivative $\\exp$.\n4. By the [[chain rule|0RXX2PT1MQ]], $x \\mapsto x^a = \\exp(u(x))$ is differentiable at every $x \u003e 0$, with derivative $e^{a \\ln x} \\cdot \\frac{a}{x} = \\frac{a \\, x^a}{x}$.\n5. It remains to show $\\frac{x^a}{x} = x^{a-1}$. By the [[exponential of a sum|0E8HNVBCSP]], $e^{\\ln x} e^{-\\ln x} = e^{0} = 1$, where $e^0 = 1$ by the [[series of the exponential|0RMVS7RXZX]].\n6. Since [[$e^{\\ln x} = x$|0RKW2X63YR]], this gives $e^{-\\ln x} = \\frac{1}{x}$.\n7. Again by the exponential of a sum, $x^{a-1} = e^{a \\ln x + (-\\ln x)} = e^{a \\ln x} e^{-\\ln x} = \\frac{x^a}{x}$.\n8. By steps 4 and 7, $\\frac{d}{dx} x^a = a x^{a-1}$ for every $x \u003e 0$.\n\n## Idea\n\nRewrite the power as an exponential, $x^a = e^{a \\ln x}$, so that the exponent becomes an ordinary function of $x$. The chain rule then differentiates it, and the factor $\\frac{1}{x}$ from $\\ln$ lowers the power by one.\n",
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