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  "body": "\n## Proof\n\n1. Let $f = \\exp \\colon \\mathbb{R} \\to \\mathbb{R}$ be the [[exponential function|0RMVS7RXZX]]. The domain $\\mathbb{R}$ is an open interval.\n2. By [[$\\exp' = \\exp$|0EX9GFJ0B3]], $f$ is differentiable at every point, with $f'(y) = e^y$.\n3. A [[differentiable function is continuous|07ZMX4ANJS]], so $f$ is continuous. By [[this result|0KN2V7840E]], $f$ is strictly increasing.\n4. By [[positivity of the exponential|0P9BNBW2W3]], $f'(y) = e^y \u003e 0$ for every real $y$. In particular $f'(y) \\ne 0$.\n5. Let $x \u003e 0$, and let $a = \\ln x$, where $\\ln$ is the [[natural logarithm|0044CPHJBV]]. By [[the inverse relation of $\\ln$ and $\\exp$|0RKW2X63YR]], $f(a) = e^{\\ln x} = x$.\n6. Every value $e^y$ is positive, and every $x \u003e 0$ equals $e^{\\ln x}$. So the image $f(\\mathbb{R})$ is $(0, \\infty)$.\n7. Also $\\ln(e^y) = y$ for every real $y$. So $\\ln \\colon (0, \\infty) \\to \\mathbb{R}$ is the inverse function $f^{-1}$.\n8. By the [[inverse function rule|0M0RZ3HSB0]] at the point $a$, the function $f^{-1} = \\ln$ is differentiable at $x = f(a)$.\n9. The same rule gives $\\ln'(x) = \\frac{1}{f'(a)} = \\frac{1}{e^{\\ln x}} = \\frac{1}{x}$.\n10. Since $x \u003e 0$ was arbitrary, $\\ln$ is differentiable on $(0, \\infty)$, and $\\frac{d}{dx} \\ln x = \\frac{1}{x}$ for $x \u003e 0$.\n\n## Idea\n\nThe logarithm undoes the exponential, so its graph is the graph of $\\exp$ reflected in the line $y = x$. Reflection turns a tangent slope $m$ into $1/m$. The slope of $\\exp$ at $\\ln x$ is $e^{\\ln x} = x$, so the slope of $\\ln$ at $x$ is $1/x$.\n",
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