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  "body": "\n## Proof\n\nWe unfold the definitions.\n\n1. Let $f \\colon U \\to \\R$ be twice differentiable on the open set $U \\subseteq \\R^n$. Then $f$ is [[differentiable|0135XZZQ8M]] at each point of $U$.\n2. By the coordinate form of the [[gradient|0NS2QYVC4Y]], $\\nabla f = (g_1, \\dots, g_n)$ on $U$, where $g_i = \\partial f / \\partial x_i$ is a [[partial derivative|05AXG3QRKR]].\n3. The map $x \\mapsto Df(x)$ is differentiable on $U$, since $f$ is twice differentiable. Its entries in the standard basis are $g_1, \\dots, g_n$, so each $g_i$ is differentiable on $U$.\n4. Hence each partial derivative $\\partial g_i / \\partial x_i$ exists on $U$. By the definition of [[higher partial derivatives|02RSXVM1ZN]], $\\partial g_i / \\partial x_i = \\partial^2 f / \\partial x_i^2$.\n5. By the definition of the [[divergence|02NSMQVQWK]], $\\operatorname{div}(\\nabla f) = \\sum_{i=1}^n \\partial g_i / \\partial x_i$.\n6. By the definition of the [[Laplacian|0XQ2ZVN1PZ]] and steps 4 and 5, $\\Delta f = \\operatorname{div}(\\nabla f) = \\frac{\\partial^2 f}{\\partial x_1^2} + \\dots + \\frac{\\partial^2 f}{\\partial x_n^2}$.\n\n## Idea\n\nThe gradient collects the first partial derivatives of $f$. The divergence then differentiates the $i$-th of them in the direction $x_i$ and adds the results. So only the unmixed second derivatives appear.\n",
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