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  "body": "\n## Proof\n\nWe compute each identity in coordinates.\n\n1. Write $(x_1, x_2, x_3)$ for $(x, y, z)$, $\\partial_i = \\partial / \\partial x_i$, and $A = (A_1, A_2, A_3)$. For a function $g$, $\\partial_i \\partial_j g$ is the [[second partial derivative|02RSXVM1ZN]] $\\partial_i (\\partial_j g)$.\n2. A function on $U$ of [[class $C^1$|0TCBZMEYGN]] is differentiable by [[continuous partial derivatives imply differentiability|0KXSK8TEEB]]. So its [[gradient|0NS2QYVC4Y]] is $(\\partial_1 g, \\partial_2 g, \\partial_3 g)$.\n3. Let $\\phi, \\psi, A$ be $C^1$. A partial derivative $\\partial_i$ at a point is the derivative of a function of one variable on an open interval. So the [[product rule|0V9A43H7WV]] gives $\\partial_i (\\phi \\psi) = \\phi \\, \\partial_i \\psi + \\psi \\, \\partial_i \\phi$ and $\\partial_i (\\psi A_k) = \\psi \\, \\partial_i A_k + A_k \\, \\partial_i \\psi$.\n4. The right sides in step 3 are continuous, so $\\phi \\psi$ and $\\psi A$ are $C^1$. By step 2, the $i$-th component of $\\nabla (\\phi \\psi)$ is $\\phi \\, \\partial_i \\psi + \\psi \\, \\partial_i \\phi$. So $\\nabla (\\phi \\psi) = \\phi \\nabla \\psi + \\psi \\nabla \\phi$.\n5. By the definition of the [[divergence|02NSMQVQWK]] and step 3, $\\nabla \\cdot (\\psi A) = \\sum_i \\partial_i (\\psi A_i) = \\psi \\sum_i \\partial_i A_i + \\sum_i A_i \\, \\partial_i \\psi = \\psi \\, \\nabla \\cdot A + \\nabla \\psi \\cdot A$.\n6. By the definition of the [[curl|0GDNHS9J2X]] and step 3, the first component of $\\nabla \\times (\\psi A)$ is $\\partial_2 (\\psi A_3) - \\partial_3 (\\psi A_2) = \\psi (\\partial_2 A_3 - \\partial_3 A_2) + (\\partial_2 \\psi \\, A_3 - \\partial_3 \\psi \\, A_2)$.\n7. The second component is $\\partial_3 (\\psi A_1) - \\partial_1 (\\psi A_3) = \\psi (\\partial_3 A_1 - \\partial_1 A_3) + (\\partial_3 \\psi \\, A_1 - \\partial_1 \\psi \\, A_3)$.\n8. The third component is $\\partial_1 (\\psi A_2) - \\partial_2 (\\psi A_1) = \\psi (\\partial_1 A_2 - \\partial_2 A_1) + (\\partial_1 \\psi \\, A_2 - \\partial_2 \\psi \\, A_1)$.\n9. In steps 6 to 8, the first brackets are the components of $\\nabla \\times A$. By the definition of the [[cross product|0SEDGXG9S3]], the second brackets are the components of $\\nabla \\psi \\times A$. So $\\nabla \\times (\\psi A) = \\psi \\, \\nabla \\times A + \\nabla \\psi \\times A$.\n10. Now let $\\phi$ and $A$ be $C^2$. Then $\\nabla \\times (\\nabla \\phi) = 0$, because the [[curl of a gradient is zero|0M2JRW1J0N]].\n11. Also $\\nabla \\cdot (\\nabla \\times A) = 0$, because the [[divergence of a curl is zero|0F56CMMC5P]].\n12. Let $B = \\nabla \\times A$, so $B = (\\partial_2 A_3 - \\partial_3 A_2, \\ \\partial_3 A_1 - \\partial_1 A_3, \\ \\partial_1 A_2 - \\partial_2 A_1)$, a $C^1$ field. The first component of $\\nabla \\times B$ is $\\partial_2 B_3 - \\partial_3 B_2 = \\partial_2 \\partial_1 A_2 + \\partial_3 \\partial_1 A_3 - \\partial_2 \\partial_2 A_1 - \\partial_3 \\partial_3 A_1$.\n13. The second component is $\\partial_3 B_1 - \\partial_1 B_3 = \\partial_3 \\partial_2 A_3 + \\partial_1 \\partial_2 A_1 - \\partial_3 \\partial_3 A_2 - \\partial_1 \\partial_1 A_2$.\n14. The third component is $\\partial_1 B_2 - \\partial_2 B_1 = \\partial_1 \\partial_3 A_1 + \\partial_2 \\partial_3 A_2 - \\partial_1 \\partial_1 A_3 - \\partial_2 \\partial_2 A_3$.\n15. Each $A_k$ is $C^2$, hence twice differentiable. By the formula for the [[Laplacian|0XQ2ZVN1PZ]] as a [[sum of second partial derivatives|05RCT1522R]], $\\Delta A_k = \\partial_1 \\partial_1 A_k + \\partial_2 \\partial_2 A_k + \\partial_3 \\partial_3 A_k$.\n16. The function $\\nabla \\cdot A = \\partial_1 A_1 + \\partial_2 A_2 + \\partial_3 A_3$ is $C^1$. So the $k$-th component of $\\nabla (\\nabla \\cdot A) - \\Delta A$ is $\\sum_{j \\ne k} \\partial_k \\partial_j A_j - \\sum_{j \\ne k} \\partial_j \\partial_j A_k$, since the terms $\\partial_k \\partial_k A_k$ cancel.\n17. For $k = 1, 2, 3$, step 16 agrees with steps 12, 13 and 14, once each $\\partial_j \\partial_k A_j$ there is replaced by $\\partial_k \\partial_j A_j$. This is allowed by [[Schwarz's theorem|0NNQY8WCW6]], since $A_j$ is $C^2$.\n18. Hence $\\nabla \\times (\\nabla \\times A) = \\nabla (\\nabla \\cdot A) - \\Delta A$.\n\n## Idea\n\nThe product rules come from the one-variable product rule, applied to each partial derivative of each component. The second-order identities are bookkeeping of mixed partial derivatives. They hold because, for $C^2$ functions, the order of differentiation does not matter.\n",
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