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  "body": "\n## Proof\n\nWe treat one component at a time, and we move from $a$ to $a + h$ one coordinate at a time with the mean value theorem.\n\n1. Let $U$, $f$ and $a$ be as in the statement. Fix $j \\in \\{1, \\dots, m\\}$ and write $g = f_j \\colon U \\to \\mathbb{R}$. Write $\\partial_i g$ for the [[partial derivative|05AXG3QRKR]] $\\partial g / \\partial x_i$.\n2. Since $U$ is [[open|01ZQQDCERP]], there is $\\rho \u003e 0$ such that the [[open ball|0DC3MPR7C9]] $B(a, \\rho)$ lies in $U$.\n3. Let $L \\colon \\mathbb{R}^n \\to \\mathbb{R}$ be the linear map $L h = \\sum_{i=1}^n \\partial_i g(a) \\, h_i$. We show that $g$ is [[differentiable|0135XZZQ8M]] at $a$ with $Dg(a) = L$.\n4. Let $h = (h_1, \\dots, h_n)$ with $\\|h\\| \u003c \\rho$. Put $p_0 = a$ and $p_k = a + h_1 e_1 + \\dots + h_k e_k$ for $1 \\le k \\le n$, where $e_k$ is the $k$-th standard basis vector. Then $p_n = a + h$.\n5. For each $k$ and each real $s$ between $0$ and $h_k$, the vector $p_{k-1} + s e_k - a$ is $(h_1, \\dots, h_{k-1}, s, 0, \\dots, 0)$. Its norm is at most $\\|h\\|$, because $s^2 \\le h_k^2$.\n6. Fix $k$. Let $J$ be the set of real $s$ with $\\|p_{k-1} + s e_k - a\\| \u003c \\rho$. It is an open interval, and by step 5 it contains the closed interval between $0$ and $h_k$. Put $\\varphi(s) = g(p_{k-1} + s e_k)$ for $s \\in J$.\n7. For $s \\in J$, the point $p_{k-1} + s e_k$ lies in $U$. By the definition of $\\partial_k g$ at that point, $\\varphi$ is differentiable at $s$ with $\\varphi'(s) = \\partial_k g(p_{k-1} + s e_k)$.\n8. Since [[differentiable functions are continuous|07ZMX4ANJS]], $\\varphi$ is continuous on $J$.\n9. Suppose that $h_k \\ne 0$. By the [[mean value theorem|0BARQY52N5]] for $\\varphi$ between $0$ and $h_k$, there is $\\theta_k$ strictly between $0$ and $h_k$ with $\\varphi(h_k) - \\varphi(0) = \\varphi'(\\theta_k) \\, h_k$.\n10. Put $q_k = p_{k-1} + \\theta_k e_k$. Then step 9 says $g(p_k) - g(p_{k-1}) = \\partial_k g(q_k) \\, h_k$. By step 5, $\\|q_k - a\\| \\le \\|h\\|$.\n11. If $h_k = 0$, then $p_k = p_{k-1}$. Put $q_k = p_{k-1}$. Then again $g(p_k) - g(p_{k-1}) = 0 = \\partial_k g(q_k) \\, h_k$ and $\\|q_k - a\\| \\le \\|h\\|$.\n12. Summing over $k$, the sum telescopes:\n$$g(a + h) - g(a) - L h = \\sum_{k=1}^n \\big(g(p_k) - g(p_{k-1})\\big) - \\sum_{k=1}^n \\partial_k g(a) h_k = \\sum_{k=1}^n \\big(\\partial_k g(q_k) - \\partial_k g(a)\\big) h_k.$$\n13. Since $|h_k| \\le \\|h\\|$, this gives $|g(a + h) - g(a) - L h| \\le \\|h\\| \\sum_{k=1}^n |\\partial_k g(q_k) - \\partial_k g(a)|$.\n14. Let $\\varepsilon \u003e 0$. Each $\\partial_k g$ is continuous at $a$. So there is $\\eta \u003e 0$ with $|\\partial_k g(q) - \\partial_k g(a)| \u003c \\varepsilon / n$ for each $k$ and each $q \\in U$ with $\\|q - a\\| \u003c \\eta$.\n15. Let $0 \u003c \\|h\\| \u003c \\min(\\rho, \\eta)$. By steps 10 and 11, $\\|q_k - a\\| \u003c \\eta$ for each $k$. By steps 13 and 14, $|g(a + h) - g(a) - L h| \u003c n (\\varepsilon / n) \\|h\\| = \\varepsilon \\|h\\|$.\n16. So $|g(a + h) - g(a) - L h| / \\|h\\| \\to 0$ as $h \\to 0$. Thus $f_j = g$ is differentiable at $a$.\n17. Step 16 holds for each $j$. By the [[componentwise criterion for differentiability|0NS0X75RS1]], $f$ is differentiable at $a$.\n\n## Idea\n\nGoing from $a$ to $a + h$ along the coordinate directions splits the change of $f$ into $n$ changes in one variable. The mean value theorem turns each of them into a partial derivative at a nearby point, times $h_k$. Continuity at $a$ makes these partial derivatives close to their values at $a$, and that is exactly the linear approximation.\n",
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