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  "body": "\n## Proof\n\nWe write both maps as linear approximations plus errors in little-o form, and we substitute one into the other.\n\n1. Let $U$, $W$, $g$, $f$ and $a$ be as in the statement. Put $b = g(a)$, $A = Dg(a)$ and $B = Df(b)$.\n2. By [[linear maps on finite-dimensional spaces are continuous|0C13E623WD]] and [[bounded linear maps are continuous|0F3K4C7XYW]], each linear map between Euclidean spaces is bounded. So there are $C_A, C_B \\ge 0$ with $\\|A h\\| \\le C_A \\|h\\|$ and $\\|B w\\| \\le C_B \\|w\\|$ for all $h \\in \\mathbb{R}^n$ and $w \\in \\mathbb{R}^m$.\n3. By step 2, being [[differentiable|0135XZZQ8M]] at a point with derivative $L$ means that $L$ is the [[Fréchet derivative|08EVNYZF42]] there, for the Euclidean norms. So the [[little-o form|0MN882VJ4P]] of the derivative applies.\n4. For $h$ with $a + h \\in U$, put $r(h) = g(a + h) - g(a) - A h$. For $w$ with $b + w \\in W$, put $s(w) = f(b + w) - f(b) - B w$.\n5. For $h$ with $a + h \\in U$, put $w(h) = g(a + h) - b = A h + r(h)$. Then $b + w(h) = g(a + h) \\in W$.\n6. By steps 4 and 5, $f(g(a + h)) = f(b + w(h)) = f(b) + B w(h) + s(w(h))$.\n7. Since $B$ is linear, $B w(h) = B A h + B r(h)$. So $(f \\circ g)(a + h) - (f \\circ g)(a) - (B \\circ A) h = B r(h) + s(w(h))$.\n8. Let $\\varepsilon \u003e 0$. By the little-o form for $f$ at $b$, there is $\\delta_2 \u003e 0$ with $\\|s(w)\\| \\le \\frac{\\varepsilon}{2 (C_A + 1)} \\|w\\|$ for $\\|w\\| \u003c \\delta_2$ and $b + w \\in W$.\n9. By the little-o form for $g$ at $a$, there is $\\delta_1 \u003e 0$ with $\\|r(h)\\| \\le \\min\\big(1, \\frac{\\varepsilon}{2 (C_B + 1)}\\big) \\|h\\|$ for $\\|h\\| \u003c \\delta_1$ and $a + h \\in U$.\n10. Put $\\delta = \\min\\big(\\delta_1, \\delta_2 / (C_A + 1)\\big)$, and let $\\|h\\| \u003c \\delta$ with $a + h \\in U$.\n11. By steps 2, 5 and 9 and the [[triangle inequality|0FJZMBX0ZZ]], $\\|w(h)\\| \\le C_A \\|h\\| + \\|h\\| = (C_A + 1) \\|h\\| \u003c \\delta_2$.\n12. So step 8 applies to $w = w(h)$, and $\\|s(w(h))\\| \\le \\frac{\\varepsilon}{2 (C_A + 1)} (C_A + 1) \\|h\\| = \\frac{\\varepsilon}{2} \\|h\\|$.\n13. By steps 2 and 9, $\\|B r(h)\\| \\le C_B \\|r(h)\\| \\le \\frac{C_B \\varepsilon}{2 (C_B + 1)} \\|h\\| \\le \\frac{\\varepsilon}{2} \\|h\\|$.\n14. By steps 7, 12 and 13, $\\|(f \\circ g)(a + h) - (f \\circ g)(a) - (B \\circ A) h\\| \\le \\varepsilon \\|h\\|$ for $\\|h\\| \u003c \\delta$ with $a + h \\in U$.\n15. The [[composition|05YBKMRVAR]] $B \\circ A$ is a linear map $\\mathbb{R}^n \\to \\mathbb{R}^k$. By step 14 and the little-o form, the [[composition|0THBK9V825]] $f \\circ g$ is differentiable at $a$ with derivative $B \\circ A$.\n16. That is, $D(f \\circ g)(a) = Df(g(a)) \\circ Dg(a)$.\n\n## Idea\n\nNear $a$, the map $g$ moves $a + h$ to $b + w$ with $w \\approx A h$, and near $b$ the map $f$ acts like $B$. So $f \\circ g$ acts like $B A$. The errors stay small: $B$ cannot blow up the error of $g$ by more than a fixed factor, and $w$ is at most a fixed multiple of $\\|h\\|$.\n",
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