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  "body": "\n## Proof\n\nWe write $f$ and $g$ as linear approximations plus errors, multiply out, and bound each error term.\n\n1. Let $U$, $f$, $g$ and $a$ be as in the statement. Put $A = Df(a)$ and $B = Dg(a)$, two linear maps $\\mathbb{R}^n \\to \\mathbb{R}$.\n2. By [[linear maps on finite-dimensional spaces are continuous|0C13E623WD]] and [[bounded linear maps are continuous|0F3K4C7XYW]], there are $C_A, C_B \\ge 0$ with $|A h| \\le C_A \\|h\\|$ and $|B h| \\le C_B \\|h\\|$ for each $h$.\n3. For $h$ with $a + h \\in U$, put $r(h) = f(a + h) - f(a) - A h$ and $s(h) = g(a + h) - g(a) - B h$. Put $\\Delta(h) = f(a + h) - f(a) = A h + r(h)$.\n4. Let $L \\colon \\mathbb{R}^n \\to \\mathbb{R}$ be the map $L h = f(a) \\, B h + g(a) \\, A h$. It is linear, since $A$ and $B$ are linear.\n5. Multiply out, using $f(a + h) = f(a) + \\Delta(h)$ and $g(a + h) = g(a) + B h + s(h)$:\n$$f(a + h) g(a + h) = f(a) g(a) + f(a) B h + f(a) s(h) + \\Delta(h) \\big(g(a) + B h + s(h)\\big).$$\n6. Since $\\Delta(h) g(a) = g(a) A h + g(a) r(h)$, step 5 gives $f(a + h) g(a + h) - f(a) g(a) - L h = R(h)$, where\n$$R(h) = f(a) \\, s(h) + g(a) \\, r(h) + \\Delta(h) \\big(B h + s(h)\\big).$$\n7. Since $f$ and $g$ are [[differentiable|0135XZZQ8M]] at $a$, there is $\\delta_0 \u003e 0$ with $|r(h)| \\le \\|h\\|$ and $|s(h)| \\le \\|h\\|$. This holds for $\\|h\\| \u003c \\delta_0$ with $a + h \\in U$.\n8. For these $h$, steps 2 and 7 give $|\\Delta(h)| \\le (C_A + 1) \\|h\\|$ and $|B h + s(h)| \\le (C_B + 1) \\|h\\|$.\n9. For $0 \u003c \\|h\\| \u003c \\delta_0$ with $a + h \\in U$, the [[triangle inequality|0FJZMBX0ZZ]] and step 8 give\n$$\\frac{|R(h)|}{\\|h\\|} \\le |f(a)| \\frac{|s(h)|}{\\|h\\|} + |g(a)| \\frac{|r(h)|}{\\|h\\|} + (C_A + 1)(C_B + 1) \\|h\\|.$$\n10. As $h \\to 0$, the first two terms tend to $0$ by the differentiability of $g$ and $f$ at $a$. The last term tends to $0$ as well.\n11. So $|R(h)| / \\|h\\| \\to 0$. By step 6, $f g$ is differentiable at $a$ with $D(f g)(a) = L$.\n12. That is, $D(f g)(a)(h) = f(a) \\, Dg(a)(h) + g(a) \\, Df(a)(h)$ for each $h \\in \\mathbb{R}^n$.\n\n## Idea\n\nMultiplying the two linear approximations gives the constant $f(a) g(a)$, the two cross terms $f(a) B h$ and $g(a) A h$, and some leftovers. Each leftover contains an error term, or a product of two quantities of size about $\\|h\\|$. So the leftovers are small compared with $\\|h\\|$.\n",
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