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  "body": "\n## Proof\n\nWe use the [[sequential characterization of limits|000MXHEY22]] and the limit laws for sequences.\n\n1. Let $H = \\set{h \\ne 0 : a + h \\in I}$. As $I$ is an open interval containing $a$, $0$ is a [[limit point|07EGCNHAVW]] of $H$.\n2. For $u \\colon I \\to \\R$ and $h \\in H$, write $q_u(h) = \\frac{u(a + h) - u(a)}{h}$. By the definition of the [[derivative|0F1WR9C83K]], $q_f(h) \\to f'(a)$ and $q_g(h) \\to g'(a)$ as $h \\to 0$.\n3. For $h \\in H$, add and subtract $f(a) g(a + h)$ in the numerator of $q_{fg}(h)$. This gives $q_{fg}(h) = q_f(h) \\, g(a + h) + f(a) \\, q_g(h)$.\n4. Let $(h_n)$ be a sequence in $H$ with $h_n \\to 0$. By the sequential characterization, $q_f(h_n) \\to f'(a)$ and $q_g(h_n) \\to g'(a)$.\n5. Claim: $g(a + h_n) \\to g(a)$. Since $g$ is differentiable at $a$, it is continuous at $a$ by [[differentiable functions are continuous|07ZMX4ANJS]].\n6. Let $\\varepsilon \u003e 0$. By the [[epsilon-delta form of continuity|0W8YHHGCJG]], there is $\\delta \u003e 0$ with $\\abs{g(x) - g(a)} \u003c \\varepsilon$ for $x \\in I$ with $\\abs{x - a} \u003c \\delta$.\n7. By the [[epsilon-N form|09G6KHS6GY]], there is $N$ with $\\abs{h_n} \u003c \\delta$ for $n \\ge N$. Then $\\abs{g(a + h_n) - g(a)} \u003c \\varepsilon$ for $n \\ge N$, which proves the claim.\n8. The constant sequence $f(a)$ converges to $f(a)$. By the [[limit of a product of sequences|0734RSVF2V]], $q_f(h_n) \\, g(a + h_n) \\to f'(a) g(a)$ and $f(a) \\, q_g(h_n) \\to f(a) g'(a)$.\n9. By the [[limit of a sum of sequences|0RAM4E07NR]] and step 3, $q_{fg}(h_n) \\to f'(a) g(a) + f(a) g'(a)$.\n10. This holds for every such sequence. By the sequential characterization, $q_{fg}(h) \\to f'(a) g(a) + f(a) g'(a)$ as $h \\to 0$.\n11. So $f g$ is [[differentiable|0A61B60B5R]] at $a$, and $(f g)'(a) = f'(a) g(a) + f(a) g'(a)$.\n\n## Idea\n\nAdding and subtracting $f(a) g(a + h)$ splits the change of $f g$ into a change of $f$ and a change of $g$. Each piece becomes a difference quotient times a factor. The factor $g(a + h)$ tends to $g(a)$ because differentiable functions are continuous.\n",
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