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  "body": "\n## Proof\n\nWe replace the difference quotient of $g$ by a function $Q$ that is also defined where the denominator vanishes.\n\n1. Put $b = f(a)$, so $b \\in J$. Define $Q \\colon J \\to \\R$ by $Q(y) = \\frac{g(y) - g(b)}{y - b}$ for $y \\ne b$, and $Q(b) = g'(b)$.\n2. For every $y \\in J$, we have $g(y) - g(b) = Q(y) (y - b)$. For $y \\ne b$ this is the definition of $Q$. For $y = b$ both sides are $0$.\n3. Claim: $Q$ is continuous at $b$. Let $\\varepsilon \u003e 0$. By the definition of the [[derivative|0F1WR9C83K]] and the [[epsilon-delta form of limits|0GBZDAFPX8]], there is $\\delta \u003e 0$ with this property.\n4. For $k \\ne 0$ with $\\abs{k} \u003c \\delta$ and $b + k \\in J$, we have $\\abs{\\frac{g(b + k) - g(b)}{k} - g'(b)} \u003c \\varepsilon$.\n5. Let $y \\in J$ with $\\abs{y - b} \u003c \\delta$. If $y \\ne b$, step 4 with $k = y - b$ gives $\\abs{Q(y) - Q(b)} \u003c \\varepsilon$. If $y = b$, the difference is $0$.\n6. By the [[epsilon-delta form of continuity|0W8YHHGCJG]], $Q$ is continuous at $b$, which proves the claim.\n7. Since $f$ is differentiable at $a$, it is continuous at $a$ by [[differentiable functions are continuous|07ZMX4ANJS]]. Also $f(I) \\subseteq J$. By [[composition of continuous functions|0NRQPQX3VA]], $Q \\circ f \\colon I \\to \\R$ is continuous at $a$.\n8. 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 $H$.\n9. For $h \\in H$, step 2 with $y = f(a + h)$ and division by $h$ give\n$$\\frac{g(f(a + h)) - g(f(a))}{h} = Q(f(a + h)) \\cdot \\frac{f(a + h) - f(a)}{h}.$$\n10. Let $(h_n)$ be a sequence in $H$ with $h_n \\to 0$. By the [[sequential characterization of limits|000MXHEY22]], $\\frac{f(a + h_n) - f(a)}{h_n} \\to f'(a)$.\n11. Claim: $Q(f(a + h_n)) \\to Q(f(a)) = g'(b)$. Let $\\varepsilon \u003e 0$, and take $\\delta \u003e 0$ from the continuity of $Q \\circ f$ at $a$ (step 7).\n12. By the [[epsilon-N form|09G6KHS6GY]], $\\abs{h_n} \u003c \\delta$ for all large $n$. For those $n$, $\\abs{Q(f(a + h_n)) - g'(b)} \u003c \\varepsilon$. This proves the claim.\n13. By the [[limit of a product of sequences|0734RSVF2V]] and step 9, the difference quotient of $g \\circ f$ at $h_n$ tends to $g'(b) f'(a)$.\n14. This holds for every such sequence. By the sequential characterization, the difference quotient of $g \\circ f$ at $a$ tends to $g'(f(a)) f'(a)$.\n15. So $g \\circ f$ is [[differentiable|0A61B60B5R]] at $a$, and $(g \\circ f)'(a) = g'(f(a)) \\, f'(a)$.\n\n## Idea\n\nThe naive proof divides by $f(a + h) - f(a)$, which may be zero. The function $Q$ is the difference quotient of $g$ at $f(a)$, patched with the value $g'(f(a))$ where the quotient is undefined. The identity $g(y) - g(b) = Q(y)(y - b)$ then holds everywhere, and continuity of $Q$ finishes the argument.\n",
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