Proof of the chain rule with an auxiliary difference quotient

Proof

We replace the difference quotient of $g$ by a function $Q$ that is also defined where the denominator vanishes.

1. 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 $Q(b) = g'(b)$.
2. 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 $0$.
3. Claim: $Q$ is continuous at $b$. Let $\varepsilon > 0$. By the definition of the derivative and the epsilon-delta form of limits, there is $\delta > 0$ with this property.
4. For $k \ne 0$ with $\abs{k} < \delta$ and $b + k \in J$, we have $\abs{\frac{g(b + k) - g(b)}{k} - g'(b)} < \varepsilon$.
5. Let $y \in J$ with $\abs{y - b} < \delta$. If $y \ne b$, step 4 with $k = y - b$ gives $\abs{Q(y) - Q(b)} < \varepsilon$. If $y = b$, the difference is $0$.
6. By the epsilon-delta form of continuity, $Q$ is continuous at $b$, which proves the claim.
7. Since $f$ is differentiable at $a$, it is continuous at $a$ by differentiable functions are continuous. Also $f(I) \subseteq J$. By composition of continuous functions, $Q \circ f \colon I \to \R$ is continuous at $a$.
8. Let $H = \set{h \ne 0 : a + h \in I}$. As $I$ is an open interval containing $a$, $0$ is a limit point of $H$.
9. For $h \in H$, step 2 with $y = f(a + h)$ and division by $h$ give $$\frac{g(f(a + h)) - g(f(a))}{h} = Q(f(a + h)) \cdot \frac{f(a + h) - f(a)}{h}.$$
10. Let $(h_n)$ be a sequence in $H$ with $h_n \to 0$. By the sequential characterization of limits, $\frac{f(a + h_n) - f(a)}{h_n} \to f'(a)$.
11. Claim: $Q(f(a + h_n)) \to Q(f(a)) = g'(b)$. Let $\varepsilon > 0$, and take $\delta > 0$ from the continuity of $Q \circ f$ at $a$ (step 7).
12. By the epsilon-N form, $\abs{h_n} < \delta$ for all large $n$. For those $n$, $\abs{Q(f(a + h_n)) - g'(b)} < \varepsilon$. This proves the claim.
13. By the limit of a product of sequences and step 9, the difference quotient of $g \circ f$ at $h_n$ tends to $g'(b) f'(a)$.
14. This holds for every such sequence. By the sequential characterization, the difference quotient of $g \circ f$ at $a$ tends to $g'(f(a)) f'(a)$.
15. So $g \circ f$ is differentiable at $a$, and $(g \circ f)'(a) = g'(f(a)) \, f'(a)$.