Proof of the product rule

Proof

We use the sequential characterization of limits and the limit laws for sequences.

1. Let $H = \set{h \ne 0 : a + h \in I}$. As $I$ is an open interval containing $a$, $0$ is a limit point of $H$.
2. 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, $q_f(h) \to f'(a)$ and $q_g(h) \to g'(a)$ as $h \to 0$.
3. For $h \in H$, add and subtract $f(a) g(a + h)$ in the numerator of $q_{fg}(h)$. This gives $q_{fg}(h) = q_f(h) \, g(a + h) + f(a) \, q_g(h)$.
4. 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 $q_g(h_n) \to g'(a)$.
5. Claim: $g(a + h_n) \to g(a)$. Since $g$ is differentiable at $a$, it is continuous at $a$ by differentiable functions are continuous.
6. Let $\varepsilon > 0$. By the epsilon-delta form of continuity, there is $\delta > 0$ with $\abs{g(x) - g(a)} < \varepsilon$ for $x \in I$ with $\abs{x - a} < \delta$.
7. By the epsilon-N form, there is $N$ with $\abs{h_n} < \delta$ for $n \ge N$. Then $\abs{g(a + h_n) - g(a)} < \varepsilon$ for $n \ge N$, which proves the claim.
8. The constant sequence $f(a)$ converges to $f(a)$. By the limit of a product of sequences, $q_f(h_n) \, g(a + h_n) \to f'(a) g(a)$ and $f(a) \, q_g(h_n) \to f(a) g'(a)$.
9. By the limit of a sum of sequences and step 3, $q_{fg}(h_n) \to f'(a) g(a) + f(a) g'(a)$.
10. This holds for every such sequence. By the sequential characterization, $q_{fg}(h) \to f'(a) g(a) + f(a) g'(a)$ as $h \to 0$.
11. So $f g$ is differentiable at $a$, and $(f g)'(a) = f'(a) g(a) + f(a) g'(a)$.