Proof of the reciprocal rule

Proof

We first find an open interval around $a$ where $1/g$ is defined. Then we use the sequential characterization of limits.

1. Since $g$ is differentiable at $a$, it is continuous at $a$ by differentiable functions are continuous. Put $\eta = \abs{g(a)} > 0$.
2. By the epsilon-delta form of continuity, there is $\delta_0 > 0$ with $\abs{g(x) - g(a)} < \eta$ for $x \in I$ with $\abs{x - a} < \delta_0$.
3. For such $x$, the reverse triangle inequality gives $\abs{g(x)} \ge \abs{g(a)} - \abs{g(x) - g(a)} > 0$. So $g(x) \ne 0$.
4. Let $I_0 = I \cap (a - \delta_0, a + \delta_0)$. It is an open interval containing $a$, and $g$ has no zero on it. So $1/g$ is defined on $I_0$, and we view it as a function $I_0 \to \R$.
5. Let $H = \set{h \ne 0 : a + h \in I_0}$. Since $I_0$ is an open interval containing $a$, $0$ is a limit point of $H$.
6. For $h \in H$, write $q_g(h) = \frac{g(a + h) - g(a)}{h}$. Over a common denominator, $$\frac{1}{h}\left( \frac{1}{g(a + h)} - \frac{1}{g(a)} \right) = \frac{- q_g(h)}{g(a + h) \, g(a)}.$$
7. Let $(h_n)$ be a sequence in $H$ with $h_n \to 0$. Each $h_n$ is a nonzero number with $a + h_n \in I$. By the definition of the derivative and the sequential characterization, $q_g(h_n) \to g'(a)$.
8. Claim: $g(a + h_n) \to g(a)$. Let $\varepsilon > 0$, and take $\delta > 0$ from the epsilon-delta form of continuity of $g$ at $a$. By the epsilon-N form, $\abs{h_n} < \delta$ for all large $n$. Then $\abs{g(a + h_n) - g(a)} < \varepsilon$.
9. The constant sequences $g(a)$ and $-1$ converge to $g(a)$ and $-1$. By the limit of a product of sequences, $g(a + h_n) \, g(a) \to g(a)^2$ and $-q_g(h_n) \to -g'(a)$.
10. We have $g(a)^2 \ne 0$, and $g(a + h_n) g(a) \ne 0$ for all $n$ by step 4. By the limit of a quotient of sequences, the right side of step 6 at $h = h_n$ tends to $-\frac{g'(a)}{g(a)^2}$.
11. This holds for every such sequence. By the sequential characterization, the difference quotient of $1/g$ at $a$ tends to $-\frac{g'(a)}{g(a)^2}$ as $h \to 0$.
12. So $1/g$ is differentiable at $a$, and $\left( \frac{1}{g} \right)'(a) = -\frac{g'(a)}{g(a)^2}$.