Proof of the quotient rule from the product and reciprocal rules

Proof

We write $f/g = f \cdot (1/g)$ near $a$ and use the product rule and the reciprocal rule.

1. Since $g$ is differentiable at $a$, it is continuous at $a$ by differentiable functions are continuous. By the epsilon-delta form of continuity with $\varepsilon = \abs{g(a)}$, there is $\delta_0 > 0$ with $\abs{g(x) - g(a)} < \abs{g(a)}$ for $x \in I$ with $\abs{x - a} < \delta_0$.
2. For such $x$, the reverse triangle inequality gives $\abs{g(x)} \ge \abs{g(a)} - \abs{g(x) - g(a)} > 0$.
3. Let $I_0 = I \cap (a - \delta_0, a + \delta_0)$. It is an open interval containing $a$, and $g$ has no zero on $I_0$. So $f/g$ is defined on $I_0$.
4. Claim: if $u \colon I \to \R$ is differentiable at $a$, then its restriction to $I_0$ is differentiable at $a$ with the same derivative.
5. Proof of the claim. Let $H = \set{h \ne 0 : a + h \in I}$ and $H_0 = \set{h \ne 0 : a + h \in I_0}$. Then $H_0 \subseteq H$, and $0$ is a limit point of $H_0$.
6. Every sequence in $H_0$ tending to $0$ is a sequence in $H$. By the sequential characterization of limits, the difference quotients of $u$ along it tend to $u'(a)$. These are also the difference quotients of the restriction. By the sequential characterization again, the claim follows.
7. By step 4, $f$ and $g$ restricted to $I_0$ are differentiable at $a$. Since $g(a) \ne 0$, the reciprocal rule on $I_0$ shows that $1/g$ is differentiable at $a$ with $(1/g)'(a) = -\frac{g'(a)}{g(a)^2}$.
8. On $I_0$ we have $\frac{f}{g} = f \cdot \frac{1}{g}$. The product rule on $I_0$ shows that $f/g$ is differentiable at $a$, with $$\left( \frac{f}{g} \right)'(a) = f'(a) \cdot \frac{1}{g(a)} + f(a) \cdot \left( -\frac{g'(a)}{g(a)^2} \right).$$
9. Over the common denominator $g(a)^2$, this equals $\frac{f'(a) g(a) - f(a) g'(a)}{g(a)^2}$.