Subsection 2.5.3 (Optional) — Proof of the Quotient Rule (Theorem 2.4.5)
We now give the proof of the quotient rule in two steps 1 We thank Serban Raianu for suggesting this approach.. We assume throughout that \(g(x) \neq 0\) and that \(f(x)\) and \(g(x)\) are differentiable, meaning that the limits defining \(f'(x)\text{,}\) \(g'(x)\) exist.
- In the first step, we prove the quotient rule under the assumption that \(f(x)/g(x)\) is differentiable.
- In the second step, we prove that \(1/g(x)\) differentiable. Once we know that \(1/g(x)\) is differentiable, the product rule implies that \(f(x)/g(x)\) is differentiable.
Step 1: the proof of the quotient rule assumng that \(\frac{f(x)}{g(x)}\) is differentiable.\(\ \ \ \) Write \(Q(x)=\frac{f(x)}{g(x)}\text{.}\) Then \(f(x) = g(x)\,Q(x)\) so that \(f'(x) = g'(x)\,Q(x) + g(x)\,Q'(x)\text{,}\) by the product rule, and
\begin{align*}
Q'(x) &= \frac{f'(x)-g'(x)\,Q(x)}{g(x)}
= \frac{f'(x)-g'(x)\,\frac{f(x)}{g(x)}}{g(x)}\\
&= \frac{f'(x)g(x)-f(x)g'(x)}{g(x)^2}
\end{align*}
Step 2: the proof that \(1/g(x)\) is differentiable.\(\ \ \ \) By definition
\begin{align*}
\diff{}{x}\frac{1}{g(x)}
&=\lim_{h\rightarrow 0}\frac{1}{h}\left[\frac{1}{g(x+h)}-\frac{1}{g(x)}\right]
=\lim_{h\rightarrow 0}\frac{g(x)-g(x+h)}{h\,g(x)\,g(x+h)}\\
&=-\lim_{h\rightarrow 0}\frac{1}{g(x)}\ \frac{1}{g(x+h)}\
\frac{g(x+h)-g(x)}{h}\\
&=-\frac{1}{g(x)}\
\lim_{h\rightarrow 0}\frac{1}{g(x+h)}\
\lim_{h\rightarrow 0}\frac{g(x+h)-g(x)}{h}\\
&=-\frac{1}{g(x)^2}g'(x)
\end{align*}