Section 14.4 The Product Rule
A rigorous proof of the Product Rule is also fairly complex, but it does not suffer from the kind of technical problems we encountered in the proof of the Chain Rule.
Theorem 14.4.1. The Product Rule for Differentiation.
If \(\alpha(x)\) and \(\beta(x)\) are differentiable at \(x\) then \(f(x)=\alpha(x)\cdot\beta(x)\) is differentiable and
\begin{equation}
f^\prime(x)=\alpha(x)\cdot\beta^\prime(x)+\beta(x)\cdot\alpha^\prime(x).\tag{14.7}
\end{equation}
Proof.
We start with two observations. The first is that
\begin{align}
f^\prime(x)=\amp{}\limit{h}{0}{\frac{\textcolor{red}{f(x+h)}-\textcolor{blue}{f(x)}}{h}}\notag\\
\amp{} = \limit{h}{0}{\frac{\textcolor{red}
{\alpha(x+h)\beta(x+h)}-\textcolor{blue}{\alpha(x)\beta(x)}}{h}}\tag{14.8}
\end{align}
\begin{align}
f^\prime(x)=\alpha(x)\amp{}\left(\limit{h}{0}{\frac{\beta(x+h)-\beta(x)}{h}}\right)\notag\\
\amp{}+\beta(x)\left(\limit{h}{0}{\frac{\alpha(x+h)-\alpha(x)}{h}}\right)\text{.}\tag{14.9}
\end{align}
It appears then that our goal is simply to reorganize
equation (14.8) until it looks like
equation (14.9). We say “simply” but it will only appear to be simple after we have succeeded. We will proceed slowly.
Observe that if we subtract
\(\alpha(x+h)\beta(x)\) from the blue part of the numerator in
equation (14.8) we get
\begin{equation*}
\alpha(x)\beta(x)-\alpha(x+h)\beta(x)=
\textcolor{blue}{-\beta(x)\left(\alpha(x+h)-\alpha(x)\right)},
\end{equation*}
whereas if we add
\(\alpha(x+h)\beta(x)\) to the red part of the numerator in
equation (14.8) we get
\begin{equation*}
\alpha(x+h)\beta(x+h)+\alpha(x+h)\beta(x) =
\textcolor{red}{\alpha(x+h)\left(\beta(x+h)-\beta(x)\right)}.
\end{equation*}
This suggests that we should both add and subtract the expression
\(\alpha(x+h)\beta(x)\) to the numerator of
equation (14.8). Doing this and factoring as we’ve indicated above we get
\begin{equation*}
f^\prime(x)=\limit{h}{0}{ \frac{ \textcolor {red}
{\alpha(x+h)\left(\beta(x+h)-\beta(x)\right)} -
\left[\textcolor{blue}{-\beta(x)\left(\alpha(x+h)-\alpha(x)\right)}\right]
}{h} },
\end{equation*}
By
Theorem 14.1.1 we can separate this into the limit of the two fractions as follows:
\begin{align*}
f^\prime(x)=\amp{}\tlimit{h}{0}{\left(
\frac{\textcolor{green}{\alpha(x+h)}\left(\textcolor{red}{\beta(x+h)-{\beta(x)}}\right)}{\textcolor{red}{h}}\right)}\\
\amp{} +
\rlimit{h}{0}{\left(\frac{\textcolor{green}{\beta(x)}\left(\textcolor{blue}{\alpha(x+h)-\alpha(x)}\right)}{\textcolor{blue}{h}}\right)}\text{.}
\end{align*}
\begin{align*}
f^\prime(x)=\amp{}
\underbrace{\left[ \limit{h}{0}{\textcolor{green}{\alpha(x+h)}}
\right]}_{=\alpha(x)}\underbrace{\left[ \limit{h}{0}{ \left(
\textcolor{red}{\frac{\beta(x+h)-\beta(x)}{h}} \right) }
\right]}_{=\beta^{\prime}(x)}\\
\amp{} + \underbrace{\left[
\limit{h}{0}{\textcolor{green}{\beta(x)}} \right]}_{\beta(x)}
\underbrace{\left[ \limit{h}{0}{ \left(
\textcolor{blue}{\frac{\alpha(x+h)-\alpha(x)}{h}} \right) }
\right]}_{\alpha^\prime(x)}\text{.}
\end{align*}
and therefore
\begin{equation*}
f^\prime(x)=\alpha(x)\beta^\prime(x)+\beta(x)\alpha^\prime(x).
\end{equation*}
