Subsection 2.5.2 Proof of the Product Rule (Theorem 2.4.3)
After the warm-up above, we will just jump straight in. Let \(P(x)=f(x)\, g(x)\text{,}\) the product of our two functions. The derivative of the product is given by
\begin{align*}
P'(x) &= \lim_{h \to 0} \frac{P(x+h)-P(x)}{h}
\end{align*}
Again we will focus on the numerator inside the limit and massage it into the form we need. To simplify these manipulations, define
\begin{align*}
F(h) &= \frac{f(x+h)-f(x)}{h} & \text{ and } &&
G(h) &= \frac{g(x+h)-g(x)}{h}.\\
\end{align*}
Then we can write
\begin{align*} f(x+h)&= f(x)+hF(h) &\text{and} && g(x+h)&=g(x)+hG(h).\\ \end{align*}We can also write
\begin{align*} f'(x) &= \lim_{h\to0} F(h) & \text{and}&& g'(x) &= \lim_{h\to0} G(h). \end{align*}So back to that numerator:
\begin{align*}
\amp P(x+h)-P(x)
=f(x+h)\cdot g(x+h)-f(x) \cdot g(x) & \text{substitute}\\
&=[f(x)+ hF(h)]\ [g(x)+hG(h)]-f(x) \cdot g(x) & \text{expand}\\
&=f(x)g(x) + f(x)\cdot hG(h) + hF(h)\cdot g(x) + h^2 F(h)\cdot G(h) -f(x) \cdot
g(x)\hskip-0.5in\\
&= f(x) \cdot hG(h) + hF(h) \cdot g(x) + h^2F(h) \cdot G(h).
\end{align*}
Armed with this we return to the definition of the derivative:
\begin{align*}
P'(x)
&=\lim_{h\to 0}\frac{P(x+h)-P(x)}{h}\\
&= \lim_{h\to 0} \frac{f(x) \cdot hG(h) + hF(h) \cdot g(x) + h^2 F(h) \cdot
G(h) }{h}\\
&= \left(\lim_{h\to 0} \frac{f(x) \cdot h G(h)}{h}\right)
+ \left(\lim_{h\to 0} \frac{h F(h) \cdot g(x)}{h}\right)
+ \left(\lim_{h\to 0} \frac{h^2 F(h) \cdot G(h) }{h}\right)\\
&= \left(\lim_{h\to 0} f(x) \cdot G(h)\right)
+ \left(\lim_{h\to 0} F(h) \cdot g(x)\right)
+ \left(\lim_{h\to 0} h F(h) \cdot G(h)\right)\\
\end{align*}
Now since \(f(x)\) and \(g(x)\) do not change as we send \(h\) to zero, we can pull them outside. We can also write the third term as the product of 3 limits:
\begin{align*} &= \left(f(x) \lim_{h\to 0} G(h)\right) + \left(g(x) \lim_{h\to 0} F(h)\right) + \left(\lim_{h\to 0} h\right) \cdot \left(\lim_{h\to0} F(h)\right) \cdot \left(\lim_{h\to0} G(h)\right)\\ &= f(x) \cdot g'(x) + g(x)\cdot f'(x) + 0 \cdot f'(x) \cdot g'(x)\\ &= f(x) \cdot g'(x) + g(x) \cdot f'(x). \end{align*}And so we recover the product rule.