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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.