A calculus of the absurd

11.4 The product rule

Proving this is a little tricky, and needs some ingenuity. The product rule gives us a way to find the derivative of a function which is the product of two functions \(f(x) = a(x) \cdot b(x)\).

The trick here is to "add zero"

\begin{align} \lim _{h \to 0} \frac {f(x+h) - f(x)}{h} &= \frac {a(x+h)b(x+h) - a(x)b(x)}{h} \\ &= \lim _{h \to 0} \frac {a(x+h)b(x+h) - a(x+h)b(x) + a(x+h)b(x) - a(x)b(x)}{h} \\ &= \lim _{h \to 0} \frac {a(x+h)(b(x+h)-b(x)) + b(x)(a(x+h) - a(x)}{h} \\ &= \lim _{h \to 0} \frac {a(x+h)(b(x+h)-b(x))}{h} + \lim _{h \to 0} \frac {b(x)(a(x+h) - a(x)}{h} \\ &= \lim _{h \to 0} a(x+h) \frac {(b(x+h)-b(x))}{h} + \lim _{h \to 0} b(x) \frac {(a(x+h) - a(x)}{h} \end{align}

If (as it does) \(h \to 0\) then \(a(x + h) \to a(x)\), we can rewrite the limit as

\begin{align} a(x) \lim _{h \to 0} \frac {b(x+h)-b(x)}{h} + b(x) \lim _{h \to 0} \frac {a(x+h) - a(x)}{h} &= a(x) \frac {d}{dx} [b(x)] + b(x) \frac {d}{dx} [a(x)] + b(x) \end{align}

Overall, we therefore can say that the derivative of a function \(f(x) = a(x)b(x)\) is

\begin{equation} \frac {df}{dx} = \frac {d}{dx}[a(x)]b(x) + a(x)\frac {d}{dx}[b(x] \end{equation}