2.8 The Product Rule

If h(x) = f(x)·g(x), then h′(x) = f′(x)g(x) + f(x)g′(x). A handy way to say it: derivative of the first times the second, plus the first times the derivative of the second.

Because the two terms are added, the order doesn't matter. f(x)g′(x) + g(x)f′(x) is the same thing.

Picture a rectangle with width f(x) and height g(x), so its area is f(x)·g(x). When x changes a little, the width grows by about f′(x)·Δx and the height by about g′(x)·Δx. The new area adds a strip along the top (width times the height change) and a strip along the side (height times the width change), plus a tiny corner piece that becomes negligible. The two strips are the two terms of the product rule.

The exam often gives values of f, g, f′ and g′ in a table and asks for the derivative of a product at a point. Plug the table values into the rule: h′(a) = f′(a)g(a) + f(a)g′(a). Each of the four values must come from the same x-value, a.

Not every product needs the full rule. Look at the factors first:

• Use it when two non-constant functions of x are multiplied: x²eˣ, x sin x, eˣ ln x.
• Skip it when one factor is a constant: 5 sin x just needs the constant multiple rule.
• Consider expanding first when both factors are polynomials, like (x² + 1)(x − 3). Either method works; pick the faster one.
• For three factors, apply it twice: (fgk)′ = f′gk + fg′k + fgk′.

After using the rule, factoring helps when you need to solve h′(x) = 0. For h(x) = x²eˣ, h′(x) = 2xeˣ + x²eˣ = xeˣ(x + 2). Since eˣ is never 0, h′(x) = 0 only at x = 0 and x = −2.

Sometimes f and g are given as graphs made of straight segments. Then f(a) is the height of the graph at x = a and f′(a) is the slope of the segment that contains a. For h(x) = f(x)g(x), read all four values at the same x and use the rule.

If a is at a corner of one of the graphs, the derivative of that function doesn't exist there, so the product rule can't be used at that point.