The product rule

Where it comes from

You already know the sum rule: the derivative of a sum is the sum of the derivatives, (f+g)′ = f′ + g′. It feels natural to guess the same pattern for products:

Kill it with one example. Take f(x) = x and g(x) = x:

So a product needs its own rule. Here is the intuition that makes the real rule obvious. Picture f and g as the sides of a rectangle, so fg is its area. Nudge x by a tiny amount h: f grows by Δf = f(x+h) − f(x), and g grows by Δg = g(x+h) − g(x). The new area is (f + Δf)(g + Δg). Expand it:

The product rule first appeared in Gottfried Wilhelm Leibniz’s 1684 paper Nova Methodus pro Maximis et Minimis — the founding paper of differential calculus. It needs no new axioms: everything above follows from the limit definition, as we’ll prove next.

Derivation

Let F(x) = f(x)·g(x), where f and g are differentiable at x (hence continuous — more on that below). We apply the limit definition of the derivative to F. Watch the second line: it is the whole proof.

Why does differentiability imply continuity? Write f(x+h) = f(x) + h·[f(x+h)−f(x)]/h. As h → 0, the bracket tends to f′(x), so f(x+h) → f(x) + 0·f′(x) = f(x). That one line is the whole proof — and it is exactly what lets us replace f(x+h) with f(x) in Step 4.

How to use it

The procedure, every time:

1. Confirm it is a product. Two expressions in x multiplied: x²·sin x. Not a composition like sin(x²) (that is the chain rule), not a quotient like (sin x)/x (that is the quotient rule).
2. Name the factors f and g. Order never matters — the rule is symmetric: f′g + fg′ is the same as fg′ + f′g.
3. Differentiate each factor separately. This is where most errors happen: differentiate f completely (chain rule inside if needed) before assembling.
4. Assemble: f′g + fg′. Count your terms — two factors, two terms. If you wrote one term, you dropped something.
5. Simplify. Factor out common terms; students lose more marks to unsimplified answers than to wrong rules.

Expand first, or rule first?

If both factors are simple polynomials, expanding first is often faster: d/dx [(x²−1)(x+4)] expands to d/dx [x³+4x²−x−4] = 3x²+8x−1. The product rule gives the same answer (check it in Example 4 below). Rule of thumb: expand when it is cheap, reach for the product rule when a factor is transcendental (sin, cos, e^x, ln x) or expansion would be ugly.

Three factors

Product meets chain

For d/dx [x²·sin(3x)] you need both rules: the product rule for the outer structure, and the chain rule inside sin(3x), whose derivative is 3 cos(3x). Full answer: 2x·sin(3x) + 3x²·cos(3x). Forgetting the ×3 is the classic error — see Example 3.

Worked examples

Four problems, easiest first. In each one, read every step — the why of each move is the lesson.

Memorization tips

• Say it aloud: “first times derivative of second, plus second times derivative of first.” The rhythm of the sentence matches the formula.
• Leibniz form: d(uv) = u dv + v du. In each term one factor “sleeps” while the other “wakes up” (gets differentiated). Picture the sleeping factor, then the waking one.
• The x² test (your 5-second self-check): f = g = x kills f′g′ instantly, since 2x ≠ 1. Run it in your head whenever you are unsure.
• The collapse check: if one factor is constant, the rule must collapse to the constant-multiple rule — (5f)′ = 5f′ + f·0 = 5f′. If your answer does not collapse, something is off.
• The rectangle picture: two strips survive, the corner square vanishes. That image is the formula — f·g′ + g·f′, one strip per factor.
• It is symmetric: f′g + fg′ — it never matters which factor you call f. If swapping the names changes your answer, you made an algebra slip.

Final challenge

Five mixed questions — basics, applications, and the traps, all in one. Score 5/5 and the product rule is yours.