The Product Rule: Study Guide (Calculus AB 2.8)

Course scope. This topic is shared by Calculus AB and Calculus BC. Everything here is examinable for both.

A note on notation

This page has no equation renderer. f′(x) and d/dx f(x) both mean “the derivative of f with respect to x”. (f · g)(x) means f(x) · g(x). Limits are written lim (h → 0).

The rule

You can now differentiate x³, sin x and eˣ on their own. But what about x³ sin x, or eˣ ln x? These are products of two functions, and you need a new rule.

Product rule. If f and g are differentiable at x, then f · g is differentiable at x and

(f · g)′(x) = f′(x) · g(x) + f(x) · g′(x)

In words: differentiate the first factor and keep the second, then add the first factor times the derivative of the second. Each term changes exactly one factor.

A quick example: for y = x³ sin x, take f(x) = x³ and g(x) = sin x. Then f′(x) = 3x² and g′(x) = cos x, so

dy/dx = 3x² sin x + x³ cos x

Why the derivative of a product is not f′ · g′

It is tempting to guess that (f · g)′ = f′ · g′. That guess is wrong. Test it on a product you can also expand.

Let f(x) = x² + 1 and g(x) = x³ − x. Expanding first:

f(x) · g(x) = x⁵ − x³ + x³ − x = x⁵ − x, so the derivative is 5x⁴ − 1.

The product rule gives the same answer:

2x(x³ − x) + (x² + 1)(3x² − 1) = 2x⁴ − 2x² + 3x⁴ − x² + 3x² − 1 = 5x⁴ − 1 ✓

The wrong guess gives f′ · g′ = 2x(3x² − 1) = 6x³ − 2x. That is a different function. At x = 2, the true derivative is 79, but the guess gives 44.

Why the rule works

The area picture

Think of f(x) and g(x) as the side lengths of a rectangle, so the product f · g is its area. Now let x increase a little. The width grows by Δf and the height by Δg.

The extra area is made of three pieces:

change in area = g · Δf + f · Δg + Δf · Δg

A numerical illustration: if f = 3 and g = 5 grow to 3.2 and 5.1, the area changes from 15 to 16.32, a change of 1.32. The pieces are g · Δf = 5(0.2) = 1.0, f · Δg = 3(0.1) = 0.3, and the corner Δf · Δg = 0.02. The corner is tiny compared with the strips.

Divide by the small change in x. The strips become g · (Δf/Δx) and f · (Δg/Δx), which tend to g · f′ and f · g′. The corner becomes Δf · (Δg/Δx), which tends to 0 · g′ = 0. What is left is the product rule.

The limit proof

The same idea in symbols uses one trick: subtract and add f(x + h) · g(x) in the numerator.

[f(x + h)g(x + h) − f(x)g(x)]/h = [f(x + h)g(x + h) − f(x + h)g(x) + f(x + h)g(x) − f(x)g(x)]/h = f(x + h) · [g(x + h) − g(x)]/h + g(x) · [f(x + h) − f(x)]/h

As h → 0, the difference quotients tend to g′(x) and f′(x). Also f(x + h) → f(x), because f is differentiable at x and so it is continuous there (Topic 2.4). The limit is f(x)g′(x) + g(x)f′(x), the product rule.

When to use it, and when not to

Situation	Best approach
Constant times a function, e.g. 5 sin x	Constant multiple rule: 5 cos x
Product of two polynomials, e.g. (x² + 1)(x³ − x)	Product rule, or expand first; both are valid
Two different kinds of function, e.g. x² eˣ, eˣ cos x, x ln x	Product rule (no way to expand)
A function whose input is not plain x, e.g. sin(3x)	Chain rule (Topic 3.1), not the product rule
A quotient, e.g. sin x / x	Quotient rule (Topic 2.9)

The product rule includes the constant multiple rule as a special case. If f(x) = c, then f′ = 0 and (c · g)′ = 0 · g + c · g′ = c · g′.

Three factors. Use the rule twice. Group u · v · w as (u · v) · w:

(u · v · w)′ = u′ · v · w + u · v′ · w + u · v · w′

Each term differentiates one factor and leaves the other two alone.

Worked example 1: a product with eˣ and a tangent line

Question. Let h(x) = (x³ − 2x)eˣ. (a) Find h′(x). (b) Find the tangent line to the graph of h at x = 0. (c) Show that the graph of h has a horizontal tangent at x = 1.

(a)

1. Name the factors. f(x) = x³ − 2x and g(x) = eˣ.
2. Differentiate each. f′(x) = 3x² − 2 and g′(x) = eˣ.
3. Apply the rule. h′(x) = (3x² − 2)eˣ + (x³ − 2x)eˣ.
4. Tidy up. Factor out eˣ: h′(x) = eˣ(x³ + 3x² − 2x − 2).

(b)

1. h(0) = (0 − 0)e⁰ = 0, so the point is (0, 0).
2. h′(0) = e⁰(0 + 0 − 0 − 2) = −2.
3. The tangent line is y = −2x.

Check. [h(0.001) − h(0)]/0.001 ≈ −2.002, close to −2.

(c) h′(1) = e¹(1 + 3 − 2 − 2) = e · 0 = 0. A slope of 0 means a horizontal tangent. The point is (1, h(1)) = (1, −e).

Tip. Factoring out eˣ helps because eˣ is never 0. So h′(x) = 0 exactly when the polynomial factor is 0.

Worked example 2: the product rule with a table

Exam questions often give values instead of formulas. The rule works the same way: substitute the numbers.

x	f(x)	f′(x)	g(x)	g′(x)
2	3	−1	4	5

Question. (a) Let P(x) = f(x) · g(x). Find P′(2). (b) Let Q(x) = x² · f(x). Find Q′(2).

(a)

1. Write the rule at x = 2: P′(2) = f′(2) · g(2) + f(2) · g′(2).
2. Substitute: P′(2) = (−1)(4) + (3)(5) = −4 + 15 = 11.

The wrong rule f′(2) · g′(2) would give (−1)(5) = −5. Also note that P(2) = 3 × 4 = 12 is the value of the product, not its slope.

(b)

1. Here the factors are x² and f(x). d/dx (x²) = 2x.
2. Rule: Q′(x) = 2x · f(x) + x² · f′(x).
3. Substitute x = 2: Q′(2) = 2(2)(3) + (2²)(−1) = 12 − 4 = 8.

Interpretation. P′(2) = 11 means that at x = 2 the product f · g is increasing at a rate of 11 units per unit of x.

Worked example 3: finding a horizontal tangent

Question. Let y = x ln x for x > 0. Find the point where the tangent line is horizontal.

1. Factors: f(x) = x, g(x) = ln x. Derivatives: f′(x) = 1, g′(x) = 1/x.
2. Rule: dy/dx = 1 · ln x + x · (1/x) = ln x + 1.
3. Set the slope to 0: ln x + 1 = 0, so ln x = −1 and x = 1/e ≈ 0.368.
4. Height: y = (1/e) · ln(1/e) = (1/e)(−1) = −1/e.

Answer. The tangent is horizontal at (1/e, −1/e).

Check the simplification. x · (1/x) = 1 for every x > 0. Students sometimes write x · (1/x) = 0 or leave it unsimplified and miss the clean answer.

Common misconceptions

• “(f · g)′ = f′ · g′.” The most common error. Each term of the true rule differentiates only one factor.
• Dropping a term. Writing only f′ · g, as if g were a constant. This happens with products like x² sin x, where sin x is treated as fixed.
• Sign slips inside the rule. For x⁴ cos x the second term is x⁴ · (−sin x). Keep the minus sign from d/dx cos x = −sin x.
• Subtracting instead of adding. The product rule has a plus sign. The minus sign belongs to the quotient rule in Topic 2.9.
• Using the product rule on a constant multiple, then making an error. For 7eˣ, just write 7eˣ. Extra steps add chances to slip.
• Confusing the value with the slope in a table question. f(a) · g(a) is the value of the product. The slope needs f′(a) and g′(a) as well.
• Using the product rule for a composition. sin(x²) is not sin times x². It needs the chain rule (Topic 3.1).
• Not simplifying. Factor out a common factor such as eˣ before solving h′(x) = 0.

Where this leads

The product rule is the first rule that combines two functions by multiplication. Next, Topic 2.9, the quotient rule, handles division, and Topic 2.10 uses it for tan x, sec x, cot x and csc x. In Unit 3 the chain rule handles compositions, and many later problems need the product rule and chain rule together. For the four derivatives you combined here, see Topic 2.7. Students in either course can return to the Calculus AB roadmap or the Calculus BC roadmap.

Try the practice questions now, then use the revision notes and the checklist to consolidate.