A. Rate of Change · Derivatives and Differentiation Rules

The Product Rule

A3.4 A3.5About 70 minutes

Learning goals

I can explain why the derivative of a product is not the product of the derivatives.
I can verify the product rule by expanding polynomial products.
I can use the product rule to differentiate products of polynomial and radical functions.
I can interpret the product rule in a context such as revenue = price × quantity.

Words to know

product rule
If p(x) = f(x)g(x), then p′(x) = f′(x)g(x) + f(x)g′(x).

factor
One of the functions being multiplied in a product.

revenue
Income from sales: price per item × number of items sold.

counterexample
A single example that shows a general statement is false.

Products are everywhere

Revenue is price × number sold. The mass of a growing crystal is density × volume. The area of a rectangle is length × width. When both factors change, how fast does the product change?

It is tempting to guess that the derivative of a product is the product of the derivatives. It is not. Try y = x · x = x². The derivative is 2x, but multiplying the derivatives of x and x gives 1 · 1 = 1.

Why you cannot just multiply derivatives: for y = x · x, with f(x) = g(x) = x, the product of the derivatives is 1 · 1 = 1, but the tangent at (3, 9) has slope 6. The product rule gives 1(3) + 3(1) = 6. ✓

The product rule

If p(x) = f(x)g(x), where f and g are differentiable, then

p′(x) = f′(x)g(x) + f(x)g′(x)

In words: derivative of the first × the second, plus the first × derivative of the second. In Leibniz notation, if u and v are functions of x: d/dx (uv) = (du/dx)v + u(dv/dx).

Why it works

Think of f(x)g(x) as the area of a rectangle with side lengths f(x) and g(x). When x increases by h, the sides grow by Δf = f(x + h) − f(x) and Δg = g(x + h) − g(x). The extra area is

Δf · g + f · Δg + Δf · Δg.

Divide by h: (Δf/h) · g + f · (Δg/h) + (Δf/h) · Δg.

As h → 0, Δf/h → f′(x), Δg/h → g′(x), and Δg → 0, so the last term vanishes. What remains is f′(x)g(x) + f(x)g′(x).

A rectangle with sides f(x) and g(x) has area f(x)g(x). When x increases by h, the sides grow by Δf and Δg. The new area is made of the two strips Δf·g and f·Δg plus a tiny corner Δf·Δg, which shrinks away faster than h. That leaves the product rule.

Worked example

Example 1: Verifying the product rule

Differentiate p(x) = (x² + 1)(x³ − 2x) in two ways: by expanding first, and with the product rule. Check that the answers agree.

Expand first: p(x) = x⁵ − 2x³ + x³ − 2x = x⁵ − x³ − 2x, so p′(x) = 5x⁴ − 3x² − 2.
Product rule: f = x² + 1, f′ = 2x; g = x³ − 2x, g′ = 3x² − 2.
p′(x) = 2x(x³ − 2x) + (x² + 1)(3x² − 2) = 2x⁴ − 4x² + 3x⁴ − 2x² + 3x² − 2.
Collect: 5x⁴ − 3x² − 2.

Both methods give p′(x) = 5x⁴ − 3x² − 2. ✓

Worked example

Example 2: A tangent line

Find the equation of the tangent to y = (3x − 2)(x² + 5) at x = 1.

y′ = 3(x² + 5) + (3x − 2)(2x).
At x = 1: y′ = 3(6) + (1)(2) = 20.
y(1) = (1)(6) = 6.
Tangent: y − 6 = 20(x − 1), so y = 20x − 14.

y = 20x − 14. You do not need to simplify y′ before substituting a number.

Worked example

Example 3: A product with a radical

Differentiate y = √x (x² − 4) and find the slope at x = 4.

Let f = √x = x^(1/2), so f′ = 1/(2√x); g = x² − 4, so g′ = 2x.
y′ = (x² − 4)/(2√x) + √x(2x).
Common denominator 2√x: y′ = [x² − 4 + 2x · 2x]/(2√x) = (5x² − 4)/(2√x). (√x · √x = x was used: 2x√x = 4x²/(2√x).)
At x = 4: (80 − 4)/(2 · 2) = 76/4 = 19.

y′ = (5x² − 4)/(2√x), and the slope at x = 4 is 19. (Check without simplifying: 12/4 + 2(8) = 3 + 16 = 19 ✓.)

Worked example

Example 4: Revenue at a ski hill

Weeks after opening day, a ski hill in the Blue Mountains area charges p(t) = 60 + 2t dollars for a lift ticket and sells n(t) = 800 − 10t tickets per day. Find how fast the daily revenue R(t) = p(t)n(t) is changing at t = 5 weeks.

p′(t) = 2 dollars per week and n′(t) = −10 tickets per day, per week.
At t = 5: p = 70, n = 750.
R′(5) = p′n + pn′ = 2(750) + 70(−10) = 1500 − 700 = 800.

Daily revenue is increasing at $800 per week at t = 5. The price increase adds $1500 per week, and the drop in tickets sold takes away $700 per week.

Just for fun

The product rule is sometimes remembered as "left d-right plus right d-left": the left factor times the derivative of the right, plus the right factor times the derivative of the left. Because addition doesn't care about order, both versions are correct, which makes it one of the friendliest rules in calculus.

Hands-on activity

Grow a rectangle

You need: a sheet of graph paper, two pencils of different shades (or a pencil and a pen), a ruler and a calculator.

Draw a 6-square by 4-square rectangle in the corner of the graph paper. Its area is 24 squares.
Grow the width by 1 square and the height by 1 square, drawing the new strips in the second pencil. Count the new squares: the strip along the side, the strip along the top, and the corner square.
Now imagine the sides grow at 0.1 square per second (width) and 0.2 square per second (height). After 1 second the sides are 6.1 and 4.2. Compute the new area and the change.
Compare with the product rule: dA/dt = (dw/dt)h + w(dh/dt) = 0.1(4) + 6(0.2).

What to look for: in step 2 there are 4 + 6 + 1 = 11 new squares. In step 3 the area is 25.62, a change of 1.62; the product rule predicts 1.6 square units per second. The difference, 0.02 = 0.1 × 0.2, is the tiny corner piece.

Your turn

Differentiate y = x²(x + 3) using the product rule, then check by expanding.

Differentiate f(x) = (2x + 1)(x² − 3).

Differentiate g(x) = (x³ − x)(4 − x²).

Find the slope of y = (x + 2)(x² − x) at x = 2.

If f(1) = 3, f′(1) = −2, g(1) = 5 and g′(1) = 4, find (fg)′(1).

Differentiate y = √x (x + 1).

Why is (fg)′ = f′g′ wrong? Give a counterexample.

More practice: warm-up

State the product rule in words.

For y = x · x³, find y′ with the product rule.

If f = 5x and g = x², find f′g + fg′.

(fg)′(0) if f(0) = 2, f′(0) = 1, g(0) = 3, g′(0) = 0?

Differentiate y = 7(x² + 1). Do you need the product rule?

More practice: on your own

Find the equation of the tangent to y = (x² − 3)(2x + 1) at x = −1.

Find the points where y = x²(x − 3) has a horizontal tangent.

Differentiate y = (x² + 1)√x and evaluate at x = 1.

A rectangle's length is L = 10 + 2t cm and width is W = 6 − 0.5t cm (t in seconds). How fast is its area changing at t = 2?

Differentiate y = x⁻¹(x² + 4) with the product rule and simplify.

Challenge

Show that for three functions, (fgh)′ = f′gh + fg′h + fgh′. Use it to differentiate y = x(x + 1)(x + 2).

If y = f(x)·x², f(2) = 3 and f′(2) = −1, find y′(2).

Use the product rule to show that d/dx [f(x)]² = 2f(x)f′(x). Then find d/dx (x³ + 1)² two ways.

Check your understanding

Quiz

Which is the product rule?

What is the derivative of y = (x + 4)(x² − 1)?

f(2) = 4, f′(2) = 3, g(2) = −1, g′(2) = 5. What is (fg)′(2)?

In the rectangle picture of the product rule, why does the corner piece Δf · Δg not matter?

What is the slope of y = x²(x − 1) at x = 2?

Price p = 50 dollars rising at $2 per month; sales n = 300 items per month, falling at 10 items per month each month. How fast is monthly revenue changing?

For parents and teachers

Big idea (A3.4, A3.5): the derivative of a product is not the product of the derivatives. It is (fg)′ = f′g + fg′. Your teen should be able to verify the rule by expanding a polynomial product and to use it when expanding is impractical (for example, with a square root).

How to help: Ask your teen to explain the rectangle picture: if both sides of a rectangle grow, the area grows by two strips (and a tiny corner). Have them say the rule out loud while working: "derivative of the first times the second, plus…".

Common difficulties:

Multiplying the derivatives (f′g′).
Dropping a factor, for example writing f′g + g′.
Algebra slips when simplifying; it is fine to substitute numbers before simplifying when only a slope is needed.
Using the product rule with a constant factor (it works, but the constant multiple rule is quicker).