The Product Rule — AP® Calculus AB Lesson

Part 1 · Hook

Why this matters

A store's revenue is price times the number of items sold. If the price rises while sales fall, how fast is revenue changing? Both factors are moving at once, and the answer is not the product of their rates. The product rule tells you exactly how two changing factors combine.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. $\frac{d}{dx}(\sin x)$ is

$\cos x$
$-\cos x$
$-\sin x$
$\sin x$

Show the answer

The slopes of sine are the values of cosine.

Correct: $\cos x$ :
$-\cos x$ :
$-\sin x$ :
$\sin x$ :

2. $\frac{d}{dx}\left( e^x \right)$ is

$e^x$
$xe^{x - 1}$
$e$
$1$

Show the answer

$e^x$ is its own derivative.

Correct: $e^x$ :
$xe^{x - 1}$ :
$e$ :
$1$ :

3. If $h(x) = 3f(x)$ , then $h'(x)$ is

$3f'(x)$
$3 + f'(x)$
$f'(x)$
$0$

Show the answer

Constant multiple rule: the factor 3 rides along.

Correct: $3f'(x)$ :
$3 + f'(x)$ :
$f'(x)$ :
$0$ :

Part 4 · See it

See it first

A rectangle with width f and height g has area f times g. When x changes a little, the width grows by delta f and the height by delta g. The new area adds a strip g delta f on the right, a strip f delta g along the bottom, and a tiny corner delta f delta g. Dividing by delta x and letting delta x go to 0 makes the corner vanish, which gives (fg) prime = f prime g + f g prime. — When both sides of a rectangle grow, its area $fg$ gains two strips, $g\,\Delta f$ and $f\,\Delta g$ , plus a tiny corner that vanishes in the limit. That is the product rule. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

Both factors of $f(x)g(x)$ change as $x$ changesthe change in the product has two main parts: $g$ times the change in $f$ , and $f$ times the change in $g$
The extra piece $\Delta f \, \Delta g$ is a product of two small changesafter dividing by $\Delta x$ it still goes to 0
Divide by $\Delta x$ and take the limit $(fg)' = f'g + fg'$
Three factorsdifferentiate one factor at a time and add: $(fgk)'$ $= f'gk + fg'k + fgk'$

Part 6 · Key ideas

Key ideas

The product rule: $\frac{d}{dx}\left( f(x)g(x) \right)$ $= f'(x)g(x)$ $+ f(x)g'(x)$ .
In words: derivative of the first times the second, plus the first times the derivative of the second.
The derivative of a product is NOT the product of the derivatives: $(x \cdot x)' = 2x$ , but $1 \cdot 1 = 1$ .
From a table: $(fg)'(a)$ $= f'(a)g(a) + f(a)g'(a)$ uses both the value columns and the derivative columns.
If one factor is a constant, the product rule reduces to the constant multiple rule.

Part 7 · Misconception

A common mistake

The wrong idea: $\frac{d}{dx}\left( f(x)g(x) \right)$ $= f'(x)g'(x)$ .

What actually happens: Try $f(x) = g(x) = x$ : the product is $x^2$ , whose derivative is $2x$ , but $f'g' = 1$ . Each factor's change is multiplied by the other factor's value, and the two contributions add: $(fg)' = f'g + fg'$ .

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Table

Values of $f$ , $g$ and their derivatives

The functions

f

and

g

have derivatives for all

x

.

Selected values of $f$ , $f'$ , $g$ and $g'$
$x$	$f(x)$	$f'(x)$	$g(x)$	$g'(x)$
0	2	3	5	$-1$
1	$-1$	2	3	4
3	4	$-2$	1	2

1. Let $h(x) = f(x)g(x)$ . Find $h'(1)$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$h'(1)$ $= f'(1)g(1)$ $+ f(1)g'(1)$ $= (2)(3)$ $+ (-1)(4)$ $= 6 - 4$ $= 2$ .

Answer: 2

2. Let $k(x) = x^2 f(x)$ . Find $k'(3)$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$k'(x)$ $= 2x\,f(x) + x^2 f'(x)$ , so $k'(3)$ $= 6(4) + 9(-2)$ $= 24 - 18$ $= 6$ .

Answer: 6

3. Which is an equation of the line tangent to the graph of $h(x) = f(x)g(x)$ at $x = 0$ ?

$y = 10 + 13x$
$y = 10 - 3x$
$y = 13 + 10x$
$y = 10 + 15x$

Show the answer

$h(0) = (2)(5) = 10$ and $h'(0)$ $= f'(0)g(0)$ $+ f(0)g'(0)$ $= (3)(5)$ $+ (2)(-1)$ $= 13$ .

Correct: $y = 10 + 13x$ : Height 10, slope 13.
$y = 10 - 3x$ : Slope $-3$ is $f'(0)g'(0)$ , the product of the derivatives.
$y = 13 + 10x$ : The height and the slope are swapped.
$y = 10 + 15x$ : Slope 15 is only the first term of the product rule.

4. Let $q(x) = e^x g(x)$ . Find $q'(0)$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$q'(x)$ $= e^x g(x) + e^x g'(x)$ , so $q'(0)$ $= (1)(5) + (1)(-1)$ $= 4$ .

Answer: 4

5. Use the product rule to differentiate $x^2 \sin x$ , a power of $x$ times a sine.

Type an expression for $\frac{d}{dx}\left( x^2 \sin x \right) =$ .

Show the answer

First $x^2$ , second $\sin x$ : $(2x)(\sin x) + (x^2)(\cos x)$ .

Answer: $\frac{d}{dx}\left( x^2 \sin x \right) =$ $2x\,\sin(x) + x^{2}\,\cos(x)$

6. Let $f(x) = x$ and $g(x) = x$ . Which comparison of $\left( f(x)g(x) \right)'$ and $f'(x)g'(x)$ is correct?

$(fg)' = 2x$ and $f'g' = 1$ ; they differ, so $(fg)' \ne f'g'$
Both equal $2x$ , so $(fg)' = f'g'$ for these two functions
Both equal $1$ , because $f' = 1$ and $g' = 1$ for these functions
$(fg)' = 1$ and $f'g' = 2x$ ; they differ, so $(fg)' \ne f'g'$

Show the answer

$fg = x^2$ , whose derivative is $2x$ ; $f'g' = 1 \cdot 1 = 1$ . By the product rule, $(fg)' = 1 \cdot x + x \cdot 1 = 2x$ .

Correct: $(fg)' = 2x$ and $f'g' = 1$ ; they differ, so $(fg)' \ne f'g'$ : One counterexample is enough.
Both equal $2x$ , so $(fg)' = f'g'$ for these two functions: $f'g' = 1 \cdot 1 = 1$ , not $2x$ .
Both equal $1$ , because $f' = 1$ and $g' = 1$ for these functions: $(fg)' = (x^2)' = 2x$ , not 1.
$(fg)' = 1$ and $f'g' = 2x$ ; they differ, so $(fg)' \ne f'g'$ : That swaps the two results.

Part 9 · Summary

Summary

When two factors both change, the product changes by each factor's change times the other factor:

(fg)' = f'g + fg'

. It is never just

f'g'

. For three factors, differentiate one at a time and add the three terms. From a table, the rule needs the values as well as the derivatives.

Part 10 · Up next

What comes next

Part 11 · Connections

Why this matters

What this builds on

Quick check before you start

See it first

How it works, step by step

Key ideas

A common mistake

Check yourself

Values of ff, gg and their derivatives

Summary

What comes next

Connections

Values of $f$ , $g$ and their derivatives