Lesson 2 of 6Derivative Rules

Unit 2 · Lesson 2 of 6

The Product Rule

The derivative of a product is NOT the product of the derivatives. The product rule, f′g + fg′, comes straight from a growing rectangle, and it handles products you could never expand.

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Key things to remember

1Product rule: $(f g)^{'} = f^{'} g + f g^{'}$ .
2 $(f g)^{'} \neq = f^{'} g^{'}$ : check with $f = g = x$ , where $(x^{2})^{'} = 2 x$ but $1 \cdot 1 = 1$ .
3Leibniz form: $\frac{d}{d x} (uv) = \frac{d u}{d x} v + u \frac{d v}{d x}$ .
4For three factors: $(f g h)^{'} = f^{'} g h + f g^{'} h + f g h^{'}$ .
5To find $f^{'} (a)$ , substitute $a$ into each of $f, g, f^{'}, g^{'}$ ; there is no need to simplify first.

The sum rule says derivatives of sums behave nicely, so it is tempting to guess that derivatives of products do too. They don't. Take $f (x) = x$ and $g (x) = x$ . Their product is $x^{2}$ , whose derivative is $2 x$ . But $f^{'} (x) \cdot g^{'} (x) = 1 \cdot 1 = 1$ . The guess is wrong for the simplest product there is. Products show up everywhere (revenue is price times quantity sold, area is length times width), so we need the correct rule.

\frac{d}{d x} [f (x) g (x)] = f^{'} (x) g (x) + f (x) g^{'} (x)

The product rule: derivative of the first times the second, plus the first times the derivative of the second.

Where the rule comes from: a growing rectangle

Fig 1A rectangle with sides f and g grows by delta f and delta g. The added area is f times delta g, plus g times delta f, plus the small corner delta f times delta g, which shrinks fastest. This leads to the product rule.

Think of the product $f \cdot g$ as the area of a rectangle with sides $f$ and $g$ . When $x$ changes by a small amount $Δ x$ , the sides change by $Δ f$ and $Δ g$ . The new area is made of the old rectangle plus three new strips: a long strip of area $f Δ g$ , another of area $g Δ f$ , and a tiny corner of area $Δ f Δ g$ .

\frac{Δ ( f g )}{Δ x} = f \frac{Δ g}{Δ x} + g \frac{Δ f}{Δ x} + Δ f \frac{Δ g}{Δ x} ⟶ f g^{'} + g f^{'} + 0 \cdot g^{'}

As delta x shrinks to 0, delta f also shrinks to 0, so the corner term vanishes and only the two strips are left.

Video3Blue1Brown animates the growing rectangle with sides f and g and shows why the tiny corner piece disappears, leaving the product rule. The second half of the video previews the chain rule.Watch on YouTube

Checking the rule by expanding

AProduct rule on (3x + 2)(2x − 1)

$f = 3 x + 2$ , so $f^{'} = 3$
$g = 2 x - 1$ , so $g^{'} = 2$
$f^{'} g + f g^{'} = 3 (2 x - 1) + (3 x + 2) (2)$
$= 6 x - 3 + 6 x + 4 = 12 x + 1$

BExpand first, then power rule

$(3 x + 2) (2 x - 1) = 6 x^{2} - 3 x + 4 x - 2$
$= 6 x^{2} + x - 2$
$\frac{d}{d x} (6 x^{2} + x - 2) = 12 x + 1$
Same derivative, as it must be.

For two small polynomials, expanding and using the power rule is perfectly good. The product rule earns its keep when expanding is long or impossible: $x^{2} x + 1$ , $x sin x$ and $x e^{x}$ cannot be multiplied out at all. It also saves work when you only need $f^{'} (a)$ : substitute $a$ into $f^{'} g + f g^{'}$ directly, without simplifying the general formula.

Set up a four-box grid

Before writing the rule, list the four pieces:

f

g

f^{'}

and

g^{'}

. Then cross-multiply:

f^{'}

pairs with

g

, and

f

pairs with

g^{'}

. Put brackets around every binomial so signs survive.

VideoJensenMath works through several product rule examples using the same f′g + fg′ pattern as this lesson.Watch on YouTube

Products that change over time

Suppose revenue is $R (t) = p (t) n (t)$ , the ticket price times the number of tickets sold. The product rule splits the change in revenue into two honest pieces: $p^{'} (t) n (t)$ is the effect of the price changing while the number of tickets is held fixed, and $p (t) n^{'} (t)$ is the effect of the number of tickets changing while the price is held fixed. If the price rises while ticket sales fall, the two pieces have opposite signs, and the rule tells you which wins.

Leibniz form: if $y = uv$ , then $\frac{d y}{d x} = \frac{d u}{d x} v + u \frac{d v}{d x}$ .
Three factors: $(f g h)^{'} = f^{'} g h + f g^{'} h + f g h^{'}$ , one term for each factor taking its turn being differentiated.
Order does not matter: $f^{'} g + f g^{'}$ and $f g^{'} + g f^{'}$ are the same, because addition and multiplication are commutative.

Worked examples

Reveal one step at a time and try to beat it to the next line.

Example 1

For

h (x) = (x^{2} + 1) (x^{3} - 2 x)

, find

h^{'} (2)

with the product rule, then check by expanding.

Example 2

For

f (x) = x (x^{2} - 3)

, find

f^{'} (4)

exactly.

Example 3

A theatre's ticket price is

p (t) = 40 + 2 t

dollars and its audience per show is

n (t) = 500 - 15 t

people, where

t

is in months. Revenue per show is

R (t) = p (t) n (t)

. How fast is revenue per show changing at

t = 4

Example 4

Find the points where

f (x) = x^{2} (x - 6)

has a horizontal tangent.

Where marks get lost

Writing $(f g)^{'} = f^{'} g^{'}$ . Test it on $x \cdot x$ : it predicts a derivative of $1$ , but $(x^{2})^{'} = 2 x$ .
Writing only one of the two terms, for example $f^{'} g$ , and forgetting $f g^{'}$ . Each factor needs its turn to be differentiated.
Dropping brackets: $3 (2 x - 1)$ written as $3 \cdot 2 x - 1$ . Bracket every binomial before multiplying.
Leaving a radical as $x$ and then differentiating it as if it were $x$ . Rewrite $x$ as $x^{\frac{1}{2}}$ before applying the power rule to that factor.

Flashcards 6

Product rule

Product rule in Leibniz form

(f g)^{'} = f^{'} g^{'}

Product rule for three factors

+ 2 more

Practice

Which is

\frac{d}{d x} [x^{3} (2 x + 1)]

For

f (x) = (2 x - 1) (x^{2} + 3)

, find

f^{'} (1)

A ticket price is

p (t) = 25 + t

dollars and the number of tickets sold per week is

n (t) = 800 - 20 t

, with

t

in weeks. Revenue is

R = p n

. Find

R^{'} (5)

in dollars per week.

dollars/week

In the growing-rectangle picture of the product rule, why does the small corner

Δ f Δ g

not appear in the final rule?

Match each product to its derivative.

A. $2 x - 1$
B. $2 x$
C. $3 x^{2} + 2 x$
D. $4 x^{3}$

x^{2} (x + 1)

x (x - 1)

(x + 1) (x - 1)

x^{3} \cdot x

Put the steps in order to find

h^{'} (a)

for

h (x) = f (x) g (x)

using the product rule.

1Differentiate each factor to get $f^{'}$ and $g^{'}$
2Substitute $x = a$ and evaluate
3Write $h^{'} (x) = f^{'} (x) g (x) + f (x) g^{'} (x)$
4Identify the two factors $f$ and $g$

f (x) = x^{2} (x - 6)

has a horizontal tangent at

x = 0

and at one other

x

-value. Find the other

x

-value.

Name the rule that says

\frac{d}{d x} (uv) = \frac{d u}{d x} v + u \frac{d v}{d x}

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