Why can't I just multiply the derivatives of each factor?

Because that gives the wrong answer. Test it with x· x = x^2: the true derivative is 2x, but multiplying derivatives 1· 1 gives 1. The product rule accounts for both factors changing, producing a sum of two terms, not a single product of derivatives.

How do I know when to use the product rule versus just simplifying?

If the two factors can be multiplied out or combined into a simpler form (like x^2· x^3 = x^5 or x(x+1)=x^2+x), simplify first and use the power rule or linearity. Reserve the product rule for genuine products of functions you cannot merge, such as x^2 e^x or x x.

Does the order of the factors matter in the product rule?

No. Since f'g+fg' = g'f+gf', you get the same result regardless of which factor you call f and which you call g. Just be consistent within a single problem and keep exactly one factor undifferentiated in each of the two terms.

How do I apply the product rule to three factors?

Differentiate one factor at a time and add the results: d/dx[fgh]=f'gh+fg'h+fgh'. Each term differentiates a single factor while leaving the other two alone. The pattern extends to any number of factors.

AP-CALCBC-2.8

U2.8 The Product Rule

Master the AP Calculus BC product rule: derive it, apply it to products of functions, and know when it's required versus simple linearity.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on U2.8 The Product Rule, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

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What this lesson covers

You already know how to differentiate sums, powers, and basic functions like

\sin x

and

e^x

. But what about a product like

x^2 e^x

? A common trap is to differentiate each factor and multiply the results — that is wrong. The derivative of a product is not the product of the derivatives.

This lesson introduces the product rule, one of the most heavily used differentiation tools on the AP exam. You will learn to state the rule precisely, apply it cleanly to two- and three-factor products, and — just as important — recognize when you actually need it versus when linearity alone finishes the job.

Stating the Product Rule

The product rule tells you how to differentiate a function written as the product of two differentiable functions. If

h(x) = f(x)\cdot g(x)

, then

h'(x) = f'(x)\,g(x) + f(x)\,g'(x).

In words: differentiate the first factor times the second, plus the first factor times the derivative of the second. The order does not matter since multiplication is commutative, but the structure — keep one factor undifferentiated in each term — is essential.

The single most common error is assuming

\frac{d}{dx}[f\cdot g] = f'\cdot g'

. This is false. A quick sanity check: let

f(x)=g(x)=x

, so

h(x)=x^2

. The correct derivative is

2x

. The product rule gives

1\cdot x + x\cdot 1 = 2x

, which matches. The wrong "multiply the derivatives" method gives

1\cdot 1 = 1

, which is clearly wrong.

Why does the rule have two terms? Intuitively, when both factors change, the product changes because of the change in

f

(weighted by the current

g

) and because of the change in

g

(weighted by the current

f

). Each factor contributes its own rate of change, so you get a sum, not a product.

When Do You Actually Need It?

A major skill the AP exam tests is deciding whether the product rule is even necessary. Sometimes an expression looks like a product but can be simplified first, letting you use the power rule or linearity instead — which is faster and less error-prone.

Expression	Best approach	Reason
$x^2 e^x$	Product rule	Two genuine functions multiplied; cannot simplify
$3x^2$	Constant multiple rule	A constant times a function, not two functions
$x^2\cdot x^3$	Simplify to $x^5$ , power rule	Combines into a single power
$x(x+1)$	Expand to $x^2+x$ , then linearity	Distributing is easier than product rule
$\sin x\cos x$	Product rule (or identity)	Two nonpolynomial factors

The rule for a constant multiple,

\frac{d}{dx}[c\cdot f(x)] = c\,f'(x)

, is not the product rule — a constant has derivative zero, so applying the product rule just recovers the constant multiple rule. Reserve the product rule for two factors that both actually depend on

x

.

On multiple-choice questions, if you can expand or simplify in one step, do it; you will usually reach the answer faster. On free-response, the product rule is often required because factors like

e^x

\ln x

, or trig functions cannot be merged.

Applying the Rule Step by Step

A reliable procedure prevents mistakes. Identify the two factors, label them

f

and

g

, compute each derivative separately, then assemble the two terms.

Consider

h(x) = x^3\sin x

. Let

f(x)=x^3

and

g(x)=\sin x

. Then

f'(x)=3x^2

and

g'(x)=\cos x

. Assemble:

h'(x) = 3x^2\sin x + x^3\cos x.

Notice you do not combine these terms — they are structurally different (one has

\sin x

, one has

\cos x

), so the answer stays as a sum.

When a coefficient is present, keep it attached to its factor. For

h(x)=5x^2 e^x

, treat the constant

5

as riding along:

f=5x^2

gives

f'=10x

, and

g=e^x

gives

g'=e^x

, so

h'(x)=10x\,e^x + 5x^2 e^x = 5x e^x(2+x)

. Factoring at the end is optional but often helps when the problem asks where

h'(x)=0

.

For three factors, apply the rule twice:

\frac{d}{dx}[fgh] = f'gh + fg'h + fgh'

. Each term differentiates exactly one factor. This pattern generalizes and is worth memorizing, since triple products occasionally appear in BC problems.

Combining the Product Rule with Other Rules

AP problems rarely use the product rule in isolation. You will pair it with the derivatives of

e^x

\ln x

, and trig functions from U2.7, and later with the chain rule. Watch for products hidden inside larger expressions.

A frequent exam setup gives you a table of values for

f

g

f'

, and

g'

at a specific point and asks for

(fg)'

at that point. You never need formulas — just plug in. For example, if

f(2)=3

f'(2)=-1

g(2)=4

g'(2)=5

, then

(fg)'(2)=f'(2)g(2)+f(2)g'(2)=(-1)(4)+(3)(5)=11

.

Another classic: finding the equation of a tangent line to

y=f(x)g(x)

requires the product rule to get the slope, then point-slope form. Or you may need to locate horizontal tangents by setting the product-rule derivative equal to zero and solving.

A subtle misconception: students sometimes forget that after applying the product rule, individual factor derivatives may themselves require another rule. If a factor is

\sin(2x)

, its derivative uses the chain rule (

2\cos(2x)

), which you will learn next. For now, keep factors simple, but stay alert that the product rule tells you the overall structure, while each

f'

and

g'

is computed with whatever rule that factor demands.

Key terms

Product Rule.: The formula $\frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x)$ for differentiating a product of two functions.
Factor.: One of the functions being multiplied in a product. In $x^2 e^x$ , the factors are $x^2$ and $e^x$ .
Constant Multiple Rule.: $\frac{d}{dx}[c\,f(x)] = c\,f'(x)$ ; used when a constant multiplies a function, distinct from the product rule.
Linearity of the Derivative.: The property that the derivative of a sum is the sum of derivatives, and constants factor out; sufficient for sums but not for products.
Differentiable.: A function has a derivative at a point; both factors in the product rule must be differentiable for the rule to apply.
Horizontal Tangent.: A point where the derivative equals zero; found by setting the product-rule result to zero and solving.

Worked example

Let

h(x) = (2x^2 - 1)e^x

. Find

h'(x)

and determine all

x

where the tangent line to

h

is horizontal.

Identify the two factors. Let

f(x) = 2x^2 - 1

and

g(x) = e^x

, since both depend on

x

and cannot be merged into a single simpler function.

Compute each derivative separately. Using the power rule and linearity,

f'(x) = 4x

. Using the exponential rule,

g'(x) = e^x

.

Apply the product rule:

h'(x) = f'(x)g(x) + f(x)g'(x) = 4x\,e^x + (2x^2 - 1)e^x

.

Factor out the common

e^x

to simplify:

h'(x) = e^x(4x + 2x^2 - 1) = e^x(2x^2 + 4x - 1)

.

For horizontal tangents, set

h'(x) = 0

. Since

e^x > 0

for all

x

, it can never be zero, so we only need

2x^2 + 4x - 1 = 0

.

Use the quadratic formula:

x = \frac{-4 \pm \sqrt{16 + 8}}{4} = \frac{-4 \pm \sqrt{24}}{4} = \frac{-4 \pm 2\sqrt{6}}{4} = \frac{-2 \pm \sqrt{6}}{2}

.

So horizontal tangents occur at

x = \frac{-2 + \sqrt{6}}{2}

and

x = \frac{-2 - \sqrt{6}}{2}

Practice questions

h(x) = x^2\cos x

, what is

h'(x)

$2x\cos x - x^2\sin x$
$2x\cos x + x^2\sin x$
$-2x\sin x$
$2x\sin x$

Answer: $2x\cos x - x^2\sin x$

Let

f(x)=x^2

f'(x)=2x

, and

g(x)=\cos x

g'(x)=-\sin x

. The product rule gives

h'(x)=f'g+fg'=2x\cos x + x^2(-\sin x)=2x\cos x - x^2\sin x

. The negative sign comes from the derivative of cosine — a frequent slip.

The functions

f

and

g

satisfy

f(1)=2

f'(1)=3

g(1)=-4

, and

g'(1)=5

. If

p(x)=f(x)g(x)

, find

p'(1)

Answer: $p'(1) = -2$

Apply the product rule at

x=1

p'(1)=f'(1)g(1)+f(1)g'(1)=(3)(-4)+(2)(5)=-12+10=-2

. No formulas for

f

g

are needed — just substitute the given values into the product rule structure.

For which expression is the product rule the most efficient method, rather than simplifying first?

$x^3 \cdot x^4$
$4x^2$
$e^x \ln x$
$x(2x - 5)$

Answer: $e^x \ln x$

x^3\cdot x^4=x^7

(power rule),

4x^2

is a constant multiple, and

x(2x-5)=2x^2-5x

expands easily. Only

e^x\ln x

is a product of two functions that cannot be combined or simplified, so the product rule is genuinely required.

FAQ

Why can't I just multiply the derivatives of each factor?: Because that gives the wrong answer. Test it with $x\cdot x = x^2$ : the true derivative is $2x$ , but multiplying derivatives $1\cdot 1$ gives $1$ . The product rule accounts for both factors changing, producing a sum of two terms, not a single product of derivatives.
How do I know when to use the product rule versus just simplifying?: If the two factors can be multiplied out or combined into a simpler form (like $x^2\cdot x^3 = x^5$ or $x(x+1)=x^2+x$ ), simplify first and use the power rule or linearity. Reserve the product rule for genuine products of functions you cannot merge, such as $x^2 e^x$ or $\sin x \ln x$ .
Does the order of the factors matter in the product rule?: No. Since $f'g+fg' = g'f+gf'$ , you get the same result regardless of which factor you call $f$ and which you call $g$ . Just be consistent within a single problem and keep exactly one factor undifferentiated in each of the two terms.
How do I apply the product rule to three factors?: Differentiate one factor at a time and add the results: $\frac{d}{dx}[fgh]=f'gh+fg'h+fgh'$ . Each term differentiates a single factor while leaving the other two alone. The pattern extends to any number of factors.

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