Product Rule: Formula & Worked Examples | AP Calculus 2.8

1. The product rule: formula and meaning

Core concept: Sums use linearity, but a product requires a different rule. Each factor can change, so the derivative accounts for both contributions. Learn the product rule, justify it with limits, and combine it with the power rule and trigonometric, exponential and logarithm derivatives.

A common error: multiplying the two derivatives

The derivative of a product is not generally the product of its derivatives. This is a statement about the rule, not a claim that the two expressions can never agree at an isolated input.

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

Example showing why this is wrong:

Let $f (x) = x^{2}$ and $g (x) = x^{3}$

Correct approach: $f (x) \cdot g (x) = x^{5}$ , so $\frac{d}{d x} [x^{5}] = 5 x^{4}$
Wrong approach: $f^{'} (x) \cdot g^{'} (x) = 2 x \cdot 3 x^{2} = 6 x^{3}$ (different answer!)

Use the product rule, or simplify the product first when that gives an easier valid method. Here, combining the powers is especially efficient.

THE PRODUCT RULE

If $u$ and $v$ are differentiable functions of $x$ , then:

\frac{d}{d x} [u \cdot v] = u \cdot \frac{d v}{d x} + v \cdot \frac{d u}{d x}

Alternative Notation:

\frac{d}{d x} [u \cdot v] = u \cdot v^{'} + v \cdot u^{'}

Or using $f$ and $g$ :

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

Shorthand notation:

(u v)^{'} = u v^{'} + v u^{'}

MEMORY AIDS FOR THE PRODUCT RULE

Method 1: The Chant

"First d Second, plus Second d First"

Translation: (First function) × (derivative of Second) + (Second function) × (derivative of First)

Method 2: The Pattern

Take the first function, keep it. Differentiate the second.

Then ADD: Take the second function, keep it. Differentiate the first.

Method 3: Left-Right Mnemonic

"Left d Right, plus Right d Left"

Product Rule in Plain English:

To find the derivative of a product of two functions:

Keep the first function unchanged
Multiply by the derivative of the second function
Add (+)
Keep the second function unchanged
Multiply by the derivative of the first function

Conditions and interpretation: If $u$ and $v$ are differentiable at the same input $a$ , their product is differentiable there and $(u v)^{'} (a) = u^{'} (a) v (a) + u (a) v^{'} (a)$ . Differentiability of both factors is a sufficient condition for this rule. It is not necessary for every product to be differentiable: $| x | \cdot | x | = x^{2}$ is differentiable at zero even though the separate factors are not.

For trigonometric factors use radians. If a factor is $\ln x$ , the real input must satisfy $x > 0$ . Every evaluation must belong to the original product's domain.

2. When to use the product rule or simplify first

Use the Product Rule when:

Two functions are multiplied: $x^{2} \cdot \sin x$ , $e^{x} \cdot \ln x$ , $(3 x + 1) \cdot \cos x$
No useful expansion: Products mixing a polynomial with a trigonometric, exponential or logarithmic function usually benefit from the product rule.
Expanding would be messier: Even if you could expand, Product Rule might be easier

Consider a shorter method when:

Expansion is convenient: $(x + 1) (x - 2) = x^{2} - x - 2$ , so the power rule is a short alternative. The product rule remains valid.
One factor is constant: $5 x^{3}$ is a product, but the constant multiple rule is shorter. The product rule also works because the derivative of five is zero.
Powers combine: $x \cdot x^{2} = x^{3}$ , so combine them before differentiating if convenient. Both methods give the same result.

Products, sums, quotients and compositions

Structure	Example	Choose a method
Sum	$x^{2} + \sin x$	Differentiate term by term using linearity.
Product	$x^{2} \sin x$	Account for each changing factor with the product rule.
Quotient	$\frac{\sin x}{x}$ , $x \neq 0$	The quotient rule is a useful next technique.
Composition	$\sin (x^{2})$	The chain rule in Unit 3 handles an inner function.

For a constant $c$ , the product rule reduces to the familiar constant multiple rule: $\frac{d}{d x} [c u] = c u^{'} + u \cdot 0 = c u^{'}$ .

3. Six worked examples with full steps

Example 1: Basic Product (Polynomial × Trig)

Problem: Find $\frac{d}{d x} [x^{2} \sin x]$

Solution:

Identify the two functions:
- First function: $u = x^{2}$
- Second function: $v = \sin x$
Find each derivative:
- $u^{'} = 2 x$
- $v^{'} = \cos x$
Apply Product Rule: First d Second + Second d First
$\frac{d}{d x} [x^{2} \sin x] = x^{2} \cdot \cos x + \sin x \cdot 2 x$
Simplify (optional):
$= x^{2} \cos x + 2 x \sin x$

Or factor: $= x (x \cos x + 2 \sin x)$

Answer: $x^{2} \cos x + 2 x \sin x$

Example 2: Exponential × Polynomial

Problem: Differentiate $f (x) = e^{x} \cdot x^{3}$

Solution:

Identify: $u = e^{x}$ , $v = x^{3}$
Derivatives: $u^{'} = e^{x}$ , $v^{'} = 3 x^{2}$
Product Rule:
$f^{'} (x) = e^{x} \cdot 3 x^{2} + x^{3} \cdot e^{x}$
Factor out $e^{x}$ :
$f^{'} (x) = e^{x} (3 x^{2} + x^{3})$

Can further factor: $= e^{x} \cdot x^{2} (3 + x)$

Answer: $e^{x} (3 x^{2} + x^{3})$ or $e^{x} x^{2} (x + 3)$

Example 3: Both Functions are Trig

Problem: Find $g^{'} (x)$ for $g (x) = \sin x \cdot \cos x$

Solution:

Identify: $u = \sin x$ , $v = \cos x$
Derivatives: $u^{'} = \cos x$ , $v^{'} = - \sin x$ (don't forget negative!)
Product Rule:
$g^{'} (x) = \sin x \cdot (- \sin x) + \cos x \cdot \cos x$
Simplify:
$g^{'} (x) = - \sin^{2} x + \cos^{2} x = \cos^{2} x - \sin^{2} x$
Bonus identity: This equals $\cos (2 x)$ ! (double angle formula)

Answer: $\cos^{2} x - \sin^{2} x$ or $\cos (2 x)$

Example 4: Polynomial × Natural Log

Problem: Find $\frac{d y}{d x}$ for $y = x^{3} \ln x$

Solution:

Identify: $u = x^{3}$ , $v = \ln x$
Derivatives: $u^{'} = 3 x^{2}$ , $v^{'} = \frac{1}{x}$
Product Rule:
$\frac{d y}{d x} = x^{3} \cdot \frac{1}{x} + \ln x \cdot 3 x^{2}$
Simplify:
- $x^{3} \cdot \frac{1}{x} = x^{2}$
- Result: $x^{2} + 3 x^{2} \ln x$
Factor (optional): $= x^{2} (1 + 3 \ln x)$

Answer: $x^{2} + 3 x^{2} \ln x$ or $x^{2} (1 + 3 \ln x)$

Example 5: More Complex Expression

Problem: Differentiate $h (x) = (2 x^{2} - 3 x + 1) \cdot e^{x}$

Solution:

Identify: $u = 2 x^{2} - 3 x + 1$ , $v = e^{x}$
Derivatives:
- $u^{'} = 4 x - 3$
- $v^{'} = e^{x}$
Product Rule:
$h^{'} (x) = (2 x^{2} - 3 x + 1) \cdot e^{x} + e^{x} \cdot (4 x - 3)$
Factor out $e^{x}$ :
$h^{'} (x) = e^{x} [(2 x^{2} - 3 x + 1) + (4 x - 3)]$
Combine like terms:
$h^{'} (x) = e^{x} (2 x^{2} + x - 2)$

Answer: $e^{x} (2 x^{2} + x - 2)$

Example 6: Compare product rule with expansion

Problem: Find $\frac{d}{d x} [(x + 2) (x - 3)]$

Option 1: Product Rule (works but more work)

$u = x + 2$ , $v = x - 3$
$u^{'} = 1$ , $v^{'} = 1$
$(x + 2) (1) + (x - 3) (1) = x + 2 + x - 3 = 2 x - 1$

Option 2: Expand First (easier!)

Expand: $(x + 2) (x - 3) = x^{2} - x - 6$
Differentiate: $\frac{d}{d x} [x^{2} - x - 6] = 2 x - 1$

Answer: $2 x - 1$ (same result, but expanding first was simpler!)

Lesson: When factors are simple polynomials, compare expansion with the product rule and choose a clear, efficient method. Neither method is compulsory.

Domain and sign check: Examples 1, 2, 3, 5 and 6 have all real inputs as their domains. Example 4 contains $\ln x$ , so the original product and derivative require $x > 0$ . Trigonometric inputs use radians. In Example 3, cosine's negative derivative creates the negative sine-square term; the connecting product-rule operation is still addition.

4. Proof using the derivative definition

Theorem: If $u$ and $v$ are differentiable functions, then $\frac{d}{d x} [u v] = u v^{'} + v u^{'}$

Proof (Using Limit Definition):

Start with the limit definition:
$\frac{d}{d x} [u (x) v (x)] = lim_{h \to 0} \frac{u (x + h) v (x + h) - u (x) v (x)}{h}$
Add and subtract $u (x + h) v (x)$ in the numerator (clever trick!):
$= lim_{h \to 0} \frac{u (x + h) v (x + h) - u (x + h) v (x) + u (x + h) v (x) - u (x) v (x)}{h}$
Factor and separate:
$= lim_{h \to 0} \frac{u (x + h) [v (x + h) - v (x)] + v (x) [u (x + h) - u (x)]}{h}$
Split into two limits:
$= lim_{h \to 0} u (x + h) \cdot \frac{v (x + h) - v (x)}{h} + lim_{h \to 0} v (x) \cdot \frac{u (x + h) - u (x)}{h}$
Evaluate limits:
- $lim_{h \to 0} u (x + h) = u (x)$ (continuity)
- $lim_{h \to 0} \frac{v (x + h) - v (x)}{h} = v^{'} (x)$ (definition of derivative)
- $lim_{h \to 0} \frac{u (x + h) - u (x)}{h} = u^{'} (x)$ (definition of derivative)
Final result:
$\frac{d}{d x} [u v] = u \cdot v^{'} + v \cdot u^{'} ✓$

Why the limit step is justified: Differentiability at the fixed input implies continuity there, so $lim_{h \to 0} u (x + h) = u (x)$ . The two difference quotients have finite limits by differentiability. These facts permit the sum and product limit laws used above. See the derivative definition and limit laws.

Adding and subtracting the intermediate product is an exact algebraic step for nonzero $h$ . It separates the change in the second factor from the change in the first; simply splitting the original numerator does not supply the two derivative terms.

5. Products of three or more functions

Product Rule for Three Functions

If you have three functions multiplied together:

\frac{d}{d x} [u v w] = u^{'} v w + u v^{'} w + u v w^{'}

Pattern: Differentiate one factor at a time, keeping the others unchanged, then sum all terms.

Example: Find $\frac{d}{d x} [x^{2} \cdot \sin x \cdot e^{x}]$

$u = x^{2}$ , $v = \sin x$ , $w = e^{x}$
$u^{'} = 2 x$ , $v^{'} = \cos x$ , $w^{'} = e^{x}$
Apply formula:
$= 2 x \cdot \sin x \cdot e^{x} + x^{2} \cdot \cos x \cdot e^{x} + x^{2} \cdot \sin x \cdot e^{x}$
Factor $e^{x}$ : $= e^{x} (2 x \sin x + x^{2} \cos x + x^{2} \sin x)$

General rule: For a product of $n$ differentiable factors, add $n$ terms. In each term, differentiate one factor and leave all other factors unchanged.

6. Use function data, tangent lines and rate units

Evaluate a derivative from a table

Suppose $p (x) = f (x) g (x)$ , and both factors are differentiable at two. The supplied values are:

Input	$f (x)$	$f^{'} (x)$	$g (x)$	$g^{'} (x)$
$2$	$3$	$- 1$	$4$	$5$

p^{'} (2) = f^{'} (2) g (2) + f (2) g^{'} (2) .

p^{'} (2) = (- 1) (4) + (3) (5) = - 4 + 15 = 11 .

All four entries must come from the same input. The product value is $p (2) = 3 \cdot 4 = 12$ , which differs from its derivative value. The tangent line is therefore

y - p (2) = p^{'} (2) (x - 2), y = 12 + 11 (x - 2) .

A changing rectangle: both dimensions contribute

Let the length $L (t)$ and width $W (t)$ be differentiable functions of time, measured in metres while time is measured in seconds. Then

A (t) = L (t) W (t), \frac{d A}{d t} = L \frac{d W}{d t} + W \frac{d L}{d t} .

At an instant with $L = 5 m$ , $W = 3 m$ , $\frac{d L}{d t} = 0.2 \frac{m}{s}$ and $\frac{d W}{d t} = - 0.1 \frac{m}{s}$ , calculate

\frac{d A}{d t} = 5 (- 0.1) + 3 (0.2) = 0.1 \frac{m^{2}}{s} .

The area is increasing at that instant despite the shrinking width. Each contribution has area per time units. Multiplying the two side rates would have incompatible units and would miss the current dimensions. Related rates are developed further in Unit 4.

7. Strategies, memory aids and common mistakes

Essential Tips

Memorize the chant: "First d Second plus Second d First" until it's automatic
Label the factors: Write $u$ and $v$ clearly before applying the rule.
Find derivatives first: Calculate $u^{'}$ and $v^{'}$ separately before substituting.
Don't forget the plus: Product Rule has TWO terms connected by addition
Simplify at the end: Factor out common terms, combine like terms
Check by expanding: For simple polynomials, expand first to see if it's easier
Order doesn't matter: $u v^{'} + v u^{'}$ and $v u^{'} + u v^{'}$ are the same (addition is commutative)

Step-by-Step Workflow

Universal Process for Product Rule:

Identify: Label the two factors $u$ and $v$ , and check that each is differentiable at the input under consideration.
Differentiate: Find $u^{'}$ and $v^{'}$ separately.
First term: Write $u v^{'}$ , keeping the first factor unchanged.
PLUS SIGN: Don't forget the addition!
Second term: Write $v u^{'}$ , keeping the second factor unchanged.
SIMPLIFY: Factor, combine like terms, clean up notation
Verify: Before simplifying, check that both contributions appear. In each term, one factor is differentiated and the other is unchanged. A factored final answer need not visibly have two terms.

Memory Devices

Mnemonic #1: The Rhythm

"First times d-Second,
Plus Second times d-First"

Say this out loud while writing! The rhythm helps it stick.

Mnemonic #2: The Song Parody

(To the tune of "Row, Row, Row Your Boat")
"First, first, dee the second,
Plus the second, first.
Merrily, merrily, merrily, merrily,
Product Rule works best!"

Visual Pattern:

$(u v)^{'} = u v^{'} + v u^{'}$

In each term, keep one factor unchanged and differentiate the other.

Common Mistakes to Avoid

Mistake 1: Multiplying the derivatives is not a valid general rule. The correct identity is $(u v)^{'} = u^{'} v + u v^{'}$ .
Mistake 2: Keeping only $u v^{'}$ omits the $u^{'} v$ contribution in general. Include both before simplifying.
Mistake 3: Wrong sign between terms → Product Rule uses PLUS, not minus
Mistake 4: Differentiating both factors in the same term. In each term, differentiate one factor and keep the other unchanged.
Mistake 5: Simplifying without retaining the original domain. For instance, $x^{3} \cdot \frac{1}{x} = x^{2}$ where $x \neq 0$ ; in the logarithm example the original domain is $x > 0$ . Simplification is helpful but is not always required for a correct derivative.
Mistake 6: Treating a constant as though it had a nonzero derivative. Both the constant multiple rule and product rule give $\frac{d}{d x} [5 x^{3}] = 15 x^{2}$ .
Mistake 7: Changing factor labels partway through a calculation. Either choice of $u$ and $v$ is valid; remain consistent.
Mistake 8: Not parenthesizing negatives → $\frac{d}{d x} [x^{2} (- \sin x)]$ , watch the signs!

8. Six practice problems with solutions

Try each original problem before opening its solution. Use radians and retain the original function's domain.

1. Linear times exponential

Find the derivative of $f (x) = x e^{x}$ .

Show the two contributions

f^{'} (x) = x (e^{x}) + e^{x} (1) = e^{x} (1 + x) .

The derivative and original function are defined for every real input.

2. Polynomial times sine

Differentiate $g (x) = (3 x^{2} + 1) \sin x$ .

Show the factor derivatives

The factors have derivatives $6 x$ and $\cos x$ .

g^{'} (x) = (3 x^{2} + 1) \cos x + 6 x \sin x .

Both domains are all real inputs.

3. Cosine times logarithm

Differentiate $h (x) = \cos x \ln x$ .

Show the sign and domain check

h^{'} (x) = \cos x \cdot \frac{1}{x} + \ln x \cdot (- \sin x) .

h^{'} (x) = \frac{\cos x}{x} - \ln x \sin x, x > 0 .

The minus sign comes from cosine's derivative, and the logarithm restricts the original domain.

4. Fourth power times exponential

Differentiate $k (x) = x^{4} e^{x}$ .

Show the factored answer

k^{'} (x) = x^{4} e^{x} + 4 x^{3} e^{x} = e^{x} x^{3} (x + 4) .

Factoring is optional; the factored answer still contains both contributions. All real inputs are allowed.

5. Compare two valid methods

Differentiate $f (x) = (x - 5) (x + 2)$ .

Show expansion and product rule

f (x) = x^{2} - 3 x - 10, f^{'} (x) = 2 x - 3 .

Alternatively, the product rule gives $f^{'} (x) = (x - 5) (1) + (x + 2) (1) = 2 x - 3$ . Both are valid for every real input.

6. Evaluate at a special angle

For $f (x) = x^{2} \cos x$ , find $f^{'} (π)$ .

Show differentiation before evaluation

f^{'} (x) = - x^{2} \sin x + 2 x \cos x .

f^{'} (π) = - π^{2} \sin π + 2 π \cos π = - 2 π .

Use $\sin π = 0$ and $\cos π = - 1$ . This is the derivative value, rather than $f (π) = - π^{2}$ .

9. Write a clear AP-style response

How to write a clear AP-style response:

Show the requested work: Write a correct product-rule setup and enough differentiation or evaluation to support your result. Naming all four factor quantities is useful working, rather than a universal scoring requirement.
Identify the method: State that you are using the product rule when it helps communicate your reasoning.
Simplify purposefully: Combine or factor terms when helpful or when requested. Algebraically equivalent correct derivative expressions can take different forms.
Choose an efficient method: Expansion or a constant multiple may be shorter, but the product rule still applies.
Check both contributions: Verify the two product-rule terms in your setup. Factoring can combine their appearance later.
Combine rules appropriately: The quotient rule follows in Topic 2.9. The chain rule is introduced in Unit 3 and is often used together with the product rule.
Avoid the false shortcut: Multiplying two derivatives does not justify a product derivative; show the correct rule or another valid calculation.

Common FRQ Formats:

Find $f^{'} (x)$ for a product of differentiable functions.
Find a tangent at $x = a$ , using both $f (a)$ and $f^{'} (a)$ .
"At what rate is the area changing..." (Product Rule in context)
Justify $f^{'} (c) = 0$ through a valid derivative calculation and evaluation.
Find critical points of $f$ : first compute its derivative, then check inputs in the original domain where the derivative is zero or does not exist.
Determine increasing intervals by analysing where $f^{'} (x) > 0$ within the original domain.

Pro Tip: On multiple choice, if you see a product, immediately think "Product Rule OR can I expand first?" This decision tree will save you time!

10. Product rule quick reference

THE PRODUCT RULE - MEMORIZE NOW!

(u v)^{'} = u v^{'} + v u^{'}

"First d Second, Plus Second d First"

Product rule is convenient	A shorter alternative
$x^{2} \sin x$	$(x + 1) (x - 2)$ → Expand first
$e^{x} \ln x$	$5 x^{3}$ → Constant multiple
$(3 x + 1) \cos x$	$x \cdot x^{2} = x^{3}$ → Simplify first

Check the general identity: $(u v)^{'} = u v^{'} + v u^{'}$ , rather than the product of the two derivatives.

Conditions: The factors must be differentiable at the input when applying the theorem. Write both contributions before combining terms; simplification does not extend the original domain.

11. Connections and next study steps

Topic 2.8 connects to:

Topic 2.9: The quotient rule differentiates quotients. It can be derived from the product rule and differentiation of a reciprocal.
Unit 3: The chain rule handles compositions and may be needed inside either factor of a product.
Unit 5: Analytical applications, including optimization, often begin by differentiating a product.
Unit 4: Related rates frequently need Product Rule
BC Unit 6: Integration by parts reverses the product-rule relationship. This integration technique is BC-only content.
BC Unit 10: Taylor polynomials may require repeated differentiation, including products when they arise.
Real-world: Area, volume, and many physics formulas are products

Remember: For differentiable factors, $\frac{d}{d x} [u v] = u v^{'} + v u^{'}$ . Keep one factor unchanged in each contribution, then add. Multiplying the derivatives is not a general shortcut. Expansion or simplification can provide an efficient alternative, while products of more than two factors have one contribution for each changing factor. Keep original domains, radians and units in every calculation.

← Topic 2.7: basic function derivatives Next: Topic 2.9 quotient rule →Review linearity and derivative rules →All Unit 2 study notes →

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