When should I use Product Rule instead of just multiplying out the product first and using Power Rule?

Multiply out first when it's simple—for instance, 3x \cdot x^2 = 3x^3, which you can then differentiate directly. Use Product Rule when the product is complex, has variables in both factors, or would be tedious to expand.

Can the order of the two terms in (fg)' = f'g + fg' be reversed?

Yes, you can write (fg)' = fg' + f'g since addition is commutative. Both orders are mathematically equivalent and will give the same final answer.

Why isn't the derivative of a product equal to the product of the derivatives?

Because when two quantities are multiplied, a change in one quantity affects the entire product—not just the part from that quantity. The Product Rule accounts for both effects: how the first function's change affects the product, plus how the second function's change affects it.

Do I need to memorize the Product Rule or is there a way to derive it?

You should know the formula (fg)' = f'g + fg', but you can derive it from the limit definition of a derivative if you need to. Many textbooks show the derivation, which helps solidify why the rule has this particular form.

Product Rule

Product Rule lets you find the derivative of a product of two functions without multiplying them out first; use it when the product is complex.

(fg)' = f'g + fg'

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What each symbol means

What Product Rule takes

Symbol	Meaning
$f$	A function being multiplied; $f'$ denotes its derivative.
$g$	A function being multiplied; $g'$ denotes its derivative.

When to use it

When you need to find the derivative of two functions being multiplied together.

Level

Usually taught in: Calculus I · Appears on: AP Calculus

Worked examples

1. Find the derivative of a polynomial product

Problem

Find the derivative of

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

$f = x^2 + 3, \quad g = 2x - 1$
Identify the two functions being multiplied.
$f' = 2x, \quad g' = 2$
Find the derivative of each function separately using the Power Rule.
$(fg)' = f'g + fg' = (2x)(2x - 1) + (x^2 + 3)(2)$
Apply the Product Rule by substituting the functions and their derivatives.
$(2x)(2x - 1) = 4x^2 - 2x \quad \text{and} \quad (x^2 + 3)(2) = 2x^2 + 6$
Expand each product separately.
$4x^2 - 2x + 2x^2 + 6 = 6x^2 - 2x + 6$
Combine like terms.

Answer: $6x^2 - 2x + 6$

The Product Rule lets us find the derivative without expanding the entire product first, which would give $2x^3 - x^2 + 6x - 3$ . By using Product Rule, we work directly with each function and its derivative, which often leads to fewer arithmetic errors and a cleaner path to the final answer.

2. Find the derivative of a product with negative coefficients

Problem

Find the derivative of

h(x) = (1 - 3x)(x^2 + 4x)

$u = 1 - 3x, \quad v = x^2 + 4x$
Identify the two functions being multiplied.
$u' = -3, \quad v' = 2x + 4$
Find the derivative of each function using the Power Rule; note that $u'$ is negative.
$(uv)' = u'v + uv' = (-3)(x^2 + 4x) + (1 - 3x)(2x + 4)$
Apply the Product Rule by substituting the functions and derivatives into the formula.
$(-3)(x^2 + 4x) = -3x^2 - 12x \quad \text{and} \quad (1 - 3x)(2x + 4) = 2x + 4 - 6x^2 - 12x$
Expand each product carefully, especially with negative signs.
$-3x^2 - 12x + 2x + 4 - 6x^2 - 12x = -9x^2 - 22x + 4$
Combine all like terms.

Answer: $-9x^2 - 22x + 4$

With negative coefficients and higher-degree terms, expanding first would be messy. The Product Rule handles this systematically: we find each derivative cleanly (including the negative in $u' = -3$ ), then apply the rule. This approach is less error-prone and shows how each part of the product contributes to its rate of change.

3. Find how phone plan revenue changes over time

Problem

A phone company's monthly revenue (in dollars) from a plan is

R(t) = (20 + 0.5t)(1000 + 100t)

, where

t

is time in months,

(20 + 0.5t)

is the monthly price per subscriber in dollars, and

(1000 + 100t)

is the number of subscribers. Find

R'(t)

, the rate of change of revenue.

$p(t) = 20 + 0.5t, \quad n(t) = 1000 + 100t$
Identify the two functions: the price per subscriber and the number of subscribers.
$p'(t) = 0.5, \quad n'(t) = 100$
Differentiate each function; the coefficients represent the rates of change of price and subscriber count.
$R'(t) = p'(t) \cdot n(t) + p(t) \cdot n'(t) = (0.5)(1000 + 100t) + (20 + 0.5t)(100)$
Apply the Product Rule to find the rate of change of revenue.
$(0.5)(1000 + 100t) = 500 + 50t \quad \text{and} \quad (20 + 0.5t)(100) = 2000 + 50t$
Distribute to expand each product.
$R'(t) = 500 + 50t + 2000 + 50t = 2500 + 100t$
Combine like terms to get the final expression for the rate of change.

Answer: $R'(t) = 2500 + 100t$

Real-world scenarios often involve products of changing quantities. The Product Rule helps us find how fast revenue is changing by considering each factor separately—how price changes with time, and how subscriber count changes with time. This is essential for business analysis and optimization.

Common mistakes

Where Product Rule usually goes wrong

The mistake	Why it is wrong	The fix
Students write $(fg)' = f'g'$ .	Students multiply the derivatives together instead of following the Product Rule formula, giving an incorrect answer that doesn't match the actual derivative.	Apply the full rule: $(fg)' = f'g + fg'$ —the two derivative terms must be multiplied by the original (non-derived) function.
Students compute only the first term and write $(x^2 + 3)(2x - 1))' = (2x)(2)$ .	Forgetting the second term $fg'$ and only computing $f'g$ leaves out half the formula, resulting in a completely wrong answer.	Always write both terms: $(fg)' = f'g + fg'$ . Here, the answer is $(2x)(2x-1) + (x^2+3)(2) = 6x^2 - 2x + 6$ .
Students apply the Product Rule to a quotient and write $(\frac{x^2}{2x+1})' = (2x)(2)$ .	Applying Product Rule to a quotient instead of using the Quotient Rule is a fundamental formula mix-up.	For quotients, use the Quotient Rule: $(\frac{f}{g})' = \frac{f'g - fg'}{g^2}$ , not the Product Rule.

Tips and when to use something else

Remember: the Product Rule has two terms, $f'g + fg'$ . A common mistake is forgetting the second term or multiplying the derivatives as $f'g'$ .
If multiplying out the product first is simpler, do that instead and use the Power Rule—for example, $5x \cdot x^2 = 5x^3$ is faster without Product Rule.
For a product of three functions like $uvw$ , apply Product Rule twice: first to get $(uv)'w + (uv)w'$ , then apply Product Rule to $(uv)'$ to get $u'v + uv'$ .
Product Rule is for multiplication. For division, use the Quotient Rule: $(f/g)' = (f'g - fg')/g^2$ .

Related formulas

Power RuleThe Power Rule states that the derivative of x^n is nx^{n-1}, making it the essential tool for differentiating any polynomial expression.Quotient RuleThe Quotient Rule gives you the derivative of a fraction by combining the derivatives and original functions of the numerator and denominator.Chain RuleThe Chain Rule tells you how to differentiate a composite function—when one function is plugged into another—by multiplying the outer and inner derivatives.Definition of the DerivativeThe Definition of the Derivative finds the instantaneous rate of change at a point using limits; the foundation for all derivative calculations.Implicit DifferentiationUse implicit differentiation to find dy/dx when an equation defines y implicitly instead of as an explicit function of x.

Frequently asked questions

When should I use Product Rule instead of just multiplying out the product first and using Power Rule?: Multiply out first when it's simple—for instance, $3x \cdot x^2 = 3x^3$ , which you can then differentiate directly. Use Product Rule when the product is complex, has variables in both factors, or would be tedious to expand.
Can the order of the two terms in $(fg)' = f'g + fg'$ be reversed?: Yes, you can write $(fg)' = fg' + f'g$ since addition is commutative. Both orders are mathematically equivalent and will give the same final answer.
Why isn't the derivative of a product equal to the product of the derivatives?: Because when two quantities are multiplied, a change in one quantity affects the entire product—not just the part from that quantity. The Product Rule accounts for both effects: how the first function's change affects the product, plus how the second function's change affects it.
Do I need to memorize the Product Rule or is there a way to derive it?: You should know the formula $(fg)' = f'g + fg'$ , but you can derive it from the limit definition of a derivative if you need to. Many textbooks show the derivation, which helps solidify why the rule has this particular form.

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Reviewed 2026-09-18