01 / 20
\( f \cdot g \)
\( \frac{d}{dx} \)

The Product Rule

Learn how to differentiate the product of two functions using one of calculus's most essential rules: \( \frac{d}{dx}[f(x) \cdot g(x)] \).

AP Calculus AB

Why Can't We Just Multiply Derivatives?

It's tempting to think that the derivative of a product is just the product of the derivatives. But let's test this with a simple example:

We need a special rule for differentiating products: the Product Rule.

What Is a Product of Functions?

A product of functions is any expression where two (or more) functions are multiplied together. You encounter these constantly in calculus:

\( x^2 \sin x \)
Polynomial times trig
\( e^x \ln x \)
Exponential times logarithm
\( (3x+1)(x^2-4) \)
Two polynomials multiplied

Geometric Intuition

Think of \( f(x) \cdot g(x) \) as the area of a rectangle with width \( f \) and height \( g \). When \( x \) changes by a tiny amount \( dx \), both sides change:

\( f \cdot g \)
\( g \cdot df \)
\( f \cdot dg \)
\( \approx 0 \)
\( f \)
\( df \)
\( g \)
\( dg \)
\( \frac{d(\text{Area})}{dx} = f'g + fg' \)

An expanding patio reveals the product rule's two strips

Workers extend two adjacent courtyard sides; an exact rectangle decomposition isolates two strips and the small corner.Increasing both side lengths adds two strips and a corner. Dividing the area change by the input change and taking a limit gives the product rule, (fg)′ = f′g + fg′.

The product rule adds two first-order area contributions.

Increasing both side lengths adds two strips and a corner. Dividing the area change by the input change and taking a limit gives the product rule, (fg)′ = f′g + fg′.Increasing both side lengths adds two strips and a corner. Dividing the area change by the input change and taking a limit gives the product rule, (fg)′ = f′g + fg′.

The Product Rule: Formal Statement

If \( f(x) \) and \( g(x) \) are both differentiable, then:

In words: "the derivative of the first times the second, plus the first times the derivative of the second."

\( f'(x) \cdot g(x) \)
Differentiate the first, keep the second unchanged
\( f(x) \cdot g'(x) \)
Keep the first unchanged, differentiate the second

Equivalent Notations

The Product Rule appears in several equivalent forms. Recognizing each helps you read different textbooks and exams:

1
Prime notation: \( (fg)' = f'g + fg' \)
2
Leibniz notation: \( \dfrac{d}{dx}(uv) = \dfrac{du}{dx} \cdot v + u \cdot \dfrac{dv}{dx} \)
3
Operator notation: \( D_x[f \cdot g] = D_x[f] \cdot g + f \cdot D_x[g] \)

Product Rule vs. Other Rules

Scenario Rule Formula
\( c \cdot f(x) \) Constant Multiple \( c \cdot f'(x) \)
\( f(x) + g(x) \) Sum Rule \( f'(x) + g'(x) \)
\( f(x) \cdot g(x) \) Product Rule \( f'g + fg' \)
\( f(g(x)) \) Chain Rule \( f'(g(x)) \cdot g'(x) \)

How to Apply the Product Rule

1
Identify \( f(x) \) and \( g(x) \): Split the expression into two factors. Choose whichever split makes differentiation easiest.
2
Differentiate each separately: Find \( f'(x) \) and \( g'(x) \) using basic rules (power, trig, exponential, etc.).
3
Assemble the formula: Compute \( f'(x) \cdot g(x) + f(x) \cdot g'(x) \).
4
Simplify: Combine like terms, factor if possible, and present a clean final answer.

Example 1: \( x^2 \sin x \)

Find \( \dfrac{d}{dx}\bigl[x^2 \sin x\bigr] \).

1
Identify: \( f(x) = x^2 \), \( g(x) = \sin x \)
2
Differentiate: \( f'(x) = 2x \), \( g'(x) = \cos x \)
3
Assemble: \( f'g + fg' = (2x)(\sin x) + (x^2)(\cos x) \)

Example 2: \( e^x \ln x \)

Find \( \dfrac{d}{dx}\bigl[e^x \ln x\bigr] \).

1
Identify: \( f(x) = e^x \), \( g(x) = \ln x \)
2
Differentiate: \( f'(x) = e^x \), \( g'(x) = \dfrac{1}{x} \)
3
Assemble: \( e^x \cdot \ln x + e^x \cdot \dfrac{1}{x} \)

Example 3: \( (3x+1)(x^2-4) \)

Find \( \dfrac{d}{dx}\bigl[(3x+1)(x^2-4)\bigr] \). We'll solve it two ways to verify.

Product Rule
\( f = 3x+1 \), \( g = x^2-4 \)
\( f' = 3 \), \( g' = 2x \)
\( = 3(x^2-4) + (3x+1)(2x) \)
\( = 3x^2-12 + 6x^2+2x \)
\( = 9x^2 + 2x - 12 \)
Expand First
\( = 3x^3 - 12x + x^2 - 4 \)
\( = 3x^3 + x^2 - 12x - 4 \)
\( \frac{d}{dx} \rightarrow 9x^2 + 2x - 12 \)

✓ Same answer!

Visualizing \( x \cdot e^x \)

Here are the graphs of \( f(x) = x \), \( g(x) = e^x \), and their product \( h(x) = xe^x \). The Product Rule tells us the slope of \( h \) at any point.

\(x\)
\(y\)
\(1\)
\(2\)
\(-1\)
\(-2\)
\(1\)
\(2\)
\(f(x)=x\)
\(g(x)=e^x\)
\(h(x)=xe^x\)

Extending to Three Functions

The Product Rule generalizes. For three functions \( f \cdot g \cdot h \):

Each factor "takes a turn" being differentiated while the others stay. For \( n \) functions, there are \( n \) terms.

Ex
Find \( \dfrac{d}{dx}[x \cdot \sin x \cdot e^x] \):
\( = (1)(\sin x)(e^x) + (x)(\cos x)(e^x) + (x)(\sin x)(e^x) \)
\( = e^x(\sin x + x\cos x + x\sin x) \)

When to Use vs. Expand

Expression Best Strategy Why
\( (2x+1)(x^3) \) Expand first Simple polynomial
\( x^2 \cos x \) Product Rule Can't expand trig times poly
\( e^x \sqrt{x} \) Product Rule Can't combine exp and radical
\( 5x^4 \) Power Rule alone Constant times single power

Common Mistakes

✕ Multiplying Derivatives
Writing \( (fg)' = f' \cdot g' \). You must use the sum of two terms: \( f'g + fg' \).
✕ Forgetting a Term
Writing only \( f'g \) and dropping the \( fg' \) piece. Both terms are required.
✕ Confusing with Chain Rule
Using Chain Rule when you see \( f \cdot g \). The Chain Rule is for \( f(g(x)) \), not \( f(x) \cdot g(x) \).

Question 1 of 5

What is \( \dfrac{d}{dx}\bigl[x^3 \cos x\bigr] \)?

Question 2 of 5

The Product Rule states that \( (fg)' \) equals:

Question 3 of 5

Find \( \dfrac{d}{dx}\bigl[e^x \cdot x^2\bigr] \) evaluated at \( x = 0 \).

Question 4 of 5

Which expression requires the Product Rule to differentiate (cannot be simplified first)?

Question 5 of 5

If \( h(x) = \sin x \cdot \cos x \), what is \( h'(x) \)?