The Product Rule

Master the technique for differentiating products of two functions: one of the most essential tools in calculus.

AP Calculus BC

Why Can't We Just Multiply Derivatives?

When you first learn differentiation, a natural guess is that the derivative of a product equals the product of the derivatives. But this is wrong.

We need a special rule that accounts for how both factors change simultaneously. That rule is called the Product Rule.

Intuition: Two Changing Quantities

Imagine a rectangle whose width is \( f(x) \) and height is \( g(x) \). The area is \( A = f(x) \cdot g(x) \). When \( x \) changes by a tiny amount \( dx \), both dimensions shift:

The Area Model Summarized

Original Area

\( f(x) \cdot g(x) \): the quantity we are differentiating.

Right Strip

\( f'(x) \cdot g(x) \): growth from the width changing while height stays.

Top Strip

\( f(x) \cdot g'(x) \): growth from the height changing while width stays.

Adding both strips gives the Product Rule. The corner rectangle \( df \cdot dg \) is infinitesimally small and drops out in the limit.

The Product Rule: Formal Statement

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

Proof from the Limit Definition

1

Start with the definition: \( \displaystyle\lim_{h\to 0}\frac{f(x+h)\,g(x+h)-f(x)\,g(x)}{h} \)

2

Add and subtract \( f(x+h)\,g(x) \) in the numerator to create two recognizable pieces.

3

Factor: \( f(x+h)\!\cdot\!\dfrac{g(x+h)-g(x)}{h} + g(x)\!\cdot\!\dfrac{f(x+h)-f(x)}{h} \)

4

Take \( h \to 0 \). Since \( f \) is continuous, \( f(x+h)\to f(x) \). The two fractions become \( g'(x) \) and \( f'(x) \).

Product Rule vs. Chain Rule

	Product Rule	Chain Rule
Structure	\( f(x) \cdot g(x) \)	\( f\!\bigl(g(x)\bigr) \)
Operation	Two functions multiplied	One function inside another
Formula	\( f'g + fg' \)	\( f'\!\bigl(g(x)\bigr)\cdot g'(x) \)
Example	\( x^2 \sin x \)	\( \sin(x^2) \)
Key cue	Two separate pieces with \( \times \)	Parentheses nesting

How to Apply the Product Rule

1

Identify the two factors. Label them \( f(x) \) and \( g(x) \).

2

Differentiate each factor individually to get \( f'(x) \) and \( g'(x) \).

3

Assemble the formula: \( f'(x)\,g(x) + f(x)\,g'(x) \).

4

Simplify. Combine like terms, factor common expressions if possible.

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

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

1

Let \( f = x^2 \) and \( g = \sin x \).

2

\( f' = 2x \), \( g' = \cos x \).

3

Apply: \( f'g + fg' = 2x\sin x + x^2\cos x \).

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

Differentiate \( h(x) = e^x \ln x \).

1

Let \( f = e^x \), \( g = \ln x \).

2

\( f' = e^x \), \( g' = \frac{1}{x} \).

3

\( h'(x) = e^x \ln x + e^x \cdot \frac{1}{x} = e^x\!\left(\ln x + \frac{1}{x}\right) \).

Example 3: \( (3x+1)^4\!\cdot\!\cos x \)

Find the derivative. This requires the Product Rule and Chain Rule together.

1

Let \( f = (3x+1)^4 \), \( g = \cos x \).

2

Chain Rule on \( f \): \( f' = 4(3x+1)^3 \cdot 3 = 12(3x+1)^3 \). Also \( g' = -\sin x \).

3

Product Rule: \( 12(3x+1)^3 \cos x + (3x+1)^4(-\sin x) \).

4

Factor: \( (3x+1)^3\bigl[12\cos x - (3x+1)\sin x\bigr] \).

Visualizing \( f \), \( g \), and \( f \cdot g \)

Consider \( f(x)=x \) and \( g(x)=\sin x \). The product \( fg = x\sin x \) captures how both functions interact:

Extension: Three or More Factors

The Product Rule extends naturally. For three factors:

The pattern: differentiate one factor at a time, leaving all others untouched, and add the results.

2 factors

\( f'g + fg' \): 2 terms

3 factors

\( f'gh + fg'h + fgh' \): 3 terms

n factors

\( n \) terms, each with exactly one derivative

Common Mistakes

✕ Multiplying Derivatives

Writing \( (fg)' = f' \cdot g' \) instead of the correct \( f'g + fg' \). This ignores how both functions change together.

✕ Forgetting a Term

Only writing \( f'g \) and dropping the \( fg' \) term. Both terms are essential: always write both before simplifying.

✕ Skipping the Chain Rule

When a factor is composite, like \( (3x+1)^4 \), students forget to apply the Chain Rule when computing that factor's derivative.

Summary: Key Derivatives with Products

Function	Derivative
\( x \cdot e^x \)	\( e^x + x\,e^x = e^x(1+x) \)
\( x^2 \sin x \)	\( 2x\sin x + x^2\cos x \)
\( e^x \ln x \)	\( e^x\!\left(\ln x + \frac{1}{x}\right) \)
\( x^3 \cos x \)	\( 3x^2\cos x - x^3\sin x \)
\( \sqrt{x}\,\tan x \)	\( \frac{\tan x}{2\sqrt{x}} + \sqrt{x}\,\sec^2 x \)

Question 1 of 5

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

With \( f=x \) and \( g=e^x \): \( f'g + fg' = 1\cdot e^x + x\cdot e^x = e^x + xe^x \).

Question 2 of 5

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

Product Rule: \( \cos x \cdot \cos x + \sin x \cdot (-\sin x) = \cos^2 x - \sin^2 x \). This also equals \( \cos 2x \).

Question 3 of 5

If \( h(x) = x^3 e^x \), what is \( h'(0) \)?

\( h'(x) = 3x^2 e^x + x^3 e^x = x^2 e^x(3+x) \). At \( x=0 \): \( h'(0) = 0^2 \cdot e^0 \cdot 3 = 0 \).

Question 4 of 5

Which expression correctly applies the Product Rule to \( (2x+5) \cdot \ln x \)?

Let \( f=2x+5 \), \( g=\ln x \). Then \( f'=2 \), \( g'=\frac{1}{x} \). Product Rule: \( 2\cdot\ln x + (2x+5)\cdot\frac{1}{x} = 2\ln x + \frac{2x+5}{x} \).

Question 5 of 5

Find \( \dfrac{d}{dx}\bigl[(x^2+1)\sin x \cos x\bigr] \) using the extended Product Rule for three factors.

Three-factor rule: \( f'gh + fg'h + fgh' \) where \( f=x^2+1 \), \( g=\sin x \), \( h=\cos x \). This gives \( 2x\sin x\cos x + (x^2+1)\cos x\cos x + (x^2+1)\sin x(-\sin x) \) = \( 2x\sin x\cos x + (x^2+1)\cos^2 x - (x^2+1)\sin^2 x \).