Product Rule

AP Calculus AB· AP Calculus AB CED — Differentiation: Definition and Fundamental Properties· 14 min read

1. Product Rule for Two Differentiable Functions★★☆☆☆⏱ 4 min

For any function $f (x) = g (x) h (x)$ , where $g (x)$ and $h (x)$ are both differentiable at $x$ , the product rule can be derived directly from the limit definition of the derivative.

🔬 Derivation

Goal:

Derive the product rule formula from first principles

Starting from:

The limit definition of $f^{'} (x)$ for $f (x) = g (x) h (x)$

1
Start with the standard limit definition:
2
$f'(x) = \lim_{\Delta x \to 0} \frac{g(x+\Delta x)h(x+\Delta x) - g(x)h(x)}{\Delta x}$
3
Add and subtract $g (x + Δ x) h (x)$ in the numerator to factor:
4
$\text{Numerator} = g(x+\Delta x)\left(h(x+\Delta x) - h(x)\right) + h(x)\left(g(x+\Delta x) - g(x)\right)$
5
Split the limit, use the fact that differentiable functions are continuous ( $lim_{Δ x \to 0} g (x + Δ x) = g (x)$ ), and apply the definition of $g^{'} (x)$ and $h^{'} (x)$ .

Result:

We arrive at the standard product rule formula for two functions.

📘 Definition

Product Rule (Two Functions)

f(x) = g(x)h(x)

If both $g$ and $h$ are differentiable at $x$ , the derivative of $f (x)$ is: $f^{'} (x) = g (x) h^{'} (x) + h (x) g^{'} (x)$

Example:

For $f (x) = x^{3} sin x$ , $f^{'} (x) = x^{3} cos x + 3 x^{2} sin x$

📐 Worked Example

Find the derivative of $f (x) = x^{3} sin x$ .

1
1. Identify the two differentiable factors:
2
$g(x) = x^3, \quad h(x) = \sin x$
3
1. Calculate the derivatives of each factor:
4
$g'(x) = 3x^2, \quad h'(x) = \cos x$
5
1. Substitute into the product rule formula:
6
$f'(x) = g(x)h'(x) + h(x)g'(x) = x^3 \cos x + (\sin x)(3x^2)$
7
1. Simplify by factoring out the common term $x^{2}$ :
8
$f'(x) = x^2\left(x \cos x + 3 \sin x\right)$

Exam tip:

Always explicitly label $g (x), g^{'} (x), h (x), h^{'} (x)$ in FRQ working to earn partial credit for small errors.

2. Extended Product Rule for Three or More Functions★★★☆☆⏱ 3 min

The product rule generalizes naturally to products of three or more differentiable functions, and this extension is commonly tested on AP exams. The pattern is simple: for $n$ factors multiplied together, the derivative will have $n$ terms, where each term is the derivative of exactly one factor multiplied by all the other original (unchanged) factors.

📘 Definition

Product Rule (Three Functions)

f(x) = u(x)v(x)w(x)

The derivative is given by: $f^{'} (x) = u^{'} (x) v (x) w (x) + u (x) v^{'} (x) w (x) + u (x) v (x) w^{'} (x)$

If you forget the extended pattern, you can always derive it by grouping two factors as a single product and applying the two-function product rule twice. This method requires no extra memorization and always gives the correct result.

📐 Worked Example

Find the derivative of $f (x) = 2 x e^{x} cos x$ .

1
1. Label the three factors and their individual derivatives:
2
$u(x) = 2x, \quad u'(x) = 2 \\ v(x) = e^x, \quad v'(x) = e^x \\ w(x) = \cos x, \quad w'(x) = -\sin x$
3
1. Apply the extended product rule:
4
$f'(x) = (2)(e^x)(\cos x) + (2x)(e^x)(\cos x) + (2x)(e^x)(-\sin x)$
5
1. Simplify by factoring out the common term $2 e^{x}$ :
6
$f'(x) = 2e^x\left(\cos x + x \cos x - x \sin x\right)$

Exam tip:

If you forget the extended pattern, group two factors and apply the two-function rule twice for a foolproof result.

3. AP Application: Finding Tangent Lines★★★☆☆⏱ 4 min

One of the most common AP exam applications of the product rule is finding the equation of a tangent line to a curve that is defined as a product of functions. This problem combines your knowledge of the product rule with the geometric definition of the derivative as the slope of the tangent line, following three core steps: 1) Use product rule to find $f^{'} (x)$ , 2) Evaluate $f^{'} (a)$ at the given $x = a$ to get the slope $m$ , 3) Find $f (a)$ to get the point $(a, f (a))$ , then use point-slope form to write the tangent line equation.

📐 Worked Example

Find the equation of the tangent line to $f (x) = (x^{2} - 4) (3 x + 2)$ at $x = 1$ .

1
1. Identify factors and their derivatives:
2
$g(x) = x^2 - 4, \quad g'(x) = 2x \\ h(x) = 3x + 2, \quad h'(x) = 3$
3
1. Apply product rule to find $f^{'} (x)$ and simplify:
4
$f'(x) = (x^2 - 4)(3) + (3x + 2)(2x) = 3x^2 - 12 + 6x^2 + 4x = 9x^2 + 4x - 12$
5
1. Calculate the slope at $x = 1$ :
6
$f'(1) = 9(1)^2 + 4(1) - 12 = 1 \implies m = 1$
7
1. Calculate the $y$ -coordinate of the tangent point:
8
$f(1) = (1 - 4)(3 + 2) = -15 \implies \text{Point} = (1, -15)$
9
1. Use point-slope form and simplify to slope-intercept:
10
$y + 15 = 1(x - 1) \implies y = x - 16$

Exam tip:

For polynomial products, expand the original product and differentiate term-by-term to quickly check your derivative result.

4. AP-Style Concept Check★★★☆☆⏱ 3 min

✓ Quick check

Test your understanding with these AP-style practice questions:

What is the derivative of $f (x) = 4 x e^{2 x}$ ?
- A) $4 e^{2 x} (2 x + 1)$
- B) $8 e^{2 x}$
- C) $4 e^{2 x} (x + 1)$
- D) $8 x e^{2 x}$
Reveal answer
A) $4e^{2x}(2x + 1)$ —
Incorrect options come from common errors: forgetting one term of the product rule, or incorrectly applying the constant multiple rule. The correct working is: $g (x) = 4 x$ , $g^{'} (x) = 4$ , $h (x) = e^{2 x}$ , $h^{'} (x) = 2 e^{2 x}$ , so $f^{'} (x) = (4 x) (2 e^{2 x}) + 4 e^{2 x} = 4 e^{2 x} (2 x + 1)$ .
Let $f (x) = (x^{2} - 5 x) cos x$ . (a) Find $f^{'} (x)$ . (b) Find the slope of the tangent line at $x = π$ . (c) Is $f (x)$ increasing or decreasing at $x = π$ ? Justify your answer.
Reveal answer
(a) $f'(x) = -(x^2 - 5x)\sin x + (2x - 5)\cos x$ (b) Slope = $5 - 2\pi \approx -1.28$ (c) $f(x)$ is decreasing, because $f'(\pi) = 5 - 2\pi < 0$. —
This is a common multi-part FRQ style question that tests both product rule application and interpretation of the derivative sign.

5. Common Pitfalls

Wrong move:

For $f (x) = g (x) h (x)$ , claim $f^{'} (x) = g^{'} (x) h^{'} (x)$

Why:

Students incorrectly extend the sum rule pattern ( $(g + h)^{'} = g^{'} + h^{'}$ ) to products

Correct move:

Always write the full product rule $f^{'} (x) = g (x) h^{'} (x) + h (x) g^{'} (x)$ before starting calculations

Wrong move:

For $f (x) = 3 x^{2} sin x$ , write $f^{'} (x) = 3 (2 x cos x) = 6 x cos x$ , forgetting the second product rule term

Why:

Confusion between the constant multiple rule and product rule, leading to stopping after differentiating only one factor

Correct move:

Factor the constant out first to apply product rule only to the non-constant factors, or treat the constant as a factor with derivative zero

Wrong move:

For $f (x) = (x + 3)^{2} = (x + 3) (x + 3)$ , calculate $f^{'} (x) = 1 \cdot 1 = 1$

Why:

Same as the first pitfall: multiplying derivatives instead of applying the full product rule

Correct move:

Apply product rule: $f^{'} (x) = (1) (x + 3) + (x + 3) (1) = 2 (x + 3)$ , which matches the chain rule result

Wrong move:

For $f (x) = uv w$ , write $f^{'} (x) = u^{'} v^{'} w + u v^{'} w^{'} + u^{'} v^{'} w^{'}$

Why:

Misremembering the extended pattern, taking the derivative of two factors per term instead of one

Correct move:

Follow the rule: each term has exactly one derivative, all other factors are unchanged: $f^{'} (x) = u^{'} v w + u v^{'} w + uv w^{'}$

Wrong move:

When finding a tangent line at $x = a$ , calculate slope $m = f (a)$ instead of $m = f^{'} (a)$

Why:

Rushing through the problem, mixing up what the original function and derivative represent

Correct move:

Explicitly label 'slope = $f^{'} (a)$ ' and 'y-coordinate = $f (a)$ ' in your working to avoid mixing them up

6. Quick Reference Cheatsheet

Category	Formula	Notes
Product Rule (Two Functions)	If $f (x) = g (x) h (x)$ , then $f^{'} (x) = g (x) h^{'} (x) + h (x) g^{'} (x)$	Applies to differentiable $g, h$ ; derivative of product $\neq =$ product of derivatives
Leibniz Notation	$\frac{d}{d x} [uv] = u \frac{d v}{d x} + v \frac{d u}{d x}$	Common in FRQ, easy to track which term is differentiated
Product Rule (Three Functions)	If $f (x) = uv w$ , then $f^{'} (x) = u^{'} v w + u v^{'} w + uv w^{'}$	Each term differentiates exactly one factor; all other factors stay unchanged
Extended Rule (n Factors)	$\frac{d}{d x} [\prod_{i = 1}^{n} f_{i}] = \sum_{i = 1}^{n} (f_{i}^{'} \prod_{j \neq = i} f_{j})$	Rarely tested for $n > 3$ on AB; group and use two-function rule if unsure
Constant Multiple Rule (Special Case)	If $f (x) = c \cdot g (x)$ , then $f^{'} (x) = c \cdot g^{'} (x)$	Derived from product rule: derivative of constant $c$ is 0
Tangent Line Slope at $x = a$	Slope $m = f^{'} (a)$	Slope always comes from the derivative, not the original function
Tangent Line Equation	$y - f (a) = m (x - a)$	Standard point-slope form; simplify to slope-intercept if requested

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · MCQ
Differentiate product of exponential and polynomial
2022 · FRQ
Find tangent line to product function

What's Next

Mastering the product rule is a non-negotiable prerequisite for every upcoming differentiation topic in AP Calculus AB. Immediately next, you will combine the product rule with the chain rule to differentiate composite functions that include products of simpler inner functions, a common source of points on both multiple-choice and free-response sections. Later, you will rely on the product rule for implicit differentiation, related rates, and optimization problems, where most functions you need to differentiate are products of two or more simpler functions. Without a solid command of the product rule, you will be unable to correctly set up and solve these more complex problems, leading to unnecessary lost points.