# Product Rule

> AP Calculus BC · Differentiation: Definition and Fundamental Properties
> Source: https://www.owlsprep.com/study/ap-calculus-bc-u2-product-rule/

This guide covers the product rule for differentiating products of two or more differentiable functions, its extension to multiple functions, tangent line applications, and common AP exam pitfalls to avoid.

**Prerequisites:** [Limit definition of the derivative](https://www.owlsprep.com/study/ap-calculus-bc-u2-derivative-definition/); [Basic power rule for polynomials](https://www.owlsprep.com/study/ap-calculus-bc-u2-power-rule/); [Derivative rules for trigonometric, exponential, and logarithmic functions](https://www.owlsprep.com/study/ap-calculus-bc-u2-derivatives-common-functions/)

## Learning objectives

- Apply the product rule to find derivatives of products of two differentiable functions
- Extend the product rule to products of three or more differentiable functions
- Use the product rule to find tangent lines to curves defined by product functions
- Identify and avoid common AP exam pitfalls when applying the product rule

## Core Definition & Two-Function Product Rule

The product rule is a fundamental differentiation technique that lets you compute the derivative of a product of two differentiable functions without expanding the product, which is impossible for non-polynomial products like $x\sin x$ or $e^x \ln x$. It accounts for 10–12% of Unit 2 points on the AP exam, and is foundational for nearly all advanced differentiation topics.

**Product Rule for Two Functions** — For two differentiable functions $f(x)$ and $g(x)$, the derivative of their product equals the first function times the derivative of the second, plus the second function times the derivative of the first.

*Notation:* \frac{d}{dx}\left[f(x)g(x)\right] = f(x)g'(x) + g(x)f'(x)

> **Memory Hook**
>
> First times derivative of the second, plus second times derivative of the first. This guarantees you never mix up the order of terms.

**Derivation:** Derive the product rule from the limit definition of the derivative

*Starting from:* Let $h(x) = f(x)g(x)$. By definition: $h'(x) = \lim_{\Delta x \to 0} \frac{f(x+\Delta x)g(x+\Delta x) - f(x)g(x)}{\Delta x}$

1. Add and subtract $f(x+\Delta x)g(x)$ to split the limit:
2. $$\lim_{\Delta x \to 0} \left[ f(x+\Delta x) \frac{g(x+\Delta x) - g(x)}{\Delta x} + g(x) \frac{f(x+\Delta x) - f(x)}{\Delta x} \right]$$
3. Take the limit, using continuity of differentiable functions, so $\lim_{\Delta x \to 0} f(x+\Delta x) = f(x)$

*Conclusion:* This simplifies to the standard product rule formula $h'(x) = f(x)g'(x) + g(x)f'(x)$

**Worked example:** Find the derivative of $h(x) = (3x^2 - 2x)(e^x + \sin x)$

1. Identify the two functions: $f(x) = 3x^2 - 2x$ and $g(x) = e^x + \sin x$
2. Compute derivatives of each function:
3. $$f'(x) = 6x - 2 \\ g'(x) = e^x + \cos x$$
4. Substitute into the product rule formula:
5. $$h'(x) = (3x^2 - 2x)(e^x + \cos x) + (e^x + \sin x)(6x - 2)$$
6. If expanded form is requested, simplify to:
7. $$h'(x) = 3x^2e^x + 4xe^x - 2e^x + 3x^2\cos x - 2x\cos x + 6x\sin x - 2\sin x$$

> **Exam tip:** On the AP exam, you do not need to expand the final derivative unless explicitly asked; factored form is almost always acceptable for full credit.

## Product Rule for Three or More Functions

The product rule extends naturally to any number of differentiable functions by applying the two-function rule repeatedly, and a clear pattern emerges: for $n$ functions, the derivative is the sum of $n$ terms, where each term differentiates exactly one function and leaves all others unchanged. AP exams most commonly test products of 3 functions in multiple-choice questions.

**Derivation:** Derive the product rule for three functions

*Starting from:* Group the product $f(x)g(x)h(x)$ as $f \cdot (gh)$ to apply the two-function rule

1. Apply the two-function product rule first:
2. $$\frac{d}{dx}[f (gh)] = f' (gh) + f \frac{d}{dx}(gh)$$
3. Apply the two-function rule again to $\frac{d}{dx}(gh)$:
4. $$= f' g h + f (g' h + g h') = f' g h + f g' h + f g h'$$

*Conclusion:* The pattern holds: $n$ functions produce exactly $n$ terms in the derivative, one for each differentiated function.

**Worked example:** Find the derivative of $y = x \cos x \ln x$

1. Label the three functions: $f_1(x) = x$, $f_2(x) = \cos x$, $f_3(x) = \ln x$
2. Compute individual derivatives:
3. $$f_1'(x) = 1 \\ f_2'(x) = -\sin x \\ f_3'(x) = \frac{1}{x}$$
4. Apply the 3-function product rule: $\frac{dy}{dx} = f_1' f_2 f_3 + f_1 f_2' f_3 + f_1 f_2 f_3'$
5. Substitute and simplify, cancelling the $x$ term in the final term:
6. $$\frac{dy}{dx} = \cos x \ln x - x \sin x \ln x + \cos x$$

> **Exam tip:** Always confirm you have exactly 3 terms for a 3-function product; missing one term is the most common mistake on these problems.

## Tangent Line Applications

A common AP exam application of the product rule is finding the equation of a tangent line to a curve defined as a product of functions. This combines the product rule with point-slope form of a line, tested frequently in both MCQ and early FRQ sections. To solve, you need two values: the tangency point $(x_0, y_0)$ where $y_0 = h(x_0)$, and the slope $m = h'(x_0)$ found via the product rule.

**Worked example:** Find the equation of the tangent line to $h(x) = (x^2 - 4)(2^x)$ at $x = 2$

1. Find the $y$-coordinate of the tangency point:
2. $$h(2) = (2^2 - 4)(2^2) = 0 \cdot 4 = 0$$
3. The tangency point is $(2, 0)$
4. Identify functions and compute their derivatives:
5. $$f(x) = x^2 - 4, \quad f'(x) = 2x \\ g(x) = 2^x, \quad g'(x) = 2^x \ln 2$$
6. Apply the product rule to get $h'(x)$:
7. $$h'(x) = (x^2 - 4)(2^x \ln 2) + 2^x (2x)$$
8. Evaluate $h'(2)$ to get the slope:
9. $$h'(2) = (4 - 4)(4 \ln 2) + 4(4) = 0 + 16 = 16$$
10. Write and simplify the tangent line equation:
11. $$y - 0 = 16(x - 2) \\ y = 16x - 32$$

> **Exam tip:** Always calculate the $y$-coordinate before the slope; it is common to accidentally plug $x_0$ into the derivative for $y_0$, costing easy points.

## AP-Style Concept Check

**Check your understanding**

Test your understanding with this AP-style multiple-choice question:

1. Which of the following is equal to $\frac{d}{dx}\left[ (x^3 - 5x^2)(\cos x) \right]$?

   - A) $(3x^2 - 10x)(-\sin x)$
   - B) $-(3x^2 - 10x)\sin x + (x^3 - 5x^2)\cos x$
   - C) $(x^3 - 5x^2)\sin x + (3x^2 - 10x)\cos x$
   - D) $-(x^3 - 5x^2)\sin x + (3x^2 - 10x)\cos x$

   *Why:* Apply the product rule: $f(x) = x^3 - 5x^2$, $g(x) = \cos x$. $f'(x) = 3x^2 - 10x$, $g'(x) = -\sin x$. Substitute into $f(x)g'(x) + g(x)f'(x)$ to get the correct result.

## Common pitfalls

- **Wrong:** Writing $\frac{d}{dx}[f(x)g(x)] = f'(x)g'(x)$ instead of $f(x)g'(x) + g(x)f'(x)$
  - Why it fails: Students confuse the derivative of a product with the product of derivatives, incorrectly assuming the derivative distributes over multiplication like it does over addition
  - Correct: Recall the mnemonic 'first derivative of second plus second derivative of first' before starting, and confirm you have two terms for a product of two functions
- **Wrong:** Forgetting the negative sign when differentiating cosine or other trigonometric functions with negative derivatives in the product rule step
  - Why it fails: Students focus on remembering the product rule structure and ignore basic derivative rules for individual functions
  - Correct: After writing all product rule terms, double-check the derivative of each individual function for sign errors before simplifying
- **Wrong:** Missing a term when differentiating a three-function product (e.g., writing $\frac{d}{dx}[fgh] = f'gh + fgh'$ omitting the $fg'h$ term)
  - Why it fails: Students try to apply two-function intuition directly to three functions and count the wrong number of terms
  - Correct: For any product of $n$ functions, confirm you have exactly $n$ terms in your derivative before proceeding, one for each differentiated function
- **Wrong:** Expanding the entire product before differentiating, leading to algebraic errors on higher-degree polynomial products
  - Why it fails: Students think expanding is easier than applying the product rule, but it introduces unnecessary extra multiplication steps
  - Correct: Always use the product rule for products of any functions, even polynomials, to avoid extra algebraic work and reduce error
- **Wrong:** Evaluating the derivative of each function at $x_0$ before applying the product rule, leading to an incorrect constant when a general derivative function is requested
  - Why it fails: Students rush to plug in the given $x$ value when asked for a slope, but questions sometimes ask for a general derivative first
  - Correct: Read the question carefully: if it asks for a general derivative, do not plug in the value until after you have found the full derivative function

## Cheatsheet

| Category | Formula | Notes |
| --- | --- | --- |
| Product Rule (2 functions) | $\frac{d}{dx}\left[f(x)g(x)\right] = f(x)g'(x) + g(x)f'(x)$ | Applies when both $f$ and $g$ are differentiable; mnemonic: first d-second plus second d-first |
| Product Rule (3 functions) | $\frac{d}{dx}\left[f(x)g(x)h(x)\right] = f'gh + fg'h + fgh'$ | One term per differentiated function; $n$ functions give $n$ terms total |
| Tangent Line (product function) | $y - y_0 = h'(x_0)(x - x_0)$ | $y_0 = h(x_0)$, $h'(x_0)$ is slope found via product rule |
| n-function product rule | $\frac{d}{dx}\left[\prod_{i=1}^n f_i(x)\right] = \sum_{i=1}^n \left(f_i'(x) \prod_{j \neq i} f_j(x)\right)$ | Works for any number of differentiable functions; rarely tested for $n>3$ |
| Constant multiple rule (special case) | $\frac{d}{dx}\left[c f(x)\right] = c f'(x)$ | Special case of product rule where one function is constant; use this instead of full product rule |
| Factored polynomial derivative | $\frac{d}{dx}\left[(a_nx^n + ...)(b_mx^m + ...)\right] = f g' + g f'$ | Faster than expanding, reduces algebraic error for higher-degree polynomials |

## What's next

Mastering the product rule is an essential prerequisite for the next core topic in Unit 2: the quotient rule, which is derived directly from the product rule and power rule. Without a solid understanding of product rule, you will struggle to correctly apply the quotient rule and avoid common sign and term errors. Beyond Unit 2, the product rule is a foundational step for the chain rule, implicit differentiation, logarithmic differentiation, and integration by parts later in the course. Nearly every complex differentiation problem you will encounter on the AP exam will require the product rule at some step, so building the habit of applying it correctly will pay off for the entire course.

- [Quotient rule](https://www.owlsprep.com/study/ap-calculus-bc-u2-quotient-rule/)
- [Chain rule](https://www.owlsprep.com/study/ap-calculus-bc-u3-chain-rule/)
- [Implicit differentiation](https://www.owlsprep.com/study/ap-calculus-bc-u3-implicit-differentiation/)
- [Integration by parts](https://www.owlsprep.com/study/ap-calculus-bc-u6-integration-by-parts/)

---

From [OwlsPrep](https://www.owlsprep.com) — free study guides for A-Level, IB, AP and IGCSE, written against the official syllabus. Canonical page: https://www.owlsprep.com/study/ap-calculus-bc-u2-product-rule/