Product rule — AP Calculus AB Study Guide

For: AP Calculus AB candidates sitting AP Calculus AB.

Covers: The product rule formula for derivatives, derivation from first principles, differentiation of products of two or more differentiable functions, and applications to finding tangent lines for AP-style exam problems.

You should already know: Limit definition of the derivative, basic derivative rules (power rule, trigonometric derivatives, constant multiple rule), algebraic factoring and expansion.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Calculus AB style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Product rule?

The product rule is a core differentiation rule in calculus, used to find the derivative of any function that can be written as the product of two or more differentiable functions. Within the AP Calculus AB CED, product rule is part of Unit 2: Differentiation: Definition and Fundamental Properties, which contributes 10–12% of the total AP exam score. Product rule questions appear in both multiple-choice (MCQ) and free-response (FRQ) sections, both as standalone questions and as a required intermediate step in larger problems covering other topics.

Before learning the product rule, you can only differentiate linear combinations of basic functions (sums, differences, constant multiples). When a function is the product of two non-constant functions, you cannot simply take the product of the individual derivatives—that approach gives an incorrect result, as we will see in common pitfalls. The product rule accounts for the fact that a small change in $x$ changes both factors in the product, so both factors contribute to the total rate of change of the function. Mastery of this rule is required for nearly all differentiation problems that come later in the course.

2. Product Rule for Two Differentiable Functions

For any function $f(x)$ that can be written as $f(x)=g(x)h(x)$, where $g(x)$ and $h(x)$ are both differentiable at $x$, the product rule gives the derivative of $f(x)$ by combining the derivatives of each individual factor. We can derive the product rule directly from the limit definition of the derivative to confirm its form: $$ f'(x) = \lim_{\Delta x \to 0} \frac{g(x+\Delta x)h(x+\Delta x) - g(x)h(x)}{\Delta x} $$ We add and subtract $g(x+\Delta x)h(x)$ in the numerator to factor: $$ \text{Numerator} = g(x+\Delta x)\left(h(x+\Delta x) - h(x)\right) + h(x)\left(g(x+\Delta x) - g(x)\right) $$ Splitting the limit and using the fact that ${\lim }_{\Delta x\to 0}g(x+\Delta x)=g(x)$ (since $g$ is differentiable, hence continuous), we get the final product rule formula: $$ \boxed{f'(x) = g(x)h'(x) + h(x)g'(x)} $$ The intuition for this formula is straightforward: when $x$ increases, the change in the product comes from two sources: the change in the first factor while the second stays constant, and the change in the second factor while the first stays constant. This gives the two terms in the formula.

Worked Example

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

1. First, identify the two factors: $g(x)=x^3$, $h(x)=\sin x$.
2. Calculate the derivatives of each factor: $g^{\prime }(x)=3x^2$, $h^{\prime }(x)=\cos x$.
3. Substitute into the product rule formula: $f^{\prime }(x)=g(x)h^{\prime }(x)+h(x)g^{\prime }(x)=x^3\cos x+(\sin x)(3x^2)$.
4. Simplify by factoring out the common term $x^2$: $f^{\prime }(x)=x^2\left(x\cos x+3\sin x\right)$.

Exam tip: Always explicitly label $g(x),g^{\prime }(x),h(x),h^{\prime }(x)$ in your FRQ working. This prevents sign and multiplication errors, and makes it clear to graders you understand the rule, which can help with partial credit if you make a small arithmetic error.

3. Product Rule for Three or More Functions

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.

For three functions $f(x)=u(x)v(x)w(x)$, the extended product rule is: $$ f'(x) = u'(x)v(x)w(x) + u(x)v'(x)w(x) + u(x)v(x)w'(x) $$ 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. For example: $$ f(x) = \left(u(x)v(x)\right)w(x) \implies f'(x) = \left(u(x)v(x)\right)'w(x) + \left(u(x)v(x)\right)w'(x) $$ Applying the two-function rule to $(uv{)}^{\prime }$ gives $u^{\prime }v+u{v}^{\prime }$, so substituting back gives the same extended formula we had earlier. This method is foolproof and requires no extra memorization.

Worked Example

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

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

Exam tip: If you get nervous about memorizing the extended rule for 3+ factors, always group two factors and apply the two-function rule twice. This will always give the same correct answer with no extra risk of misremembering the pattern.

4. Finding Tangent Lines with the Product Rule

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. The process follows three core steps: 1) Use product rule to find $f^{\prime }(x)$, 2) Evaluate $f^{\prime }(a)$ 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 $f(x)=(x^2-4)(3x+2)$ at $x=1$.

1. Identify factors: $g(x)=x^2-4$, $h(x)=3x+2$. Their derivatives are $g^{\prime }(x)=2x$, $h^{\prime }(x)=3$.
2. Apply product rule to find $f^{\prime }(x)$:$$f^{\prime }(x)=(x^2-4)(3)+(3x+2)(2x)=3x^2-12+6x^2+4x=9x^2+4x-12$$
3. Calculate the slope at $x=1$: $f^{\prime }(1)=9(1{)}^2+4(1)-12=1$, so $m=1$.
4. Calculate the $y$-coordinate of the point: $f(1)=(1-4)(3+2)=-15$, so the point is $(1,-15)$.
5. Write the tangent line in slope-intercept form: $y+15=1(x-1)⟹y=x-16$.

Exam tip: If the product is of two polynomials, expand the original product and differentiate term-by-term to check your derivative. For this example, expanding $(x^2-4)(3x+2)=3x^3+2x^2-12x-8$, which gives derivative $9x^2+4x-12$, matching our product rule result. This is a quick, reliable check for polynomial product problems.

5. Common Pitfalls (and how to avoid them)

• Wrong move: For $f(x)=g(x)h(x)$, writing $f^{\prime }(x)=g^{\prime }(x)h^{\prime }(x)$. Why: Students extend the $(g+h{)}^{\prime }=g^{\prime }+h^{\prime }$ pattern for sums incorrectly to products. Correct move: Always write the full product rule $f^{\prime }=g{h}^{\prime }+h{g}^{\prime }$ before starting, and remind yourself that derivative of a product is not the product of derivatives.
• Wrong move: For $f(x)=3x^2\sin x$, writing $f^{\prime }(x)=3(2x\cos x)=6x\cos x$, forgetting the second term of the product rule. Why: Confusion between constant multiple rule and product rule, students incorrectly pull the constant out and stop. Correct move: Either factor the constant out first so you only apply product rule to $x^2\sin x$, or treat the constant as a factor with derivative zero, which will give the correct result.
• Wrong move: For $f(x)=(x+3{)}^2=(x+3)(x+3)$, calculating $f^{\prime }(x)=1\cdot 1=1$. Why: Same as the first pitfall, students multiply derivatives instead of applying product rule. Correct move: Apply product rule: $f^{\prime }(x)=(1)(x+3)+(x+3)(1)=2(x+3)$, which matches the chain rule result.
• Wrong move: For $f(x)=uvw$, writing $f^{\prime }(x)=u^{\prime }{v}^{\prime }w+u{v}^{\prime }{w}^{\prime }+u^{\prime }{v}^{\prime }{w}^{\prime }$. Why: Misremembering the extended pattern, taking 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, so $f^{\prime }=u^{\prime }vw+u{v}^{\prime }w+uv{w}^{\prime }$.
• Wrong move: When finding a tangent line at $x=a$, calculating $m=f(a)$ instead of $m=f^{\prime }(a)$. Why: Rushing through the problem, mixing up what the original function and derivative represent. Correct move: Explicitly label "slope = $f^{\prime }(a)$" and "y-coordinate = $f(a)$" in your working to avoid mixing them up.

6. Practice Questions (AP Calculus AB Style)

Question 1 (Multiple Choice)

What is the derivative of $f(x)=4x{e}^{2x}$? A) $4e^{2x}(2x+1)$ B) $8e^{2x}$ C) $4e^{2x}(x+1)$ D) $8x{e}^{2x}$

Worked Solution: First, identify $f(x)$ as the product of $g(x)=4x$ and $h(x)=e^{2x}$. Calculate derivatives: $g^{\prime }(x)=4$, and by the chain rule $h^{\prime }(x)=2e^{2x}$. Substitute into the product rule: $$ f'(x) = g(x)h'(x) + g'(x)h(x) = (4x)(2e^{2x}) + (4)(e^{2x}) = 8x e^{2x} + 4e^{2x} $$ Factor out $4e^{2x}$ to get $f^{\prime }(x)=4e^{2x}(2x+1)$. The correct answer is A.

Question 2 (Free Response)

Let $f(x)=(x^2-5x)\cos x$. (a) Find $f^{\prime }(x)$. (b) Find the slope of the tangent line to $f(x)$ at $x=\pi$. (c) Is $f(x)$ increasing or decreasing at $x=\pi$? Justify your answer.

Worked Solution: (a) Let $g(x)=x^2-5x$ and $h(x)=\cos x$. Then $g^{\prime }(x)=2x-5$ and $h^{\prime }(x)=-\sin x$. Applying product rule: $$ f'(x) = (x^2 - 5x)(-\sin x) + (2x - 5)\cos x = -(x^2 - 5x)\sin x + (2x - 5)\cos x $$

(b) Substitute $x=\pi$, using $\sin \pi =0$ and $\cos \pi =-1$: $$ f'(\pi) = -(\pi^2 - 5\pi)(0) + (2\pi - 5)(-1) = 5 - 2\pi \approx -1.28 $$ The slope of the tangent line is $5-2\pi$.

(c) $f(x)$ is decreasing at $x=\pi$. Justification: $f^{\prime }(\pi )=5-2\pi <0$, and a function is decreasing at a point when its first derivative at that point is negative.

Question 3 (Application / Real-World Style)

A local bakery sells loaves of sourdough bread. The number of loaves sold per day depends on the price $p$ per loaf: $N(p)=120-3p$. Over 12 weeks, the bakery raises prices gradually, so price after $t$ weeks is $p(t)=3+0.25t$, for $0\le t\le 12$. Daily revenue is $R(t)=N(p(t))\cdot p(t)$, in dollars. Find the rate of change of revenue with respect to time at $t=8$, and interpret your result in context.

Worked Solution: First rewrite $R(t)$ as a product of functions of $t$: $$ N(p(t)) = 120 - 3(3 + 0.25t) = 111 - 0.75t, \quad R(t) = (111 - 0.75t)(3 + 0.25t) $$ Apply product rule, with $g(t)=111-0.75t$, $h(t)=3+0.25t$, so $g^{\prime }(t)=-0.75$, $h^{\prime }(t)=0.25$: $$ R'(t) = (111 - 0.75t)(0.25) + (-0.75)(3 + 0.25t) $$ Evaluate at $t=8$: $$ R'(8) = (111 - 6)(0.25) - 0.75(3 + 2) = (105)(0.25) - 0.75(5) = 26.25 - 3.75 = 22.5 $$ Interpretation: After 8 weeks of gradual price increases, the bakery's daily revenue is increasing at a rate of $\$22.50$ per week.

7. Quick Reference Cheatsheet

8. 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 MCQ and FRQ. 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. The pattern of the product rule also lays conceptual groundwork for integration by parts, a core technique in AP Calculus BC that extends directly from what you learn here.

• Chain rule
• Quotient rule
• Tangent lines to curves
• Implicit differentiation