Product rule — AP Calculus BC Study Guide

For: AP Calculus BC candidates sitting AP Calculus BC.

Covers: The product rule formula for products of two differentiable functions, extension to three or more functions, tangent line applications, and algebraic simplification of product rule derivatives for AP exam questions.

You should already know: Limit definition of the derivative, basic power rule for polynomials, derivative rules for trigonometric, exponential, and logarithmic functions.

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 BC 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 technique that allows you to compute the derivative of a product of two or more differentiable functions, without needing to expand the product (which is often impossible for non-polynomial products like $x\sin x$ or $e^x\ln x$). This rule is derived directly from the limit definition of the derivative, so it is consistent with all other fundamental differentiation properties you have learned. Per the AP Calculus CED, Unit 2 (Differentiation: Definition and Fundamental Properties) accounts for 10–12% of the total AP exam score, and product rule problems appear in both multiple-choice (MCQ) and free-response (FRQ) sections. You will also see product rule used as a foundational step for more advanced topics later in the course, so mastering it early is critical. Sometimes it is called the Leibniz product rule, after its discoverer, but you will almost always see it referred to as the product rule on the AP exam.

2. Product Rule for Two Functions

If we have two differentiable functions $f(x)$ and $g(x)$, their product $h(x)=f(x)\cdot g(x)$ is also differentiable, and its derivative is given by the product rule formula. To build intuition for why the formula has its form, we can derive it quickly from the limit definition of the derivative: $$ h'(x) = \lim_{\Delta x \to 0} \frac{f(x+\Delta x)g(x+\Delta x) - f(x)g(x)}{\Delta x} $$ Add and subtract $f(x+\Delta x)g(x)$ in the numerator to split the limit into two separate terms: $$ \lim_{\Delta x \to 0} \left[ f(x+\Delta x) \frac{g(x+\Delta x) - g(x)}{\Delta x} + g(x) \frac{f(x+\Delta x) - f(x)}{\Delta x} \right] $$ Taking the limit as $\Delta x\to 0$, this simplifies to the standard product rule: $h^{\prime }(x)=f(x)g^{\prime }(x)+g(x)f^{\prime }(x)$. The intuition is that the derivative of the product accounts for the change in both functions: the first function times the change in the second, plus the second function times the change in the first. A common mnemonic to remember the order is "first times derivative of the second, plus second times derivative of the first".

Worked Example

Find the derivative of $h(x)=(3x^2-2x)(e^x+\sin x)$.

1. Identify the two functions: let $f(x)=3x^2-2x$ (first function) and $g(x)=e^x+\sin x$ (second function).
2. Compute the derivative of each function: $f^{\prime }(x)=6x-2$, $g^{\prime }(x)=e^x+\cos x$.
3. Apply the product rule formula: $h^{\prime }(x)=f(x)g^{\prime }(x)+g(x)f^{\prime }(x)$.
4. Substitute and simplify (expand only if required):$$h^{\prime }(x)=(3x^2-2x)(e^x+\cos x)+(e^x+\sin x)(6x-2)$$
5. If expanded form is requested:$$h^{\prime }(x)=3x^2{e}^x+4x{e}^x-2e^x+3x^2\cos x-2x\cos x+6x\sin x-2\sin x$$

Exam tip: On the AP exam, you do not always need to expand the final derivative unless the question explicitly asks for expanded form; leaving it in factored product rule form is usually acceptable for full credit.

3. Product Rule for Three or More Functions

The product rule can be extended to any number of differentiable functions by applying the two-function rule repeatedly, and a clear pattern emerges that makes it easy to remember. For three functions $f(x),g(x),h(x)$, group the product as $f\cdot (g\cdot h)$ and apply the two-function rule: $$ \frac{d}{dx}[f g h] = f' (g h) + f \frac{d}{dx}(g h) = f' g h + f (g' h + g h') = f' g h + f g' h + f g h' $$ The general pattern for $n$ functions is simple: the derivative is the sum of $n$ terms, where each term is the product of all original functions except we take the derivative of exactly one function. AP exams rarely ask for products of more than 3 functions, but 3-function products commonly appear in MCQ problems.

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 the derivative of each individual function: $f_1^{\prime }(x)=1$, $f_2^{\prime }(x)=-\sin x$, $f_3^{\prime }(x)=\frac{1}{x}$.
3. Apply the 3-function product rule: $\frac{dy}{dx}=f_1^{\prime }{f}_2{f}_3+f_1{f}_2^{\prime }{f}_3+f_1{f}_2{f}_3^{\prime }$.
4. Substitute and simplify:$$\frac{dy}{dx}=(1)(\cos x)(\ln x)+(x)(-\sin x)(\ln x)+(x)(\cos x)\left(\frac{1}{x}\right)$$ The $x$ terms cancel in the final term, so the simplified derivative is:$$\frac{dy}{dx}=\cos x\ln x-x\sin x\ln x+\cos x$$

Exam tip: When working with 3-function products, always double-check that you have exactly three terms in your derivative; missing one term is the most common mistake on these problems.

4. Tangent Lines to Product Functions

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 the point-slope form of a line, a skill tested frequently in both MCQ and early FRQ parts, and it serves as a foundation for later topics like linear approximation. To solve these problems, you need two pieces of information: the point of tangency $(x_0,y_0)$ (where $y_0=h(x_0)$ for product function $h(x)$) and the slope of the tangent line at $x_0$, which equals $h^{\prime }(x_0)$ found via the product rule. Once you have these values, substitute into point-slope form to get the tangent line equation.

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: $h(2)=(2^2-4)(2^2)=0\cdot 4=0$, so the point is $(2,0)$.
2. Identify the two functions: $f(x)=x^2-4$, $g(x)=2^x$, and compute their derivatives: $f^{\prime }(x)=2x$, $g^{\prime }(x)=2^x\ln 2$.
3. Apply the product rule to get $h^{\prime }(x)$: $h^{\prime }(x)=(x^2-4)(2^x\ln 2)+2^x(2x)$.
4. Evaluate $h^{\prime }(2)$ to get the slope: $h^{\prime }(2)=(4-4)(4\ln 2)+4(4)=0+16=16$, so $m=16$.
5. Write the tangent line equation: $y-0=16(x-2)$, or simplified to slope-intercept form: $y=16x-32$.

Exam tip: Always compute the $y$-coordinate of the tangency point before you compute the slope; it is common to accidentally plug $x_0$ into the derivative for $y_0$, which costs easy points.

5. Common Pitfalls (and how to avoid them)

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

6. Practice Questions (AP Calculus BC Style)

Question 1 (Multiple Choice)

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) $-(x^3-5x^2)\sin x+(3x^2-10x)\cos x$

Worked Solution: Let $f(x)=x^3-5x^2$ and $g(x)=\cos x$. By the two-function product rule, $\frac{d}{dx}[f(x)g(x)]=f(x)g^{\prime }(x)+g(x)f^{\prime }(x)$. Compute the individual derivatives: $f^{\prime }(x)=3x^2-10x$ and $g^{\prime }(x)=-\sin x$. Substitute into the formula to get: $(x^3-5x^2)(-\sin x)+(\cos x)(3x^2-10x)=-(x^3-5x^2)\sin x+(3x^2-10x)\cos x$. This matches option D. The correct answer is D.

Question 2 (Free Response)

Let $h(x)=2x{e}^x\sin x$. (a) Find $h^{\prime }(x)$, the general derivative of $h(x)$. (b) Find the slope of the tangent line to $h(x)$ at $x=\pi$. (c) Is the function $h(x)$ increasing or decreasing at $x=\pi$? Justify your answer.

Worked Solution: (a) Label the three differentiable functions: $f(x)=2x$, $g(x)=e^x$, $k(x)=\sin x$. Their derivatives are $f^{\prime }(x)=2$, $g^{\prime }(x)=e^x$, $k^{\prime }(x)=\cos x$. Apply the 3-function product rule: $$ h'(x) = f'(x)g(x)k(x) + f(x)g'(x)k(x) + f(x)g(x)k'(x) $$ Substitute and factor to simplify: $$ h'(x) = 2 e^x \sin x + 2x e^x \sin x + 2x e^x \cos x = 2e^x\left(\sin x + x \sin x + x \cos x\right) $$

(b) Evaluate $h^{\prime }(\pi )$ to get the slope. We know $\sin \pi =0$ and $\cos \pi =-1$, so: $$ h'(\pi) = 2e^\pi\left(0 + \pi(0) + \pi(-1)\right) = -2\pi e^\pi $$ The slope of the tangent line at $x=\pi$ is $-2\pi e^{\pi }$.

(c) The function $h(x)$ is decreasing at $x=\pi$. Justification: For a differentiable function, the sign of the derivative at a point determines whether the function is increasing or decreasing. Since $h^{\prime }(\pi )=-2\pi e^{\pi }<0$, $h(x)$ is decreasing at $x=\pi$.

Question 3 (Application / Real-World Style)

A biologist is modeling the population of bacteria in a petri dish as a function of time $t$ (measured in hours) given by $P(t)=t(10+2e^{-t})$, where $P(t)$ is measured in thousands of bacteria. Find the rate of change of the population at $t=2$ hours, and interpret your result in context.

Worked Solution: We need $P^{\prime }(2)$ to find the instantaneous rate of change. Let $f(t)=t$ and $g(t)=10+2e^{-t}$. Apply the product rule: $$ P'(t) = f(t)g'(t) + g(t)f'(t) = t(-2e^{-t}) + (10 + 2e^{-t})(1) = 10 + 2e^{-t} - 2t e^{-t} $$ Evaluate at $t=2$: $$ P'(2) = 10 + 2e^{-2} - 4e^{-2} = 10 - 2e^{-2} \approx 10 - 2(0.135) = 9.73 $$ Since $P(t)$ is measured in thousands of bacteria, the rate of change is approximately 9.73 thousand bacteria per hour. Interpretation: At 2 hours after the start of the experiment, the population of bacteria is increasing at a rate of approximately 9730 bacteria per hour.

7. Quick Reference Cheatsheet

8. 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 the habit of applying it correctly will pay off for the entire course.

• Quotient rule
• Chain rule
• Implicit differentiation
• Integration by parts