AP®MathsCollege BoardCalculus ABStudy GuidesUnit 2: Differentiation Definition & Fundamental PropertiesFundamental Properties of DifferentiationThe Product Rule

The Product Rule (College Board AP® Calculus AB): Study Guide

Jamie Wood

Written by: Jamie Wood

Reviewed by: Dan Finlay

Updated on

College Board Calculus ABCollege Board Calculus BC

Derivatives of products

How do I differentiate the product of two functions?

  • The derivative of the product of two functions can be found by using the product rule

  • The product rule states that

    • If f open parentheses x close parentheses equals g open parentheses x close parentheses times h open parentheses x close parentheses,

    • then f to the power of apostrophe open parentheses x close parentheses equals g to the power of apostrophe open parentheses x close parentheses times h open parentheses x close parentheses space plus space g open parentheses x close parentheses times h to the power of apostrophe open parentheses x close parentheses

  • This is also commonly written as

    • If y equals u times v,

    • then fraction numerator d y over denominator d x end fraction equals fraction numerator d u over denominator d x end fraction times v space plus space u times fraction numerator d v over denominator d x end fraction

    • Or in a more concise form: y to the power of apostrophe equals u to the power of apostrophe v space plus space u v to the power of apostrophe

Examiner Tips and Tricks

  • Don't confuse the product of two functions with a composite function

    • The product of two functions like f open parentheses x close parentheses times g open parentheses x close parentheses is two functions multiplied together

    • A composite function like f open parentheses g open parentheses x close parentheses close parentheses is a function of a function

    • For how to differentiate composite functions, see the "Chain rule" study guide

Worked Example

Find the derivative of the following functions.

(a) f open parentheses x close parentheses equals e to the power of 2 x end exponent open parentheses x to the power of 5 plus 3 x close parentheses

Answer:

Assign u and v to each function

u equals e to the power of 2 x end exponent

v equals x to the power of 5 plus 3 x

Find the derivatives of u and v

u to the power of apostrophe equals 2 e to the power of 2 x end exponent

v to the power of apostrophe equals 5 x to the power of 4 plus 3

Apply the product rule, y to the power of apostrophe equals u to the power of apostrophe v space plus space u v to the power of apostrophe

y to the power of apostrophe equals 2 e to the power of 2 x end exponent open parentheses x to the power of 5 plus 3 x close parentheses plus e to the power of 2 x end exponent open parentheses 5 x to the power of 4 plus 3 close parentheses

If you have used a different notation such as y to the power of apostrophe when working, you should write your final answer in a format that matches how the question was asked

f to the power of apostrophe open parentheses x close parentheses equals 2 e to the power of 2 x end exponent open parentheses x to the power of 5 plus 3 x close parentheses plus e to the power of 2 x end exponent open parentheses 5 x to the power of 4 plus 3 close parentheses

This answer can also be factorized

f to the power of apostrophe open parentheses x close parentheses equals e to the power of 2 x end exponent open parentheses 2 x to the power of 5 plus 5 x to the power of 4 plus 6 x plus 3 close parentheses

(b) g open parentheses x close parentheses equals ln space 3 x times sin space 2 x

Answer:

Assign u and v to each function

u equals ln space 3 x

v equals sin space 2 x

Find the derivatives of u and v

u to the power of apostrophe equals 1 over x

v to the power of apostrophe equals 2 cos space 2 x

Apply the product rule, y to the power of apostrophe equals u to the power of apostrophe v space plus space u v to the power of apostrophe

y to the power of apostrophe equals 1 over x times sin space 2 x space plus space ln space 3 x times 2 cos space 2 x

Simplify

g to the power of apostrophe open parentheses x close parentheses equals fraction numerator sin space 2 x over denominator x end fraction plus 2 ln open parentheses 3 x close parentheses cos open parentheses 2 x close parentheses

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Author
Reviewer
Jamie Wood

Author: Jamie Wood

Expertise: Maths Content Creator

Jamie graduated in 2014 from the University of Bristol with a degree in Electronic and Communications Engineering. He has worked as a teacher for 8 years, in secondary schools and in further education; teaching GCSE and A Level. He is passionate about helping students fulfil their potential through easy-to-use resources and high-quality questions and solutions.

Dan Finlay

Reviewer: Dan Finlay

Expertise: Maths Subject Lead

Dan graduated from the University of Oxford with a First class degree in mathematics. As well as teaching maths for over 8 years, Dan has marked a range of exams for Edexcel, tutored students and taught A Level Accounting. Dan has a keen interest in statistics and probability and their real-life applications.

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