\( \newcommand{\abs}[1]{\left| \, {#1} \, \right| } \) \( \newcommand{\cm}{\mathrm{cm} } \) \( \newcommand{\sec}{ \, \mathrm{sec} \, } \) \( \newcommand{\vhat}[1]{\,\hat{#1}} \) \( \newcommand{\vhati}{\,\hat{i}} \) \( \newcommand{\vhatj}{\,\hat{j}} \) \( \newcommand{\vhatk}{\,\hat{k}} \) \( \newcommand{\vect}[1]{\boldsymbol{\vec{#1}}} \) \( \newcommand{\norm}[1]{\|{#1}\|} \) \( \newcommand{\arccot}{ \, \mathrm{arccot} \, } \) \( \newcommand{\arcsec}{ \, \mathrm{arcsec} \, } \) \( \newcommand{\arccsc}{ \, \mathrm{arccsc} \, } \) \( \newcommand{\sech}{ \, \mathrm{sech} \, } \) \( \newcommand{\csch}{ \, \mathrm{csch} \, } \) \( \newcommand{\arcsinh}{ \, \mathrm{arcsinh} \, } \) \( \newcommand{\arccosh}{ \, \mathrm{arccosh} \, } \) \( \newcommand{\arctanh}{ \, \mathrm{arctanh} \, } \) \( \newcommand{\arccoth}{ \, \mathrm{arccoth} \, } \) \( \newcommand{\arcsech}{ \, \mathrm{arcsech} \, } \) \( \newcommand{\arccsch}{ \, \mathrm{arccsch} \, } \)

17Calculus Derivatives - Product Rule

Limits

Using Limits

Limits FAQs

Derivatives

Graphing

Applications

Derivatives FAQs

Integrals

Improper Integrals

Trig Integrals

Length-Area-Volume

Applications - Tools

Infinite Series

Applications

Tools

Parametrics

Conics

Polar Coordinates

Practice

Practice Problems

Practice Exams

Tools

Calculus Tools

Additional Tools

Articles

Derivatives

Graphing

Applications

Derivatives FAQs

SV Calculus

MV Calculus

Practice

Practice Problems

Practice Exams

Tools

Calculus Tools

Additional Tools

Articles

Product Rule

The Product Rule is pretty straight-forward. If you have a function with two main parts that are multiplied together, for example \(h(x)=f(x) \cdot g(x)\), the derivative is

\(\displaystyle{ \frac{dh}{dx} = }\) \(\displaystyle{ \frac{d}{dx}[f \cdot g] = }\) \(\displaystyle{ f'\cdot g + f \cdot g' }\)

An interesting thing to notice about the product rule is that the constant multiple rule is just a special case of the product rule. For example, if you have function \(f(x) = cg(x)\), the product rule says \(f'(x) = (c)' g(x) + c g'(x) =\) \( 0 + c g'(x) = c g'(x)\). Notice that we use the constant rule to say that \(d[c]/dx = 0\).

And that's all you need to know to use the product rule. A common mistake many students make is to think that the product rule allows you to take the derivative of both terms and multiply them together. WRONG! If this confuses you, go back to the top of the page and reread the product rule and then go through some examples in your textbook.

Before you start using the product rule, it is important to know where it comes from. So take a few minutes to watch this video showing the proof of the product rule.

PatrickJMT - Product Rule Proof [6min-6secs]

video by PatrickJMT

Okay, practice problem time. Once you are finished with those, the quotient rule is the next logical step.

Practice

Unless otherwise instructed, calculate the derivatives of these functions using the product rule, giving your final answers in simplified, factored form.

Basic

Here are some problems that use only the product rule, the power rule and the other basic rules on the main derivatives page.

\( f(x) = x(25-x) \)

Problem Statement

Calculate the derivative of this function using the product rule, giving your final answer in simplified, factored form. \( f(x) = x(25-x) \)

Solution

938 video

video by Krista King Math

close solution

Log in to rate this practice problem and to see it's current rating.

\( f(x) = (x-2)(x+3) \)

Problem Statement

Calculate the derivative of this function using the product rule, giving your final answer in simplified, factored form. \( f(x) = (x-2)(x+3) \)

Final Answer

\(f'(x)=2x+1\)

Problem Statement

Calculate the derivative of this function using the product rule, giving your final answer in simplified, factored form. \( f(x) = (x-2)(x+3) \)

Solution

1928 video

video by PatrickJMT

Final Answer

\(f'(x)=2x+1\)

close solution

Log in to rate this practice problem and to see it's current rating.

\( f(x) = (2x+3)(3x-2) \)

Problem Statement

Calculate the derivative of this function using the product rule, giving your final answer in simplified, factored form. \( f(x) = (2x+3)(3x-2) \)

Solution

939 video

video by Krista King Math

close solution

Log in to rate this practice problem and to see it's current rating.

\(h(x)=(x^2+5x+7)(x^3+2x-4)\)

Problem Statement

Calculate the derivative of this function using the product rule, giving your final answer in simplified, factored form. \(h(x)=(x^2+5x+7)(x^3+2x-4)\)

Final Answer

\(h'(x)=5x^4+20x^3+27x^2+12x-6\)

Problem Statement

Calculate the derivative of this function using the product rule, giving your final answer in simplified, factored form. \(h(x)=(x^2+5x+7)(x^3+2x-4)\)

Solution

Use the product rule.

\(\displaystyle{ h'(x) = (x^2 + 5x + 7)\frac{d}{dx}[x^3 + 2x - 4] + }\) \(\displaystyle{ (x^3 + 2x - 4) \frac{d}{dx}[x^2 + 5x + 7] }\)

Take the derivative of the appropriate terms.

\( h'(x) = (x^2 + 5x + 7)(3x^2 + 2) + \) \( (x^3 + 2x - 4)(2x + 5) \)

Check for common terms. In this case, there are not any, so multiply out. Leaving it in this form is not usually considered proper form but as usual check with your instructor to see what they expect.

\( h'(x) = (x^2)(3x^2 + 2) + 5x(3x^2 + 2) + \) \( 7(3x^2 + 2) + (x^3)(2x+5) + \) \( 2x(2x+5) - 4(2x+5) \)

\( h'(x) = 3x^4 + 2x^2 + 15x^3 + \) \( 10x + 21x^2 + 14 + 2x^4 + \) \(5x^3 + 4x^2 + 10x - 8x - 20 \)

Collect common terms.

\( h'(x) = 5x^4 + 20x^3 + 27x^2 + 12x - 6 \)

Final Answer

\(h'(x)=5x^4+20x^3+27x^2+12x-6\)

close solution

Log in to rate this practice problem and to see it's current rating.

\( g(x) = (x^3-7x^2+4)(3x^2+14) \)

Problem Statement

Calculate the derivative of this function using the product rule, giving your final answer in simplified, factored form. \( g(x) = (x^3-7x^2+4)(3x^2+14) \)

Final Answer

\( g'(x) = x(15x^3-84x^2+42x-172) \)

Problem Statement

Calculate the derivative of this function using the product rule, giving your final answer in simplified, factored form. \( g(x) = (x^3-7x^2+4)(3x^2+14) \)

Solution

First, we apply the product rule.

\( \displaystyle{ \frac{dg}{dx} = }\) \(\displaystyle{ (x^3-7x^2+4)\cdot \frac{d}{dx}[3x^2+14]+ }\) \(\displaystyle{ (3x^2+14)\cdot \frac{d}{dx}[x^3-7x^2+4] }\)

Now, we evaluate the derivatives in the above equation.

\( \displaystyle{ \frac{dg}{dx} = }\) \((x^3-7x^2+4)(6x)\) \(+\) \((3x^2+14)(3x^2-14x)\)

Before we simplify by multiplying out, let's look for common factors that we can factor out. Notice that there is an \(x\) in both of the terms. So we factor that out.

\(g'(x) = x[6x^3-42x^2+24+ \) \((3x^2+14)(3x-14)]\)

Now, we multiply out the terms inside the brackets.

\(g'(x) = x(15x^3-84x^2+42x-172)\)

Factoring right after taking the derivative, before multiplying out, makes it easier to see common terms.

Final Answer

\( g'(x) = x(15x^3-84x^2+42x-172) \)

close solution

Log in to rate this practice problem and to see it's current rating.

\( g(x) = (3x^4+2x-1)(x^5-2x^2) \)

Problem Statement

Calculate the derivative of this function using the product rule, giving your final answer in simplified, factored form. \( g(x) = (3x^4+2x-1)(x^5-2x^2) \)

Solution

1311 video

video by PatrickJMT

close solution

Log in to rate this practice problem and to see it's current rating.

These problems require you to know how to take the derivative of exponential functions.

\(\displaystyle{ f(x) = x^3e^x }\)

Problem Statement

Calculate the derivative of this function using the product rule, giving your final answer in simplified, factored form. \(\displaystyle{ f(x) = x^3e^x }\)

Final Answer

\(\displaystyle{ \frac{d}{dx} \left[ x^3e^x \right] = x^2e^x(x+3) }\)

Problem Statement

Calculate the derivative of this function using the product rule, giving your final answer in simplified, factored form. \(\displaystyle{ f(x) = x^3e^x }\)

Solution

\(\displaystyle{ \frac{d}{dx}\left[x^3e^x\right] }\)

Use the product rule.

\(\displaystyle{ x^3\cdot\frac{d[e^x]}{dx}+e^x\cdot\frac{d[x^3]}{dx} }\)

Take the derivative of the two terms.

\( x^3\cdot e^x+e^x\cdot3x^2 \)

Simplify by factoring.

\( x^2e^x(x+3) \)

Final Answer

\(\displaystyle{ \frac{d}{dx} \left[ x^3e^x \right] = x^2e^x(x+3) }\)

close solution

Log in to rate this practice problem and to see it's current rating.

\(\displaystyle{y=7e^{2x}}\)

Problem Statement

Calculate the derivative of this function using the product rule, giving your final answer in simplified, factored form. \(\displaystyle{y=7e^{2x}}\)

Final Answer

\(\displaystyle{ \frac{d}{dx} \left[7e^{2x} \right] = 14e^{2x} }\)

Problem Statement

Calculate the derivative of this function using the product rule, giving your final answer in simplified, factored form. \(\displaystyle{y=7e^{2x}}\)

Solution

\(\displaystyle{ \frac{d}{dx} \left[ 7e^{2x} \right] }\)

\(\displaystyle{ 7 \frac{d}{dx}[ e^x \cdot e^x ] }\)

\(\displaystyle{ 7e^x \frac{d[e^x]}{dx} + 7e^x \frac{d[e^x]}{dx} }\)

\( 7e^x (e^x) + 7e^x (e^x) \)

\( 14e^{2x} \)

Note: The derivative is more easily calculated using the chain rule. However, since you are not required to know the chain rule for this page, this is a valid way to work it without the chain rule.

Final Answer

\(\displaystyle{ \frac{d}{dx} \left[7e^{2x} \right] = 14e^{2x} }\)

close solution

Log in to rate this practice problem and to see it's current rating.

\(e^x\sqrt{x}\)

Problem Statement

Calculate the derivative of this function using the product rule, giving your final answer in simplified, factored form. \(e^x\sqrt{x}\)

Final Answer

\(\displaystyle{ \frac{e^x (1 + 2x)}{2x^{1/2}} }\)

Problem Statement

Calculate the derivative of this function using the product rule, giving your final answer in simplified, factored form. \(e^x\sqrt{x}\)

Solution

\(\displaystyle{ \frac{d}{dx}\left[ e^x \sqrt{x} \right] }\)

Rewrite \(\sqrt{x}\) as \(x^{1/2}\) so that we can use the power rule.

\(\displaystyle{ \frac{d}{dx}\left[ e^x \left( x^{1/2} \right) \right] }\)

Use the product rule.

\(\displaystyle{ (e^x) \frac{d[x^{1/2}]}{dx} + \left( x^{1/2} \right) \frac{d[e^x]}{dx} }\)

\( e^x (1/2)x^{-1/2} + x^{1/2} e^x \)

\(\displaystyle{ \left[ \frac{e^x}{2x^{1/2}} \right] (1 + 2x) }\)

Notes -
After we used the product rule, we just used algebra to simplify and factor. Although there are many ways to write the final answer, we usually want all factors written with positive exponents, except possibly exponential terms. Your instructor may or may not enforce this but most textbooks will write it this way. Writing your answer as \(e^x \left[ (1/2)x^{-1/2} + x^{1/2} \right]\) is not considered completely factored because both factors inside the parentheses have an x term.

Final Answer

\(\displaystyle{ \frac{e^x (1 + 2x)}{2x^{1/2}} }\)

close solution

Log in to rate this practice problem and to see it's current rating.

These problems require you to know how to take the derivative of trig functions.

\(x^4\tan(x)\)

Problem Statement

Calculate the derivative of this function using the product rule, giving your final answer in simplified, factored form. \(x^4\tan(x)\)

Final Answer

\(\displaystyle{ x^3[x \sec^2(x)+4 \tan(x)] }\)

Problem Statement

Calculate the derivative of this function using the product rule, giving your final answer in simplified, factored form. \(x^4\tan(x)\)

Solution

\(\displaystyle{ \frac{d}{dx}[ x^4 \tan(x) ] }\)

apply the product rule

\( \displaystyle{ x^4 \frac{d}{dx}[ \tan(x) ] + \tan(x) \frac{d}{dx}[ x^4 ] }\)

take derivatives

\( x^4 \sec^2(x) + 4x^3 \tan(x) \)

factor out common terms

\( x^3[x \sec^2(x)+4 \tan(x)] \)

Note: Not factoring out the \(x^3\) term, may cost you points. Check with your instructor to see what they expect.

Final Answer

\(\displaystyle{ x^3[x \sec^2(x)+4 \tan(x)] }\)

close solution

Log in to rate this practice problem and to see it's current rating.

\(\displaystyle{f(x)=\cos(x)\sin(x)}\)

Problem Statement

Calculate the derivative of this function using the product rule, giving your final answer in simplified, factored form. \(\displaystyle{f(x)=\cos(x)\sin(x)}\)

Final Answer

\( f'(x) = \cos^2(x) - \sin^2(x) \)

Problem Statement

Calculate the derivative of this function using the product rule, giving your final answer in simplified, factored form. \(\displaystyle{f(x)=\cos(x)\sin(x)}\)

Solution

\(\displaystyle{ f'(x) = \frac{d}{dx}\left[ \cos(x) \sin(x) \right] }\)

\(\displaystyle{ f'(x) = \cos(x) \frac{d}{dx}[ \sin(x)] + }\) \(\displaystyle{ \sin(x) \frac{d}{dx}[ \cos(x)] }\)

\( f'(x) = \cos(x)[ \cos(x)] + \sin(x)[-\sin(x)] \)

\( f'(x) = \cos^2(x) - \sin^2(x) \)

Final Answer

\( f'(x) = \cos^2(x) - \sin^2(x) \)

close solution

Log in to rate this practice problem and to see it's current rating.

Intermediate

\(\displaystyle{g(x)=(x^2+3x-5)\sqrt[3]{x}}\)

Problem Statement

Calculate the derivative of this function using the product rule, giving your final answer in simplified, factored form. \(\displaystyle{g(x)=(x^2+3x-5)\sqrt[3]{x}}\)

Final Answer

\(\displaystyle{ \frac{d}{dx}\left[(x^2+3x-5)\sqrt[3]{x}\right] = \frac{7x^2+12x-5}{3x^{2/3}}}\)

Problem Statement

Calculate the derivative of this function using the product rule, giving your final answer in simplified, factored form. \(\displaystyle{g(x)=(x^2+3x-5)\sqrt[3]{x}}\)

Solution

Rewrite \(\sqrt[3]{x}\) as \( x^{1/3} \) to use the power rule.

\(\displaystyle{ g'(x) = \frac{d}{dx}\left[ (x^2+3x-5)(x^{1/3}) \right] }\)

Use the product rule.

\(\displaystyle{ g'(x) = (x^2 + 3x - 5) \cdot \frac{d[x^{1/3}]}{dx} + }\) \(\displaystyle{ \left( x^{1/3} \right) \cdot \frac{d[x^2+3x-5]}{dx} }\)

Take the two derivatives.

\(g'(x) = (x^2 + 3x - 5) \cdot (1/3)x^{-2/3} + \) \( x^{1/3} \cdot (2x+3) \)

\(\displaystyle{ g'(x) = \frac{ x^2 + 3x - 5}{ 3x^{2/3}} + }\) \(\displaystyle{ \left[ x^{1/3} \cdot (2x+3) \right] \left[\frac{3x^{2/3}}{3x^{2/3}} \right] }\)

\(\displaystyle{ g'(x) = \frac{ x^2 + 3x - 5 + (3x)(2x+3) }{3x^{2/3}} }\)

\(\displaystyle{ g'(x) = \frac{7x^2 + 12x - 5}{3x^{2/3}} }\)

Notes -
After we used the product rule, we just used algebra to simplify and factor. Although there are many ways to write the final answer, we usually want all factors written with positive exponents. Your instructor may or may not enforce this but most textbooks will write it this way.
Also, you may see \(x^{2/3}\) written as \(\sqrt[3]{x^2}\). Although this is correct notation, in calculus we usually leave our answer in fractional exponent form. Ask your instructor which form he/she prefers or check your textbook to see which form it shows the most.

Final Answer

\(\displaystyle{ \frac{d}{dx}\left[(x^2+3x-5)\sqrt[3]{x}\right] = \frac{7x^2+12x-5}{3x^{2/3}}}\)

close solution

Log in to rate this practice problem and to see it's current rating.

\(\displaystyle{f(x)=\left[\frac{1}{x^2}-\frac{3}{x^4}\right]\cdot (x+5x^3) }\)

Problem Statement

Calculate the derivative of this function using the product rule, giving your final answer in simplified, factored form. \(\displaystyle{f(x)=\left[\frac{1}{x^2}-\frac{3}{x^4}\right]\cdot (x+5x^3) }\)

Solution

935 video

video by Krista King Math

close solution

Log in to rate this practice problem and to see it's current rating.

\( H(u)=(u-\sqrt{u}) (u+\sqrt{u}) \)

Problem Statement

Calculate the derivative of this function using the product rule, giving your final answer in simplified, factored form. \( H(u)=(u-\sqrt{u}) (u+\sqrt{u}) \)

Solution

936 video

video by Krista King Math

close solution

Log in to rate this practice problem and to see it's current rating.

\(f(x)=x(2x-1)(2x+1)\)

Problem Statement

Calculate the derivative of this function using the product rule, giving your final answer in simplified, factored form. \(f(x)=x(2x-1)(2x+1)\)

Solution

940 video

video by Krista King Math

close solution

Log in to rate this practice problem and to see it's current rating.

\(f(x)=(2x+1)(4-x^2)(1+x^2)\)

Problem Statement

Calculate the derivative of this function using the product rule, giving your final answer in simplified, factored form. \(f(x)=(2x+1)(4-x^2)(1+x^2)\)

Solution

937 video

video by Krista King Math

close solution

Log in to rate this practice problem and to see it's current rating.

Here is one with a trig derivative.

\(\displaystyle{\frac{3}{x}\cot(x)}\)

Problem Statement

Calculate the derivative of this function using the product rule, giving your final answer in simplified, factored form. \(\displaystyle{\frac{3}{x}\cot(x)}\)

Final Answer

\(\displaystyle{ \frac{d}{dx}\left[ \frac{3}{x} \cot(x) \right] = \frac{-3}{x^2}\left[ x \csc^2(x) + \cot(x) \right] }\)

Problem Statement

Calculate the derivative of this function using the product rule, giving your final answer in simplified, factored form. \(\displaystyle{\frac{3}{x}\cot(x)}\)

Solution

\(\displaystyle{ \frac{d}{dx}\left[ \frac{3}{x} \cot(x) \right] }\)

\(\displaystyle{ \frac{3}{x}\frac{d}{dx}[\cot(x)] + \cot(x) \frac{d}{dx}[3/x] }\)

\(\displaystyle{ \frac{3}{x} (-\csc^2(x)) + \cot(x) \frac{d}{dx}[3x^{-1}] }\)

\(\displaystyle{ \frac{-3}{x} \csc^2(x) + \cot(x) [3(-1)x^{-2}] }\)

\(\displaystyle{ \frac{-3}{x} \csc^2(x) - \frac{3}{x^2} \cot(x) }\)

\(\displaystyle{ \frac{-3}{x^2}[ x \csc^2(x) + \cot(x) ] }\)

Final Answer

\(\displaystyle{ \frac{d}{dx}\left[ \frac{3}{x} \cot(x) \right] = \frac{-3}{x^2}\left[ x \csc^2(x) + \cot(x) \right] }\)

close solution

Log in to rate this practice problem and to see it's current rating.

Advanced

Here is a problem that has both trig and exponential derivatives. Give it a try.

\(f(x)=4x^2e^x\) \(\sin(x)\sec(x)\)

Problem Statement

Calculate the derivative of this function using the product rule, giving your final answer in simplified, factored form. \(f(x)=4x^2e^x\) \(\sin(x)\sec(x)\)

Solution

934 video

video by Krista King Math

close solution

Log in to rate this practice problem and to see it's current rating.

You CAN Ace Calculus

Topics You Need To Understand For This Page

basic derivative rules

power rule

Some of the practice problems require you to know the following rules also (in their basic form, not including the chain rule). If you don't know one or more of these rules, no worries. You can skip those problems and come back to them later.

exponential derivative

derivatives of trig functions

You do NOT need to know the chain rule for anything on this page, including practice problems. For practice problems using the product rule and chain rule, see the chain rule page.

Related Topics and Links

external links you may find helpful

product rule youtube playlist

WikiBooks - Product Rule

To bookmark this page and practice problems, log in to your account or set up a free account.

Topics Listed Alphabetically

Single Variable Calculus

Multi-Variable Calculus

Differential Equations

Precalculus

Search Practice Problems

Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.

how to read math books

Get great tutoring at an affordable price with Chegg. Subscribe today and get your 1st 30 minutes Free!

The 17Calculus and 17Precalculus iOS and Android apps are no longer available for download. If you are still using a previously downloaded app, your app will be available until the end of 2020, after which the information may no longer be available. However, do not despair. All the information (and more) is now available on 17calculus.com for free.

Calculus, Better Explained: A Guide To Developing Lasting Intuition

Shop eBags.com, the leading online retailer of luggage, handbags, backpacks, accessories, and more!

Shop Amazon - Rent eTextbooks - Save up to 80%

Math Word Problems Demystified

Shop eBags.com, the leading online retailer of luggage, handbags, backpacks, accessories, and more!

Shop Amazon - New Textbooks - Save up to 40%

Practice Instructions

Unless otherwise instructed, calculate the derivatives of these functions using the product rule, giving your final answers in simplified, factored form.

Do NOT follow this link or you will be banned from the site!

When using the material on this site, check with your instructor to see what they require. Their requirements come first, so make sure your notation and work follow their specifications.

DISCLAIMER - 17Calculus owners and contributors are not responsible for how the material, videos, practice problems, exams, links or anything on this site are used or how they affect the grades or projects of any individual or organization. We have worked, to the best of our ability, to ensure accurate and correct information on each page and solutions to practice problems and exams. However, we do not guarantee 100% accuracy. It is each individual's responsibility to verify correctness and to determine what different instructors and organizations expect. How each person chooses to use the material on this site is up to that person as well as the responsibility for how it impacts grades, projects and understanding of calculus, math or any other subject. In short, use this site wisely by questioning and verifying everything. If you see something that is incorrect, contact us right away so that we can correct it.

Links and banners on this page are affiliate links. We carefully choose only the affiliates that we think will help you learn. Clicking on them and making purchases help you support 17Calculus at no extra charge to you. However, only you can decide what will actually help you learn. So think carefully about what you need and purchase only what you think will help you.

We use cookies on this site to enhance your learning experience.

17calculus

Copyright © 2010-2020 17Calculus, All Rights Reserved     [Privacy Policy]     [Support]     [About]

mathjax.org
Real Time Web Analytics
17Calculus
We use cookies to ensure that we give you the best experience on our website. By using this site, you agree to our Website Privacy Policy.