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

17Calculus Subjects Listed Alphabetically

Single Variable Calculus

Multi-Variable Calculus

Differential Equations

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.

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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.

 
quotient rule →

Practice

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

Conversion Between A-B-C Level (or 1-2-3) and New Numbered Practice Problems

Please note that with this new version of 17calculus, the practice problems have been relabeled but they are MOSTLY in the same order. Here is a list converting the old numbering system to the new.

Derivative Product Rule - Practice Problems Conversion

[A01-938] - [A02-1928] - [A03-939] - [A04-925] - [A05-927] - [A06-1311] - [A07-928] - [A08-929] - [A09-930]

[A10-931] - [A11-932] - [B01-926] - [B02-935] - [B03-936] - [B04-940] - [B05-937] - [B06-933] - [C01-934]

Please update your notes to this new numbering system. The display of this conversion information is temporary.

GOT IT. THANKS!

Basic Problems

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 \( f(x) = x(25-x) \) using the product rule.

Solution

938 solution video

video by Krista King Math

close solution

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

Problem Statement

Calculate the derivative of \( f(x) = (x-2)(x+3) \) using the product rule.

Final Answer

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

Problem Statement

Calculate the derivative of \( f(x) = (x-2)(x+3) \) using the product rule.

Solution

1928 solution video

video by PatrickJMT

Final Answer

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

close solution

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

Problem Statement

Calculate the derivative of \( f(x) = (2x+3)(3x-2) \) using the product rule.

Solution

939 solution video

video by Krista King Math

close solution

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

Problem Statement

Calculate the derivative of \( h(x) = (x^2+5x+7)(x^3+2x-4) \) using the product rule.

Final Answer

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

Problem Statement

Calculate the derivative of \( h(x) = (x^2+5x+7)(x^3+2x-4) \) using the product rule.

Solution

Use the product rule.

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

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

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

Problem Statement

Calculate the derivative of \( g(x) = (x^3-7x^2+4)\) \((3x^2+14) \) using the product rule.

Final Answer

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

Problem Statement

Calculate the derivative of \( g(x) = (x^3-7x^2+4)\) \((3x^2+14) \) using the product rule.

Solution

First, we apply the product rule.

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

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

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

Problem Statement

Calculate the derivative of \( g(x) = (3x^4+2x-1)\) \((x^5-2x^2) \) using the product rule.

Solution

1311 solution video

video by PatrickJMT

close solution

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 \(\displaystyle{f(x)=x^3e^x}\) using the product rule.

Final Answer

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

Problem Statement

Calculate the derivative of \(\displaystyle{f(x)=x^3e^x}\) using the product rule.

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

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

Problem Statement

Calculate the derivative of \(\displaystyle{y=7e^{2x}}\) using the product rule.

Final Answer

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

Problem Statement

Calculate the derivative of \(\displaystyle{y=7e^{2x}}\) using the product rule.

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

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

Problem Statement

Calculate the derivative of \(\displaystyle{e^x\sqrt{x}}\) using the product rule.

Final Answer

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

Problem Statement

Calculate the derivative of \(\displaystyle{e^x\sqrt{x}}\) using the product rule.

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

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

\(\displaystyle{x^4\tan(x)}\)

Problem Statement

Calculate the derivative of \(\displaystyle{x^4\tan(x)}\) using the product rule.

Final Answer

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

Problem Statement

Calculate the derivative of \(\displaystyle{x^4\tan(x)}\) using the product rule.

Solution

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

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

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

Problem Statement

Calculate the derivative of \(\displaystyle{f(x)=\cos(x)\sin(x)}\) using the product rule.

Final Answer

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

Problem Statement

Calculate the derivative of \(\displaystyle{f(x)=\cos(x)\sin(x)}\) using the product rule.

Solution

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

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

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

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

Final Answer

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

close solution

Intermediate Problems

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

Problem Statement

Calculate the derivative of \(\displaystyle{g(x)=(x^2+3x-5)\sqrt[3]{x}}\) using the product rule.

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 \(\displaystyle{g(x)=(x^2+3x-5)\sqrt[3]{x}}\) using the product rule.

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.

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

Take the two derivatives.

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

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

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

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

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

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

Problem Statement

Calculate the derivative of \(\displaystyle{f(x)=\left[\frac{1}{x^2}-\frac{3}{x^4}\right]\cdot}\) \((x+5x^3)\) using the product rule.

Solution

935 solution video

video by Krista King Math

close solution

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

Problem Statement

Calculate the derivative of \( H(u)=(u-\sqrt{u}) \) \( (u+\sqrt{u}) \) using the product rule.

Solution

936 solution video

video by Krista King Math

close solution

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

Problem Statement

Calculate the derivative of \(f(x)=x(2x-1)\) \((2x+1)\) using the product rule.

Solution

940 solution video

video by Krista King Math

close solution

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

Problem Statement

Calculate the derivative of \(f(x)=(2x+1)\) \((4-x^2)(1+x^2)\) using the product rule.

Solution

937 solution video

video by Krista King Math

close solution

Here is one with a trig derivative.

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

Problem Statement

Calculate the derivative of \(\displaystyle{\frac{3}{x}\cot(x)}\) using the product rule.

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 \(\displaystyle{\frac{3}{x}\cot(x)}\) using the product rule.

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

Advanced Problems

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 \(f(x)=4x^2e^x\) \(\sin(x)\sec(x)\) using the product rule.

Solution

934 solution video

video by Krista King Math

close solution
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