Call Now to Set Up Tutoring:

(888) 888-0446

AP Calculus AB : Computation of the Derivative

Study concepts, example questions & explanations for AP Calculus AB

Create An Account

All AP Calculus AB Resources

3 Diagnostic Tests 164 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #56 : Calculus I — Derivatives

What is the second derivative of x^2+5x ?

Possible Answers:

2x+5

3x^2+5

$\frac{1}{3}x^3+\frac{5}{2}x^2+c$

Correct answer:

Explanation:

To get the second derivative, first we need to find the first derivative.

To do that, we can use the power rule. That means we lower the exponent of the variable by one and multiply the variable by that original exponent.

$\frac{\mathrm{d} }{\mathrm{d} x}(x^2+5x)=(2*x^{2-1})+(1*5x^{1-0})$

$\frac{\mathrm{d} }{\mathrm{d} x}(x^2+5x)=2x+5x^0$

Remember that anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(x^2+5x)=2x+5*1$

$\frac{\mathrm{d} }{\mathrm{d} x}(x^2+5x)=2x+5$

Now we do the same process again, but using 2x+5 as our expression:

$\frac{\mathrm{d} }{\mathrm{d} x}(2x+5x^0)=(1*2x^{1-1})+(0*5x^{0-1})$

Notice that $(0*5x^{0-1})=0$ , as anything times zero will be zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(2x+5x^0)=(1*2x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(2x+5x^0)=(2x^{0})$

Anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(2x+5x^0)=(2*1)$

$\frac{\mathrm{d} }{\mathrm{d} x}(2x+5x^0)=2$

Report an Error

Example Question #57 : Calculus I — Derivatives

What is the second derivative of 5x+8 ?

Possible Answers:

$\frac{1}{5x}$

Undefined

$\frac{5}{x}$

Correct answer:

Explanation:

To get the second derivative, first we need to find the first derivative.

To do that, we can use the power rule. That means we lower the exponent of the variable by one and multiply the variable by that original exponent.

We're going to treat as 8x^0 , as anything to the zero power is one.

That means this problem will look like this:

$\frac{\mathrm{d} }{\mathrm{d} x}(5x+8)=(5*x^{1-1})+(0*8x^{0-1})$

Notice that $(0*8x^{0-1})=0$ as anything times zero will be zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(5x+8)=(5*x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(5x+8)=(5*x^{0})$

Remember, anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(5x+8)=5*1$

$\frac{\mathrm{d} }{\mathrm{d} x}(5x+8)=5$

Now to get the second derivative we repeat those steps, but instead of using 5x+8 , we use 5x^0 .

$\frac{\mathrm{d} }{\mathrm{d} x}(5x^0)=(0*5x^{0-1})$

Notice that $(0*5x^{0-1})=0$ as anything times zero will be zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(5x^0)=0$

Report an Error

Example Question #58 : Calculus I — Derivatives

What is the second derivative of x^3+2x+5 ?

Possible Answers:

18x^4+13x^2+c

12x+5

3x^2+2

Correct answer:

Explanation:

To get the second derivative, first we need to find the first derivative.

To do that, we can use the power rule. That means we lower the exponent of the variable by one and multiply the variable by that original exponent.

We're going to treat as 5x^0 , as anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x+5)=(3*x^{3-1})+(1*2x^{1-1})+(0*5x^{0-1})$

Notice that $(0*5x^{0-1})=0$ , as anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x+5)=(3*x^{3-1})+(1*2x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x+5)=(3*x^{2})+(2x^{0})$

$\frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x+5)=3x^2+2$

Now we repeat the process using 3x^2+2 as the expression.

Just like before, we're going to treat as 2x^0 .

$\frac{\mathrm{d} }{\mathrm{d} x}(3x^2+2)=(2*3x^{2-1})+(0*2x^{0-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}3x^2+2=(2*3x^{2-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(3x^2+2)=6x^{1}$

$\frac{\mathrm{d} }{\mathrm{d} x}(3x^2+2)=6x$

Report an Error

Example Question #51 : Derivatives

If f(x)=6x^2+8 , what is f''(x) ?

Possible Answers:

12x

Correct answer:

Explanation:

The question is asking us for the second derivative of the equation. First, we need to find the first derivative.

To do that, we can use the power rule. To use the power rule, we lower the exponent on the variable and multiply by that exponent.

We're going to treat as 8x^0 since anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(6x^2+8)=(2*6x^{2-1})+(0*8x^{0-1})$

Notice that $0*8x^{0-1}=0$ since anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(6x^2+8)=(2*6x^{2-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(6x^2+8)=(2*6x^{1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(6x^2+8)=12x$

Now we do the exact same process but using 12x as our expression.

$\frac{\mathrm{d} }{\mathrm{d} x}(12x)=1*12x^{1-1}$

$\frac{\mathrm{d} }{\mathrm{d} x}(12x)=12x^{0}$

As stated earlier, anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(12x)=12*1$

$\frac{\mathrm{d} }{\mathrm{d} x}(12x)=12$

Report an Error

Example Question #52 : Derivatives

What is the second derivative of x+1 ?

Possible Answers:

$-\sqrt x$

Undefined

Correct answer:

Explanation:

To find the second derivative, first we need to find the first derivative.

To do that, we can use the power rule. To use the power rule, we lower the exponent on the variable and multiply by that exponent.

We're going to treat as 1x^0 since anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(x+1)=(1*x^{1-1})+(0*x^{0-1})$

Notice that $(0*x^{0-1})=0$ since anything times zero is zero.

That leaves us with $\frac{\mathrm{d} }{\mathrm{d} x}(x+1)=(1*x^{1-1})$ .

Simplify.

$\frac{\mathrm{d} }{\mathrm{d} x}(x+1)=(1*x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(x+1)=(x^{0})$

As stated earlier, anything to the zero power is one, leaving us with:

$\frac{\mathrm{d} }{\mathrm{d} x}(x+1)=1$

Now we can repeat the process using or x^0 as our equation.

$\frac{\mathrm{d} }{\mathrm{d} x}(x^0)=0*x^{0-1}$

As pointed out before, anything times zero is zero, meaning that $\frac{\mathrm{d} }{\mathrm{d} x}(x^0)=0$ .

Report an Error

Example Question #53 : Derivatives

What is the second derivative of 12x^2+13x+4 ?

Possible Answers:

24 x

24x+13

144

Correct answer:

Explanation:

To find the second derivative, first we need to find the first derivative.

To do that, we can use the power rule. To use the power rule, we lower the exponent on the variable and multiply by that exponent.

We're going to treat as 4x^0 since anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(12x^2+13x+4)=(2*12x^{2-1})+(1*13x^{1-1})+(0*4x^{0-1})$

Notice that $(0*4x^{0-1})=0$ since anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(12x^2+13x+4)=(2*12x^{2-1})+(1*13x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(12x^2+13x+4)=(2*12x^{1})+(1*13x^{0})$

$\frac{\mathrm{d} }{\mathrm{d} x}(12x^2+13x+4)=(24x^{1})+(13x^{0})$

Just like it was mentioned earlier, anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(12x^2+13x+4)=(24x)+(13*1)$

$\frac{\mathrm{d} }{\mathrm{d} x}(12x^2+13x+4)=24x+13$

Now we repeat the process using 24x+13 as our expression.

$\frac{\mathrm{d} }{\mathrm{d} x}(24x+13)=(1*24x^{1-1})+(0*13x^{0-1})$

Like before, anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(24x+10)=(1*24x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(24x+13)=(24x^{0})$

Anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(24x+13)=(24*1)$

$\frac{\mathrm{d} }{\mathrm{d} x}(24x+13)=24$

Report an Error

Example Question #301 : Computation Of The Derivative

Define $g(x) = 7x^{5} - 6x^{3} - 17x$ .

What is g ''(x) ?

Possible Answers:

$g'' (x) = 70x^{3} - 36x$

$g'' (x) = 140x^{3} - 36x -17$

$g'' (x) = 70x^{3} - 18x$

$g'' (x) = 140x^{3} - 36x$

$g'' (x) = 140x^{3} - 18x$

Correct answer:

$g'' (x) = 140x^{3} - 36x$

Explanation:

Take the derivative of , then take the derivative of .

$g(x) = 7x^{5} - 6x^{3} - 17x$

$g' (x) = 5\cdot 7x^{5-1} - 3\cdot 6x^{3-1} - 17$

$g' (x) = 35x^{4} - 18x^{2} - 17$

$g'' (x) = 4\cdot 35x^{4-1} - 2\cdot 18x^{2-1} - 0$

$g'' (x) = 140x^{3} - 36x$

Report an Error

View AP Calculus AB Tutors

Sharif
Certified Tutor

NCCU, Bachelors, Physics and Mathematics.

View AP Calculus AB Tutors

Muhammad
Certified Tutor

The University of Texas at Dallas, Bachelor of Science, Mechanical Engineering.

View AP Calculus AB Tutors

Megan
Certified Tutor

University of Wisconsin-Madison, Bachelor of Science, Chemical Engineering.

All AP Calculus AB Resources

3 Diagnostic Tests 164 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Statistics Tutors in Los Angeles, Reading Tutors in Philadelphia, Statistics Tutors in Chicago, Computer Science Tutors in Dallas Fort Worth, Reading Tutors in Miami, French Tutors in Denver, ACT Tutors in Chicago, Algebra Tutors in Boston, Biology Tutors in San Francisco-Bay Area, MCAT Tutors in New York City

Popular Courses & Classes

LSAT Courses & Classes in San Diego, LSAT Courses & Classes in Miami, Spanish Courses & Classes in Philadelphia, ACT Courses & Classes in Atlanta, SAT Courses & Classes in Atlanta, ACT Courses & Classes in Chicago, GMAT Courses & Classes in Boston, GMAT Courses & Classes in Atlanta, MCAT Courses & Classes in San Diego, GRE Courses & Classes in Philadelphia

Popular Test Prep

GRE Test Prep in Miami, MCAT Test Prep in San Francisco-Bay Area, SAT Test Prep in Atlanta, SAT Test Prep in Houston, GMAT Test Prep in Atlanta, ISEE Test Prep in Los Angeles, ISEE Test Prep in Washington DC, MCAT Test Prep in Los Angeles, GRE Test Prep in San Francisco-Bay Area, SAT Test Prep in San Diego

DMCA Complaint

If you believe that content available by means of the Website (as defined in our Terms of Service) infringes one or more of your copyrights, please notify us by providing a written notice (“Infringement Notice”) containing the information described below to the designated agent listed below. If Varsity Tutors takes action in response to an Infringement Notice, it will make a good faith attempt to contact the party that made such content available by means of the most recent email address, if any, provided by such party to Varsity Tutors.

Your Infringement Notice may be forwarded to the party that made the content available or to third parties such as ChillingEffects.org.

Please be advised that you will be liable for damages (including costs and attorneys’ fees) if you materially misrepresent that a product or activity is infringing your copyrights. Thus, if you are not sure content located on or linked-to by the Website infringes your copyright, you should consider first contacting an attorney.

Please follow these steps to file a notice:

You must include the following:

A physical or electronic signature of the copyright owner or a person authorized to act on their behalf; An identification of the copyright claimed to have been infringed; A description of the nature and exact location of the content that you claim to infringe your copyright, in \ sufficient detail to permit Varsity Tutors to find and positively identify that content; for example we require a link to the specific question (not just the name of the question) that contains the content and a description of which specific portion of the question – an image, a link, the text, etc – your complaint refers to; Your name, address, telephone number and email address; and A statement by you: (a) that you believe in good faith that the use of the content that you claim to infringe your copyright is not authorized by law, or by the copyright owner or such owner’s agent; (b) that all of the information contained in your Infringement Notice is accurate, and (c) under penalty of perjury, that you are either the copyright owner or a person authorized to act on their behalf.

Send your complaint to our designated agent at:

Charles Cohn Varsity Tutors LLC
101 S. Hanley Rd, Suite 300
St. Louis, MO 63105

Or fill out the form below:

Do not fill in this field

Contact Information

* Full legal name

Company name

* Phone number

* Email address

Address

* Address1

Address2

* City

* State

* Zipcode

* Country

Complaint Details

* URL of copyrighted work (where your original material is located)

* Please describe the copyrighted work so that it may be easily identified

I am the owner, or an agent authorized to act on behalf of the owner of an exclusive right that is allegedly infringed.

I have a good faith belief that the use of the material in the manner complained of is not authorized by the copyright owner, its agent, or the law.

This notification is accurate.

I acknowledge that there may be adverse legal consequences for making false or bad faith allegations of copyright infringement by using this process.

* Signature (your digital signature is legally binding)

HIGH SCHOOL

GRADUATE SCHOOL

K-8

Search 50+ Tests

math tutoring

science tutoring

foreign languages

elementary tutoring

other

Search 350+ Subjects

AP Calculus AB : Computation of the Derivative

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #56 : Calculus I — Derivatives

Example Question #57 : Calculus I — Derivatives

Example Question #58 : Calculus I — Derivatives

Example Question #51 : Derivatives

Example Question #52 : Derivatives

Example Question #53 : Derivatives

Example Question #301 : Computation Of The Derivative

All AP Calculus AB Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: