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 #291 : Computation Of The Derivative

Find the derivative of $15x^5-2x^4+11x^2+e^x-\ln x$ .

Possible Answers:

$15x^4-2x^3+11x+e^x-\ln x$

$75x^4+8x^3-22x-e^x+\tfrac{1} {x}$

$75x^4-8x^3+22x+e^x-\tfrac{1} {x}$

55x^4-6x^3+22x

$25x^4+2x^3+11x+e^x+\ln x$

Correct answer:

$75x^4-8x^3+22x+e^x-\tfrac{1} {x}$

Explanation:

Take the derivative of each term.

$\tfrac{\mathrm{d} }{\mathrm{d} x}(15x^5)=75x^4$

$\tfrac{\mathrm{d} }{\mathrm{d} x}(-2x^4)=-8x^3$

$\tfrac{\mathrm{d} }{\mathrm{d} x}(11x^2)=22x$

$\tfrac{\mathrm{d} }{\mathrm{d} x}(e^x)=e^x$

$\tfrac{\mathrm{d} }{\mathrm{d} x}(-\ln x)=-\tfrac{1}{x}$

Add them:

$75x^4-8x^3+22x+e^x-\tfrac{1} {x}$

Report an Error

Example Question #292 : Computation Of The Derivative

Find the derivative of $2\sin x\cos x$ .

Possible Answers:

$2\cos2x$

$-\sin^2x+cos^2x$

$-2\cos x\sin x$

$2\sin2x$

Correct answer:

$2\cos2x$

Explanation:

There are two ways to solve this problem.

First, you can use a trig identity to replace $2\sin x\cos x$ with $\sin 2x$ . Using the chain rule, $(\sin 2x)'=2\cos2x$ .

Alternatively, you could use the product rule.

$2(\sin x(-\sin x)+\cos x(\cos x))=2(\cos^2x-\sin^2x)$

Since $\cos^2x-\sin^2x=\cos2x$ , our final answer is still $2\cos2x$ .

Report an Error

Example Question #293 : Computation Of The Derivative

Find the derivative of $\frac{e^x}{\sin x}$ .

Possible Answers:

$\frac{-e^x\tan x}{\sin x}$

$\frac{e^x\sin x-e^x\cos x}{\sin x}$

$\frac{e^x-e^x\tan x}{\sin x}$

$\frac{e^x\tan x-e^x}{\cos x}$

$\frac{e^x}{\cos x}$

Correct answer:

$\frac{e^x-e^x\tan x}{\sin x}$

Explanation:

For this problem, we need to use the quotient rule.

$(\frac{f}{g})'=\frac{gf'-fg'}{g^2}$

$(\frac{e^x}{\sin x})'=\frac{\sin x(e^x)-e^x(\cos x)}{\sin^2x}$

Simplifying:

$e^x (\frac{\sin x-\cos x}{\sin^2x})=\frac{e^x-e^x\tan x}{\sin x}$

Report an Error

Example Question #461 : Derivatives

Find the derivative of y=(2x^4+5x)(cos(x))

Possible Answers:

y'=8x^3cos(x)+5cos(x)

y'=-2x^4sin(x)+8x^3cos(x)-5xsin(x)+5cos(x)

y'=2x^4cos(x)-2x^4sin(x)+5xcos(x)-5xsin(x)

None of the other answers

Correct answer:

y'=-2x^4sin(x)+8x^3cos(x)-5xsin(x)+5cos(x)

Explanation:

$\frac{d}{dx}[2x^4+5x]=8x^3+5$

$\frac{d}{dx}[cos(x)]=-sin(x)$

Product rule states: $\frac{d}{du}[f(u) \cdot g(u)]=f'(u)g(u)+g'(u)f(u)$

Therefore:

y'=(8x^3+5)(cos(x))-sin(x)(2x^4+5x)

=-2x^4sin(x)+8x^3cos(x)-5xsin(x)+5cos(x)

Report an Error

Example Question #471 : Derivatives

Use the chain rule to differentiate the following function: $h(x)=sin(x^{2}e^{x})$

Possible Answers:

$h^{'}(x)=sin(x^{2}e^{x})xe^{x}(2+x)$

$h^{'}(x)=-cos(x^{2}e^{x})xe^{x}(2+x)$

$h^{'}(x)=-sin(x^{2}e^{x})xe^{x}(2+x)$

$h^{'}(x)=cos(x^{2}e^{x})xe^{x}(2+x)$

Correct answer:

$h^{'}(x)=cos(x^{2}e^{x})xe^{x}(2+x)$

Explanation:

$h(x)=sin(x^{2}e^{x})$

By the chain rule: $h^{'}(x)=cos(x^{2}e^{x})\frac{\mathrm{d} }{\mathrm{d} x}(x^{2}e^{x})$

Differentiate $(x^{2}e^{x})$ using the product rule: $\frac{\mathrm{d} }{\mathrm{d} x}(x^{2}e^{x})=\frac{\mathrm{d} }{\mathrm{d} x}(x^{2})e^{x}+x^{2}\frac{\mathrm{d} }{\mathrm{d} x}e^{x}=2xe^{x}+x^{2}e^{x}$

Substitute this derivative for $\frac{\mathrm{d} }{\mathrm{d} x}(x^{2}e^{x})$ in the first equation: $h^{'}(x)=cos(x^{2}e^{x})(2xe^{x}+x^{2}e^{x})$

Factor the equation: $h^{'}(x)=cos(x^{2}e^{x})xe^{x}(2+x)$

Report an Error

Example Question #771 : Ap Calculus Ab

f(p)=cos(2p-1)-(2p^3+4p)

Find f'(p) .

Possible Answers:

f'(p)=cos(2p-1)+sin(2p-1)-6p^2+4

f'(p)=-2cos(2p-1)(p^3+3p^2+2p+2)

f'(p)=sin(2p-1)-sin(2p^3+4p)

f'(p) is undefined.

f'(p)=-2(sin(2p-1)+3p^2+2)

Correct answer:

f'(p)=-2(sin(2p-1)+3p^2+2)

Explanation:

$\frac{d}{dx}[cos(2p-1)]=2 \cdot -sin(2p-1)=-2sin(2p-1)$

$\frac{d}{dx}[2p^3+4p]=6p^2+4$

Therefore:

f'(p)=-2sin(2p-1)-(6p^2+4)

=-2sin(2p-1)-6p^2-4

=-2(sin(2p-1)+3p^2+2)

f'(p)=-2(3p^2+sin(2p-1)+2)

Report an Error

Example Question #1 : Finding Second Derivative Of A Function

Let f(x)=sin(x)-cos(x) .

Find the second derivative of f(x) .

Possible Answers:

f''(x)=sin(x)-cos(x)

f''(x)=sin(x)+cos(x)

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

f''(x)=sin^2(x)+cos^2(x)

f''(x)=cos(x)-sin(x)

Correct answer:

f''(x)=cos(x)-sin(x)

Explanation:

The second derivative is just the derivative of the first derivative. So first we find the first derivative of f(x) . Remember the derivative of $\sin x$ is $\cos x$ , and the derivative for $\cos x$ is $-\sin x$ .

f'(x)= cos(x)+sin(x)

Then to get the second derivative, we just derive this function again. So

f''(x)=-sin(x)+cos(x)=cos(x)-sin(x)

Report an Error

Example Question #2 : Finding Second Derivative Of A Function

Define $f (x) = 6x^{3} - 12x^{2} + 4x -8$ .

What is f ' ' (x) ?

Possible Answers:

f ' '(x) = 6x - 1 2

f ' '(x) = 36x - 24

f ''(x) = 18x - 24

f ' '(x) = 36x

f ' '(x) = 18

Correct answer:

f ' '(x) = 36x - 24

Explanation:

Take the derivative of , then take the derivative of .

$f (x) = 6x^{3} - 12x^{2} + 4x -8$

$f '(x) = 3 \cdot 6x^{3-1} - 2 \cdot 12x^{2-1} + 4 -0$

$f '(x) = 18x^{2} - 24x+ 4$

$f ' '(x) = 2 \cdot 18x^{2-1} - 24+ 0$

f ' '(x) = 36x - 24

Report an Error

Example Question #662 : Derivatives

Define $g(x) = \cos \left (x^{2} \right )$ .

What is g''(x) ?

Possible Answers:

$g''(x) =4x^2 \cos(x^2)$

$g''(x) = - 2 \cos(x^2)-4x^2 \sin(x^2) \right ]$

$g''(x) = - 2 \sin(x^2)-4x^2 \cos(x^2) \right ]$

$g''(x) =-4x^2 \cos(x^2)$

$g''(x) = -4x^2 \sin(x^2)$

Correct answer:

$g''(x) = - 2 \sin(x^2)-4x^2 \cos(x^2) \right ]$

Explanation:

Take the derivative of , then take the derivative of .

$g(x) = \cos \left (x^{2} \right )$

$g'(x) = 2x \cdot [-\sin (x^2)]$

$g'(x) = - 2x \sin (x^2)$

$g''(x) = - 2\left [ 1 \cdot \sin(x^2)+ x \cdot 2x \cos(x^2) \right ]$

$g''(x) = - 2\left [ \sin(x^2)+ 2x^2 \cos(x^2) \right ]$

$g''(x) = - 2 \sin(x^2)-4x^2 \cos(x^2) \right ]$

Report an Error

Example Question #663 : Derivatives

Define $f(x) = \frac{1}{e^{4x}}$ .

What is f ''(x) ?

Possible Answers:

$f''(x) = \frac{1 }{e^{4x+2}}$

$f''(x) = \frac{1 }{e^{4x-2}}$

$f''(x) =- \frac{16 }{e^{4x}}$

$f''(x) = \frac{16 }{e^{4x}}$

$f(x) = \frac{1}{16e^{4x}}$

Correct answer:

$f''(x) = \frac{16 }{e^{4x}}$

Explanation:

Rewrite:

$f(x) = \frac{1}{e^{4x}}$

$f(x) = e^{-4x}$

Take the derivative of , then take the derivative of .

$f'(x) = -4 e^{-4x}$

$f''(x) = -4 \cdot \left (-4 \right )e^{-4x} = 16e^{-4x} = \frac{16 }{e^{4x}}$

Report an Error

View AP Calculus AB Tutors

Juhyeon
Certified Tutor

Vanderbilt University, Master of Science, Environmental Engineering.

View AP Calculus AB Tutors

Nevryk
Certified Tutor

University of Nevada-Las Vegas, Bachelor of Science, Computer Science.

View AP Calculus AB Tutors

Donald
Certified Tutor

University of Georgia, Bachelors, Latin.

All AP Calculus AB Resources

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

Popular Subjects

Spanish Tutors in Philadelphia, Biology Tutors in Chicago, ACT Tutors in Los Angeles, GRE Tutors in Atlanta, English Tutors in New York City, Physics Tutors in Los Angeles, LSAT Tutors in New York City, French Tutors in San Francisco-Bay Area, GMAT Tutors in Miami, SAT Tutors in Denver

Popular Courses & Classes

Spanish Courses & Classes in Dallas Fort Worth, Spanish Courses & Classes in San Diego, ISEE Courses & Classes in Chicago, SAT Courses & Classes in New York City, GMAT Courses & Classes in Denver, ISEE Courses & Classes in New York City, LSAT Courses & Classes in Phoenix, MCAT Courses & Classes in Chicago, ISEE Courses & Classes in Atlanta, SAT Courses & Classes in Washington DC

Popular Test Prep

SSAT Test Prep in Atlanta, ISEE Test Prep in New York City, ACT Test Prep in Houston, ISEE Test Prep in Houston, SAT Test Prep in Houston, LSAT Test Prep in Phoenix, ACT Test Prep in San Diego, GRE Test Prep in Seattle, LSAT Test Prep in San Diego, LSAT Test Prep in Chicago

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 #291 : Computation Of The Derivative

Example Question #292 : Computation Of The Derivative

Example Question #293 : Computation Of The Derivative

Example Question #461 : Derivatives

Example Question #471 : Derivatives

Example Question #771 : Ap Calculus Ab

Example Question #1 : Finding Second Derivative Of A Function

Example Question #2 : Finding Second Derivative Of A Function

Example Question #662 : Derivatives

Example Question #663 : Derivatives

All AP Calculus AB Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: