Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 2 : First and Second Derivatives of Functions

Study concepts, example questions & explanations for Calculus 2

Create An Account

All Calculus 2 Resources

9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #121 : First And Second Derivatives Of Functions

Find the derivative of f(x)=-3x^3-9x^2+4x-1 .

Possible Answers:

$f{}'(x)=-27x^2+18x+4$

$f{}'(x)=-27x^2-18x+4x$

$f{}'(x)=-27x^2-18x+4$

$f{}'(x)=-27x^2-18x$

$f{}'(x)=27x^2-18x+4$

Correct answer:

$f{}'(x)=-27x^2-18x+4$

Explanation:

To take the derivative, remember to multiply the exponent of an x term by the coefficient in front of the x term and then subtract one from the exponent:

$f{}'(x)=-27x^2-18x+4$ .

Report an Error

Example Question #1442 : Calculus Ii

What is the acceleration function if x(t)=-16t^2+4t-1?

Possible Answers:

a(t)=-32t

a(t)=-3

a(t)=-32

a(t)=-32t-1

a(t)=-32t^2

Correct answer:

a(t)=-32

Explanation:

Recall that the acceleration function is the second derivative of the position function. First, find the velocity function since that is the first derivative. To take the derivative, multiply the exponent by the coefficient in front of the x term and then subtract one from the exponent:

v(t)=-32t+4

Now, take the derivative of the velocity function to get acceleration:

a(t)=-32

Report an Error

Example Question #1443 : Calculus Ii

Find the first derivative of the function:

$f(x)=\tan(2xe^x)$

Possible Answers:

$\sec^2(2xe^x)[4e^x]$

$\sec^2(2xe^x)[2e^x+xe^x]$

$\sec^2(2xe^x)[2e^x+2xe^x]$

$\sec^2(2xe^x)$

Correct answer:

$\sec^2(2xe^x)[2e^x+2xe^x]$

Explanation:

The first derivative of the function is equal to

$f'(x)=\sec^2(2xe^x)[2e^x+2xe^x]$

and was found using the following rule:

$\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \tan(x)=\sec^2(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{d} u}{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

Report an Error

Example Question #1444 : Calculus Ii

Find the derivative of the following:

$f(x)=\frac{\cos(2x)}{e^{2x}}$

Possible Answers:

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

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

$-2(\sin(2x)-\cos(2x))$

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

Correct answer:

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

Explanation:

The derivative of the function is equal to

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

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} \frac{f(x)}{g(x)}=\frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{du} }{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$

Report an Error

Example Question #322 : Derivatives

Find the first derivative of the following function:

$f(x)=\frac{x^4}{\cos(x)}+\csc(3x)$

Possible Answers:

$\frac{4x^3\cos(x)-x^4\sin(x)}{\cos^2(x)}-3\csc(3x)\cot(3x)$

$f'(x)=\frac{4x^3\cos(x)+x^4\sin(x)}{\cos^2(x)}+3\csc(3x)\cot(3x)$

$\frac{4x^3\cos(x)-x^4\sin(x)}{\cos^2(x)}-\csc(3x)\cot(3x)$

$\frac{4x^3\cos(x)+x^4\sin(x)}{\cos^2(x)}-3\csc(3x)\cot(3x)$

Correct answer:

$\frac{4x^3\cos(x)+x^4\sin(x)}{\cos^2(x)}-3\csc(3x)\cot(3x)$

Explanation:

The first derivative of the function is equal to

$f'(x)=\frac{4x^3\cos(x)+x^4\sin(x)}{\cos^2(x)}-3\csc(3x)\cot(3x)$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} \frac{f(x)}{g(x)}=\frac{f(x)}{g(x)}=\frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \csc(x)=-\csc(x)\cot(x)$

Report an Error

Example Question #323 : Derivatives

Find the derivative of the function:

$f(x)=(x^2+3x)^{5/2}$

Possible Answers:

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

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

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

$\frac{5}{2}(2x+3)\sqrt{x^2+3x}$

Correct answer:

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

Explanation:

The derivative of the function is equal to

$f'(x)=\frac{5}{2}(x^2+3x)^{\frac{3}{2}}(2x+3)$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

Note that the exponent is the acting as the outer function in the chain rule.

Report an Error

Example Question #1452 : Calculus Ii

Find the derivative of the function $f(x)= e^\sin(x)$

Possible Answers:

$-e^\sin(x)$

$-\cos(x)e^\cos(x)$

$\cos(x)e^\sin(x)$

$e^\sin(x)$

Correct answer:

$\cos(x)e^\sin(x)$

Explanation:

To find the derivative, you must first use the chain rule. The derivative of $e^\sin(x)$ is $e^\sin(x)$ , and you multiply that by the derivative of the exponent, which is $\cos(x)$ . Putting that all together, you get the final answer as $\cos(x)e^\sin(x)$ .

Report an Error

Example Question #1453 : Calculus Ii

Find the second derivative of the function $f(x)= 3x^3+\sin(x)$ .

Possible Answers:

$9x^2+\cos(x)$

$18x+\sin(x)$

9x+cos(x)

$18x-\sin(x)$

Correct answer:

$18x-\sin(x)$

Explanation:

Taking the first derivative of the function, using the rule $\frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}$ and the fact that $\sin'(x)=\cos(x)$ , we get $f'(x)= 9x^2+\cos(x)$ . Using the same rule and the fact that $\cos'(x)=-\sin(x)$ , we get $f''(x)=18x-\sin(x)$

Report an Error

Example Question #1454 : Calculus Ii

Find the derivative of the function $y(t)= \sin(t)+\frac{5}{t^2}$

Possible Answers:

$\cos(t)+8$

$\cos(t)-\frac{10}{t^3}$

$\sin(t)+1$

$-\sin(t)+10t$

Correct answer:

$\cos(t)-\frac{10}{t^3}$

Explanation:

To find the derivative of the function, we first take the derivative of $\sin(t)$ which is just $\cos(t)$ . Taking the derivative of the second term gets us $\frac{-10}{t^3}$ , using the rule $\frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}$ , where in this case, n=-2 . Using all of the information above and applying, $y'(t)= \cos(t)-\frac{10}{t^3}$

Report an Error

Example Question #1455 : Calculus Ii

Find the second derivative of the function $f(x)=-\cos(x)$ . Evaluate at $x=\pi/2$ .

Possible Answers:

$-\frac{1}{2}$

Correct answer:

Explanation:

We first take the first derivative of the function which is $f'(x)=\sin(x)$ . We then take the derivative of this new function, which is $f''(x)=\cos(x)$ . Evaluating at $x=\pi/2$ , we get $\cos(\pi/2)=0$ .

Report an Error

View Calculus 2 Tutors

Sharif
Certified Tutor

University of South Florida-Main Campus, Bachelor of Science, Cellular and Molecular Biology.

View Calculus 2 Tutors

Meijing
Certified Tutor

Rutgers University-New Brunswick, Bachelor of Science, Accounting.

View Calculus 2 Tutors

Charles
Certified Tutor

Pennsylvania State University-Main Campus, Bachelor of Science, Mathematics.

All Calculus 2 Resources

9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Computer Science Tutors in San Diego, Reading Tutors in Chicago, Math Tutors in New York City, Chemistry Tutors in Dallas Fort Worth, Biology Tutors in Los Angeles, MCAT Tutors in Boston, ACT Tutors in Washington DC, ACT Tutors in New York City, Spanish Tutors in New York City, MCAT Tutors in Miami

Popular Courses & Classes

LSAT Courses & Classes in Dallas Fort Worth, Spanish Courses & Classes in Boston, LSAT Courses & Classes in Philadelphia, ACT Courses & Classes in Phoenix, Spanish Courses & Classes in Houston, GRE Courses & Classes in Atlanta, ISEE Courses & Classes in Dallas Fort Worth, Spanish Courses & Classes in San Diego, ACT Courses & Classes in San Francisco-Bay Area, ACT Courses & Classes in Los Angeles

Popular Test Prep

SAT Test Prep in Chicago, SAT Test Prep in Washington DC, ISEE Test Prep in Phoenix, ACT Test Prep in Phoenix, MCAT Test Prep in Boston, ACT Test Prep in San Francisco-Bay Area, MCAT Test Prep in Denver, SSAT Test Prep in Seattle, ISEE Test Prep in Boston, ACT Test Prep in Los Angeles

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

Calculus 2 : First and Second Derivatives of Functions

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #121 : First And Second Derivatives Of Functions

Example Question #1442 : Calculus Ii

Example Question #1443 : Calculus Ii

Example Question #1444 : Calculus Ii

Example Question #322 : Derivatives

Example Question #323 : Derivatives

Example Question #1452 : Calculus Ii

Example Question #1453 : Calculus Ii

Example Question #1454 : Calculus Ii

Example Question #1455 : Calculus Ii

All Calculus 2 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: