Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 2 : Derivatives

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 #3 : Derivative Defined As Limit Of Difference Quotient

What is the equation of the line tangent to the graph of the function

$f(x) = x^{3} + x - 17$

at the point (3, 13) ?

Possible Answers:

y = 27 x- 68

y = 10x -17

y = 26 x- 65

y = 11x -20

y = 28 x- 71

Correct answer:

y = 28 x- 71

Explanation:

The slope of the line tangent to the graph of $f(x) = x^{3} + x - 17$ at the point (3, 13) is f'(3) , which can be evaluated as follows:

$f(x) = x^{3} + x - 17$

$f'(x) = \frac{\mathrm{d} }{\mathrm{d} x}\left ( x^{3} + x - 17 \right )$

$f'(x) = 3x^{2} + 1$

$f'(3) = 3\cdot 3^{2} + 1 = 28$

The line with slope 28 through (3, 13) has equation:

$y - y_{1} = m (x-x_{1})$

y -13= 28 (x-3)

$y -13= 28 x- 28 \cdot 3$

y -13= 28 x- 84

y = 28 x- 71

Report an Error

Example Question #71 : Derivatives

Let the initial approximation of the ninth root of 100 be

$x_{1}= 2$ .

Use one iteration of Newton's method to find approximation $x_{2}$ . Give your answer to the nearest thousandth.

Possible Answers:

2.021

1.821

1.971

1.871

1.921

Correct answer:

1.821

Explanation:

This is equivalent to finding a solution of the equation

x^9 = 100 ,

or the zero of the polynomial

P (x) = x^9 - 100 .

Using Newton's method, we can find $x_{2}$ from the formula

$x_{2} = x_{1} - \frac{P(x_{1})}{P'(x_{1})}$ .

P '(x) = 9 x^8 , so

$P (x_{1}) = P(2)= 2^9 - 100 = 512 - 100 = 412$

$P '(x_{1}) = P' (2)= 9 \cdot 2^8 = 9 \cdot 256 = 2,304$

$x_{2} = 2 - \frac{P(2)}{P'(2)} = 2 - \frac{412}{2,304} \approx 2 - 0.1788 \approx 1.821$

Report an Error

Example Question #72 : Derivatives

$g(x) = e ^{x^{2}+x}$

Evaluate g '(3 ) .

Possible Answers:

$12 e^{11}$

$7 e^{9}$

$7 e^{12}$

$e^{7}$

$12 e^{12}$

Correct answer:

$7 e^{12}$

Explanation:

To find g'(x) , substitute $u = x ^{2} + x$ and use the chain rule:

$g(x) = e ^{x^{2}+x}$

$g'(x) = \frac{\mathrm{d} }{\mathrm{d} x} e ^{x^{2}+x}$

$= \frac{\mathrm{d} }{\mathrm{d} x} e ^{u}$

$= \frac{\mathrm{d}u }{\mathrm{d} x} \cdot \frac{\mathrm{d} }{\mathrm{d} u}e ^{u}$

$= \frac{\mathrm{d}u }{\mathrm{d} x} \cdot e ^{u}$

$= \frac{\mathrm{d} }{\mathrm{d} x}\left ( x ^{2} + x \right ) \cdot \frac{\mathrm{d} }{\mathrm{d} u}e ^{u}$

$= \left ( 2x + 1 \right ) \cdot e ^{u}$

$= \left ( 2x + 1 \right ) \cdot e ^{ x ^{2} + x}$

$g'(x)= \left ( 2x + 1 \right ) e ^{ x ^{2} + x}$

Plug in 3:

$g'(3)= \left ( 2 \cdot 3 + 1 \right ) \cdot e ^{ 3 ^{2} + 3}$

$= \left ( 6 + 1 \right ) \cdot e ^{ 9+ 3} =7e ^{ 12}$

Report an Error

Example Question #2 : Chain Rule And Implicit Differentiation

$g (x) = \frac{1}{4x - 7}$

Evaluate g ' (1 ) .

Possible Answers:

$- \frac{4}{9}$

$-\frac{4}{3}$

$\frac{4}{3}$

Undefined

$\frac{4}{9}$

Correct answer:

$- \frac{4}{9}$

Explanation:

To find g'(x) , substitute u =4x - 7 and use the chain rule:

$g (x) = \frac{1}{4x - 7}$

$g '(x) = \frac{\mathrm{d} }{\mathrm{d} x}\cdot \frac{1}{4x - 7}$

$= \frac{\mathrm{d} }{\mathrm{d} x}\cdot \frac{1}{u}$

$= \frac{\mathrm{d} u}{\mathrm{d} x}\cdot \frac{\mathrm{d} }{\mathrm{d} u} \frac{1}{u}$

$= \frac{\mathrm{d} }{\mathrm{d} x} (4x-7) \cdot \frac{\mathrm{d} }{\mathrm{d} u} u ^{-1}$

$=4 \cdot (-1 ) u ^{-1-1}$

$=-4 u ^{-2} = - \frac{4}{u^{2}}= - \frac{4}{(4x-7)^{2}}$

$g'(x)= - \frac{4}{(4x-7)^{2}}$

and

$g'(1)= - \frac{4}{(4 \cdot 1 -7)^{2}}$

$= - \frac{4}{(4 -7)^{2}}$

$= - \frac{4}{(-3)^{2}} = - \frac{4}{9}$

Report an Error

Example Question #3 : Chain Rule And Implicit Differentiation

$g (x) = \tan \left ( x +\frac{ \pi }{4}\right )$

Evaluate $g' \left ( \pi )$ .

Possible Answers:

$\sqrt{2}$

$-\sqrt{2}$

Undefined

Correct answer:

Explanation:

To find g'(x) , substitute $u = x +\frac{ \pi }{4}$ and use the chain rule:

$g (x) = \tan \left ( x +\frac{ \pi }{4}\right )$

$g' (x) = \frac{\mathrm{d} }{\mathrm{d} x}\tan \left ( x +\frac{ \pi }{4}\right )$

$g' (x) = \frac{\mathrm{d} }{\mathrm{d} x}\tan u$

$=\frac{\mathrm{d} u}{\mathrm{d} x} \cdot \frac{\mathrm{d} }{\mathrm{d} u} \tan u$

$=\frac{\mathrm{d} }{\mathrm{d} x} \left ( x +\frac{ \pi }{4} \right )\cdot \frac{\mathrm{d} }{\mathrm{d} u} \tan u$

$=\frac{\mathrm{d} }{\mathrm{d} x} \left ( x +\frac{ \pi }{4} \right ) \cdot \sec ^{2} u$

$=1 \cdot \sec ^{2}\left ( x +\frac{ \pi }{4}\right ) = \sec ^{2}\left ( x +\frac{ \pi }{4}\right )$

$g'(x) = \sec ^{2}\left ( x +\frac{ \pi }{4}\right )$

$g'(\pi ) = \sec ^{2}\left ( \pi +\frac{ \pi }{4}\right )$

$g'(\pi ) = \sec ^{2}\frac{ 5 \pi }{4}$

$= \left ( - \sqrt{2 }\right )^{2} = 2$

Report an Error

Example Question #4 : Chain Rule And Implicit Differentiation

$g (x) = 3 ^{x^{2}}$

Evaluate g ' (-1 ) .

Possible Answers:

$-6 \ln 3$

$-\frac{3 \ln 3}{2}$

Correct answer:

$-6 \ln 3$

Explanation:

To find g'(x) , substitute $u = x ^{2}$ and use the chain rule:

$g (x) = 3 ^{x^{2}}$

$g ' (x) = \frac{\mathrm{d} }{\mathrm{d} x} 3 ^{x^{2}}$

$= \frac{\mathrm{d} }{\mathrm{d} x} 3 ^{u}$

$= \frac{\mathrm{d}u }{\mathrm{d} x} \cdot \frac{\mathrm{d} }{\mathrm{d} u} 3 ^{u}$

$= \frac{\mathrm{d}}{\mathrm{d} x} x^{2} \cdot \frac{\mathrm{d} }{\mathrm{d} u} 3 ^{u}$

$= 2x \cdot \ln 3 \cdot 3 ^{u}$

$= 2x \ln 3 \cdot 3 ^{x^2}$

So $g'(x)= 2x \ln 3 \cdot 3 ^{x^2}$

and

$g'(-1)= 2(-1) \ln 3 \cdot 3 ^{(-1)^2}$

$= -2 \ln 3 \cdot 3 ^{1} = -2 \ln 3 \cdot 3 = -6 \ln 3$

Report an Error

Example Question #5 : Chain Rule And Implicit Differentiation

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

Evaluate $g ' \left ( \frac{1}{2} \right )$ .

Possible Answers:

$- \frac{ \pi \sqrt{2}}{2}$

$- \frac{ \sqrt{2}}{2 \pi}$

$\pi \sqrt{2}$

$- \pi \sqrt{2}$

$\frac{ \pi \sqrt{2}}{2}$

Correct answer:

$- \frac{ \pi \sqrt{2}}{2}$

Explanation:

To find g'(x) , substitute $u = \pi x^{2}$ and use the chain rule:

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

$g'(x)= \frac{\mathrm{d} }{\mathrm{d} x} \cos \left ( \pi x^{2}\right )$

$= \frac{\mathrm{d} }{\mathrm{d} x} \cos u$

$=\frac{\mathrm{d} u}{\mathrm{d} x} \cdot \frac{\mathrm{d} }{\mathrm{d} u} \cos u$

$=\frac{\mathrm{d} }{\mathrm{d} x} \pi x^{2} \cdot \frac{\mathrm{d} }{\mathrm{d} u} \cos u$

$= \pi \cdot 2 x\cdot \left ( - \sin u \right )$

$= \pi \cdot 2 x\cdot \left [- \sin \left ( \pi x^{2} ) \right ]$

$=-2 \pi x \sin \left ( \pi x^{2} )$

So $g'(x)=-2 \pi x \sin \left ( \pi x^{2} )$

and

$g ' \left ( \frac{1}{2} \right )=-2 \pi\cdot \frac{1}{2} \sin \left [\pi \cdot \left ( \frac{1}{2} \right )^{2} \right ]$

$=- \pi \sin \left ( \frac{\pi}{4} \right )$

$=- \pi \cdot \frac{\sqrt{2}}{2} = - \frac{ \pi \sqrt{2}}{2}$

Report an Error

Example Question #71 : Derivatives

What is the equation of the line tangent to the graph of the function

f(x)= x^3-2

at (2,6) ?

Possible Answers:

y = 24 x-42

y = 10 x-14

y = 10 x+14

y = 12 x+18

y = 12 x-18

Correct answer:

y = 12 x-18

Explanation:

The slope of the line tangent to the graph of f(x)= x^3-2 at (2,6) is

f'(2) , which can be evaluated as follows:

f(x)= x^3-2

$f'(x)= 3x^{3-1}-0$

$f'(x)= 3x^{2}$

Then $f'(2)= 3 \cdot 2^{2} = 3 \cdot4 = 12$ , which is the slope of the line.

The equation of the line with slope 12 through (2,6) is:

$y - y_{1} = m (x-x_{1})$

y - 6 = 12 (x-2)

y - 6 = 12 x-24

y - 6 + 6= 12 x-24 + 6

y = 12 x-18

Report an Error

Example Question #73 : Derivatives

Let the initial approximation of the seventh root of 1,000 be

$x_{1}= 3$ .

Use one iteration of Newton's method to find approximation $x_{2}$ . Give your answer to the nearest thousandth.

Possible Answers:

2.727

2.787

2.707

2.767

2.687

Correct answer:

2.767

Explanation:

This is equivalent to finding a solution of the equation

x^7 = 1,000 ,

or the zero of the polynomial

P (x) = x^7 - 1,000 .

Using Newton's method, we can find $x_{2}$ from the formula

$x_{2} = x_{1} - \frac{P(x_{1})}{P'(x_{1})}$ .

P'(x) = 7x^6 , so

$P (x_{1}) =P(3)= 3^7 - 1,000 = 2,187 - 1,000 = 1,187$

$P'(x_{1}) = P'(3)= 7 \cdot 3^6 = 7 \cdot 729 = 5,103$

$x_{2} = 3 - \frac{P(3)}{P'(3)} = 3 - \frac{1,187}{5,103} \approx 3 - 0.2326 \approx 2.767$

Report an Error

Example Question #71 : Derivatives

Let the initial approximation of a solution of the equation

x^3 = x + 19

be $x_{1} = 2.5$ .

Use one iteration of Newton's method to find an approximation of $x_{2}$ . Give your answer to the nearest thousandth.

Possible Answers:

2.881

2.681

2.781

2.831

2.731

Correct answer:

2.831

Explanation:

Rewrite the equation to be solved for as

x^3 - x - 19 = 0 .

Let f(x) = x^3 - x - 19 .

Then,

$f'(x) = \frac{\mathrm{d} }{\mathrm{d} x} \left ( x^3 - x - 19 \right )$ .

f'(x) =3x^2 - 1

The problem amounts to finding a zero of f(x ) . By Newton's method, the second approximation can be derived from the first using the equation

$x_{2} = x_{1} - \frac{f(x_{1})}{f'(x_{1})}$ .

$x_{1} = 2.5$ , so

f(2.5) = 2.5^3 - 2.5 - 19 = 15.625 - 2.5 - 19 = -5.875

and

$f'(2.5) =3 \cdot 2.5^2 - 1 = 3 \cdot 6.25 - 1 = 18.75 - 1 = 17.75$

$x_{2} = 2.5 - \frac{f(2.5)}{f'(2.5)}= 2.5 - \frac{ -5.875}{17.75}\approx 2.5 - (-0.3310) \approx 2.831$

Report an Error

View Calculus 2 Tutors

Yuanhao
Certified Tutor

Auburn University, Bachelor of Science, Civil Engineering. Carnegie Mellon University, Master of Science, Engineering Mechani...

View Calculus 2 Tutors

Gregory
Certified Tutor

Pennsylvania State University-Main Campus, Bachelor of Science, Chemical Engineering. Carnegie Mellon University, Doctor of P...

View Calculus 2 Tutors

Muhammad
Certified Tutor

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

All Calculus 2 Resources

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

Popular Subjects

Chemistry Tutors in Los Angeles, ACT Tutors in Phoenix, GRE Tutors in Miami, Math Tutors in Miami, GRE Tutors in Philadelphia, MCAT Tutors in Atlanta, GMAT Tutors in New York City, French Tutors in Phoenix, Reading Tutors in Denver, Spanish Tutors in Dallas Fort Worth

Popular Courses & Classes

SAT Courses & Classes in Los Angeles, MCAT Courses & Classes in Seattle, GRE Courses & Classes in Dallas Fort Worth, Spanish Courses & Classes in Dallas Fort Worth, Spanish Courses & Classes in Miami, Spanish Courses & Classes in Los Angeles, GRE Courses & Classes in Chicago, Spanish Courses & Classes in Denver, GRE Courses & Classes in Los Angeles, ACT Courses & Classes in Phoenix

Popular Test Prep

MCAT Test Prep in Houston, ACT Test Prep in Phoenix, ISEE Test Prep in San Diego, ISEE Test Prep in Houston, MCAT Test Prep in Philadelphia, GRE Test Prep in Dallas Fort Worth, MCAT Test Prep in San Francisco-Bay Area, MCAT Test Prep in Seattle, ACT Test Prep in Dallas Fort Worth, SSAT Test Prep in Philadelphia

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 : Derivatives

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #3 : Derivative Defined As Limit Of Difference Quotient

Example Question #71 : Derivatives

Example Question #72 : Derivatives

Example Question #2 : Chain Rule And Implicit Differentiation

Example Question #3 : Chain Rule And Implicit Differentiation

Example Question #4 : Chain Rule And Implicit Differentiation

Example Question #5 : Chain Rule And Implicit Differentiation

Example Question #71 : Derivatives

Example Question #73 : Derivatives

Example Question #71 : Derivatives

All Calculus 2 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: