We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 2 : Derivative Review

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 #1621 : Calculus Ii

Differentiate $f(y)=\frac{e^y}{y}$ .

Possible Answers:

e^y

e^y(ln(y))

$\frac{(y-1)e^y}{y^2}$

$\frac{e^y-ye^y}{y^2}$

$\frac{e^y-ye^y}{y}$

Correct answer:

$\frac{(y-1)e^y}{y^2}$

Explanation:

Using the Quotient rule for derivatives, we know that the derivative will equal : $\frac{y(e^y)-e^y(1)}{y^2}$ .

If you simplify this, you will get:

$\frac{(y-1)e^y}{y^2}$

Report an Error

Example Question #1621 : Calculus Ii

Differentiate $\frac{2}{x}+\frac{1}{4x^3}-\frac{6}{x^5}$ .

Possible Answers:

2lnx-3ln(4x)-30lnx

$-2x^{-2}-\frac{1}{4}x^{-4}-30x^{-6}$

$-2x^{-2}+\frac{1}{4}x^{-4}+30x^{-6}$

$2+\frac{1}{8}x^{-2}-\frac{6}{4}x^{-4}$

$2-\frac{1}{8}x^{-2}+\frac{6}{4}x^{-4}$

Correct answer:

$-2x^{-2}-\frac{1}{4}x^{-4}-30x^{-6}$

Explanation:

Rewriting the original equation gives us $f(x)=2(x^{-1})+\frac{1}{4}x^{-3}-6(x^{-5})$ .

Then, one can just use the power rule to get: $-2x^{-2}-\frac{3}{12}x^{-4}-30x^{-6}$ .

Simplifying this gives us: $-2x^{-2}-\frac{1}{4}x^{-4}-30x^{-6}$ .

Report an Error

Example Question #1621 : Calculus Ii

Express the derivative of $y(x)=\sqrt {7x^3-6x^2+9x}$ in simplest terms.

Possible Answers:

$\large y(x)=\bigg[\frac {7x^2-4x+3}{\sqrt{(7x^3-6x^2+9x)}}\bigg]^{\frac {2}{3}}$

$\large y'(x)=\frac {7x^2-4x+3}{\sqrt [3] {(7x^3-6x^2+9x)^2}}$

$\large y'(x)=-\frac {7x^2-4x+3}{\sqrt [3] {(7x^3-6x^2+9x)^2}}$

$\large y'(x)=\frac {7x^2-4x+3}{\sqrt {(7x^3-6x^2+9x)^3}}$

Correct answer:

$\large y'(x)=\frac {7x^2-4x+3}{\sqrt [3] {(7x^3-6x^2+9x)^2}}$

Explanation:

Screen shot 2016 01 01 at 10.25.54 pm

Report an Error

Example Question #502 : Derivative Review

Find the derivative of $\large y(x)=(2x+x^{-3})^\frac {1}{4}$ in simplest form.

Possible Answers:

$\large \large y'(x)=\frac {x^\frac {-7}{4}(2x^4-3)}{4(2x^4+1)^\frac {3}{4}}$

$\large \large y'(x)=\frac {(2x^4-3)}{4x^\frac {7}{4}(2x^4+1)^\frac {3}{4}}$

$\large \large y'(x)=\frac {x^{\frac {1}{4}}(2x^4-3)}{4(2x^4+1)^\frac {-3}{4}}$

$\large \large y'(x)=\frac {5(2x^4-3)}{4x^\frac {7}{4}(2x^4+1)^\frac {3}{4}}$

Correct answer:

$\large \large y'(x)=\frac {(2x^4-3)}{4x^\frac {7}{4}(2x^4+1)^\frac {3}{4}}$

Explanation:

Screen shot 2016 01 02 at 8.27.30 am

Report an Error

Example Question #1621 : Calculus Ii

What is the first derivative of $\large y(x)=tan^4\sqrt {cot(3x)}$ ?

Possible Answers:

$\large \large -\frac {6csc^2(3x)\big (tan \sqrt {cot(3x))}\big )^3 \big (sec^2 \sqrt {cot(3x)})}{\sqrt {cot(3x)}}$

$\large -\frac {6csc^2(3x)\big (tan^3 \sqrt {cot(3x))}) \big (sec^2 \sqrt {cot(3x)})}{\sqrt {cot(3x)}}$

$\large - {6csc^2(3x)\big (tan^3 \sqrt {cot(3x))}) \big (sec^2)}$

$\large \frac {6csc^2(3x)\big (tan^3 \sqrt {cot(3x))}) \big (sec^2 \sqrt {cot(3x)})}{\sqrt {cot(3x)}}$

Correct answer:

$\large -\frac {6csc^2(3x)\big (tan^3 \sqrt {cot(3x))}) \big (sec^2 \sqrt {cot(3x)})}{\sqrt {cot(3x)}}$

Explanation:

Screen shot 2016 01 02 at 9.36.43 am

Report an Error

Example Question #501 : Derivative Review

Which of the following IS NOT a rule used when finding derivatives of any function?

Possible Answers:

Chain Rule

Product Rule

Exponential Rule of Logarithms

Power Rule

Correct answer:

Exponential Rule of Logarithms

Explanation:

Step 1: Recall any rules that are used in derivatives...

Power Rule
Quotient Rule
Product Rule
Chain Rule

Step 2: Look at the choices in the question and compare the ones listed in Step 1.

We see Power Rule, Chain Rule, and Product Rule.

There is no such rule called the Addition rule, so this is the incorrect answer.

Report an Error

Example Question #502 : Derivatives

Suppose , , and any composite function between them are defined and differentiable everywhere. Given the derivatives for and :

h'(x)=2x

g'(x)=4

Find the derivative of f(x) where,

f(x)=h(x-g(x))+g(x)

Possible Answers:

f'(x)=-6x

f'(x)=6x+6g'(x)+4

2x+4

f'(x)=6g(x)+4

f'(x)=6x-6g(x)+4

Correct answer:

f'(x)=6x-6g(x)+4

Explanation:

f(x)=h(x-g(x))+g(x)

This is a conceptual problem. First notice that the function f(x) is the sum of two functions h(x-g(x)) and g(x) . We must differentiate each term.

To differentiate the first term, notice that h(x-g(x)) is a function of the function x-g(x) , so we must use the chain rule to differentiate with respect to . We cannot conclude that h'(x-g(x))=2x .

We were given h'(x)=2x . What this equation is telling us is that the function h(x) will have a derivative that is 2 times whatever is inside the parenthesis, with respect to the whatever is inside the parenthesis. If we wish to use to differentiate the composite h(x-g(x)) we can start with the derivative with respect to x-g(x) , which will be 2(x-g(x)) .

Now if we want to differentiate with respect to , we take the derivative with respect to x-g(x) and multiply by the derivative of x-g(x) with respect to (the chain rule).

$h'(x-g(x))=2(x-g(x))\times(1-g'(x))$

We were given g'(x)=4

$h'(x-g(x))=2(x-g(x))\times(-3)$

h'(x-g(x))= 6x-6g(x)

Now we can write the derivative of f(x) ,

f'(x)=h'(x-g(x))+g'(x)

f'(x)=6x-6g(x)+4

Report an Error

Example Question #501 : Derivative Review

Find the equation of the tangent line to the curve,

$y=\ln(x)+3x^2$

at x = 2

Find the equation of the tangent line to the curve,

$y=\ln(x)+3x^2$

at x = 2

Possible Answers:

$y = -\frac{25}{2}x+\ln(2) +13$

$y = \frac{25}{2}x -12$

$y = \frac{25}{2}x+\ln(2) -13$

$y = -\frac{25}{2}x+10$

$y = \frac{25}{2}x+\ln(2)$

Correct answer:

$y = \frac{25}{2}x+\ln(2) -13$

Explanation:

Find the equation of the tangent line to the curve corresponding to

$f(x)=\ln(x)+3x^2$

The first step is to compute the derivative for the function and then evaluate the derivative at x = 2

$f'(x)=\frac{1}{x}+6x\: \: \: \:\: \: \:\Rightarrow \: \: \: \: \: \: \: \:f'(2)=\frac{1}{2}+6(2)=\frac{25}{2}$

Therefore $\frac{25}{2}$ will be the slope of the tangent line at x = 2 . Write an equation for the line,

y = mx+b

$y = \frac{25}{2}x+b$

Now we need to find the y-intercept. Use the original function to find when x=2 .

$f(2)=\ln(2)+3(2^2)=\ln(2)+12$

This gives us the point at which the tangent line meets the curve,

$(2\: \: ,\: \, \ln(2)+12)$

Now use this point to find the y-intercept,

$b = y-\frac{25}{2}x$

$b = \ln(2)+12-\frac{25}{2}(2)$

$b=\ln(2)-13$ .

The equation of the line is therefore,

$y = \frac{25}{2}x+\ln(2) -13$

Plot

Report an Error

Example Question #502 : Derivative Review

Differentiate the function:

f(x)=xln(x)-x^2

Possible Answers:

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

f'(x)=ln(x)-2x+1

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

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

f'(x)=ln(x)+2x-1

Correct answer:

f'(x)=ln(x)-2x+1

Explanation:

First we see from the sum rule that:

f'(xln(x)-x^2)=f'(xln(x))-f'(x^2)

The first term we use the product rule to differentiate:

(fg)'=f'g+g'f

$(x\cdot ln(x))'=1\cdot ln(x)+\frac{x}{x}=ln(x)+1$

The second term is:

f'(x^2)=2x

Therefore:

f'(xln(x)-x^2)=ln(x)+1-2x

=ln(x)-2x+1

Report an Error

Example Question #503 : Derivative Review

Differentiate the following function:

y=ln(cos(x))

Possible Answers:

y'=-tan(x)

$y'=-\frac{1}{sin(x)}$

$y'=\frac{1}{sin(x)}$

y'=tan(x)

$y'=\frac{1}{cos(x)}$

Correct answer:

y'=-tan(x)

Explanation:

To differentiate the function y=ln(cos(x)) we have to use the chain rule

$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$

let u=cos(x) therefore y=ln(u) and

$\frac{dy}{du}=\frac{1}{u}=\frac{1}{cos(x)}$

and

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

Therefore:

$\frac{dy}{dx}=-\frac{sin(x)}{cos(x)}=-tan(x)$

Report an Error

View Calculus 2 Tutors

Mohammad
Certified Tutor

Rutgers University-New Brunswick, Bachelor of Science, Mathematics. Rutgers University-New Brunswick, Master of Science, Stat...

View Calculus 2 Tutors

Jason
Certified Tutor

Brigham Young University, Bachelor, Neuroscience.

View Calculus 2 Tutors

Reza
Certified Tutor

University of Sherbrooke, Doctor of Philosophy, Mathematics. University of Manitoba, Master of Science, Mathematics.

All Calculus 2 Resources

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

Popular Subjects

Reading Tutors in New York City, GRE Tutors in San Diego, English Tutors in Los Angeles, Math Tutors in Boston, Reading Tutors in San Francisco-Bay Area, GMAT Tutors in Los Angeles, ISEE Tutors in San Francisco-Bay Area, GMAT Tutors in Phoenix, Computer Science Tutors in Seattle, GMAT Tutors in Denver

Popular Courses & Classes

ISEE Courses & Classes in Phoenix, LSAT Courses & Classes in Denver, GMAT Courses & Classes in San Francisco-Bay Area, GRE Courses & Classes in Los Angeles, Spanish Courses & Classes in Atlanta, GMAT Courses & Classes in Seattle, GMAT Courses & Classes in Los Angeles, Spanish Courses & Classes in San Francisco-Bay Area, SSAT Courses & Classes in Houston, SSAT Courses & Classes in Washington DC

Popular Test Prep

SAT Test Prep in Miami, ISEE Test Prep in New York City, LSAT Test Prep in Atlanta, SSAT Test Prep in Philadelphia, SAT Test Prep in Los Angeles, SSAT Test Prep in Dallas Fort Worth, GMAT Test Prep in Seattle, MCAT Test Prep in New York City, ACT Test Prep in Washington DC, LSAT Test Prep in Phoenix

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 : Derivative Review

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #1621 : Calculus Ii

Example Question #1621 : Calculus Ii

Example Question #1621 : Calculus Ii

Example Question #502 : Derivative Review

Example Question #1621 : Calculus Ii

Example Question #501 : Derivative Review

Example Question #502 : Derivatives

Example Question #501 : Derivative Review

Example Question #502 : Derivative Review

Example Question #503 : Derivative Review

All Calculus 2 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: