Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 1 : How to find differential functions

Study concepts, example questions & explanations for Calculus 1

Create An Account

All Calculus 1 Resources

10 Diagnostic Tests 438 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #261 : Differential Functions

Find the derivative of the function

$f(x) = \frac{\cos(\sin(x))}{\sin(x)}$

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

To find the derivative of this function we must use the Chain Rule and the Quotient Rule. Applying the Chain Rule to the numerator gives

$h'(x) = -\sin(\sin(x))\cos(x)$

Now using the Quotient Rule for the function, we find the derivative to be

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

Report an Error

Example Question #81 : Other Differential Functions

Find the derivative of

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

To find the derivative of this function, we must use the Quotient Rule which is

$f'(x) = \frac{h'(x)g(x) - h(x)g'(x)}{(g(x))^2}$

Applying this to the function we are given, with h(x) = x^2 and $g(x) = \sin(x)$ gives us

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

Report an Error

Example Question #81 : Other Differential Functions

Find the derivative of the following function

$f(x) = \frac{x\sin(x)}{\cos(x)}$

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

To find the derivative of this function we must use the Product Rule and the Quotient Rule. Appling the Product Rule to the numerator of the function gives us

$h'(x) = \sin(x)+x\cos(x)$

Using this with the Quotient Rule, we find

$f'(x) = \frac{[\sin(x)+x\cos(x)]\cos(x)-x\sin(x)(-\sin(x))}{\cos^2(x)} \rightarrow$

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

Report an Error

Example Question #81 : Other Differential Functions

Find the derivative of

$f(x) = \sin(x)\cos(2x)$

Possible Answers:

$f'(x) = \cos(x)\cos(2x)+\sin(2x)\sin(x)$

$f'(x) = \cos(x)\cos(2x)-\sin(2x)\sin(x)$

$f'(x) = \cos(x)\cos(2x)-2\sin(2x)\sin(x)$

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

Correct answer:

$f'(x) = \cos(x)\cos(2x)-2\sin(2x)\sin(x)$

Explanation:

To find the derivative of this function we must use the Product Rule and the Chain Rule. If $h(x) = \sin(x)$ and $g(x) = \cos(2x)$ then

$h'(x) = \cos(x)$

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

Applying these derivatives to the Product Rule gives us

$f'(x) = \cos(x)\cos(2x)-2\sin(2x)\sin(x)$

Report an Error

Example Question #271 : Differential Functions

Differentiate:

$f(x) = \left(\frac{1}{x}\right)\cos^2(x)$

Possible Answers:

$f'(x) = \frac{-1}{x^2}\cos(x)-\frac{2}{x}\cos(x)\sin(x)$

$f'(x) = \frac{-1}{x^2}\cos(x)+\frac{2}{x}\cos(x)\sin(x)$

$f'(x) = \frac{-1}{x^2}\cos^2(x)-\frac{2}{x}\cos(x)\sin(x)$

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

Correct answer:

$f'(x) = \frac{-1}{x^2}\cos^2(x)-\frac{2}{x}\cos(x)\sin(x)$

Explanation:

To find the derivative of this function we must use the Product Rule and the Chain Rule. First we set

$h(x) = \frac{1}{x}$

and

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

Now differentiating both of these functions gives

$h'(x) = \frac{-1}{x^2}$

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

Applying this to the Product Rule gives us,

$f'(x) = \frac{-1}{x^2}\cos^2(x)-\frac{2}{x}\cos(x)\sin(x)$

Report an Error

Example Question #272 : Differential Functions

Find the derivative of

$f(x) = \sin(x)\sin\left(\frac{1}{x}\right)$

Possible Answers:

$f'(x) = \cos(x)\sin\left(\frac{1}{x}\right)-\frac{\cos(\frac{1}{x})\sin(x)}{x^2}$

None of these answers are correct.

$f'(x) = \cos(x)\sin\left(\frac{1}{x}\right)-\frac{\cos(x)\sin(x)}{x^2}$

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

Correct answer:

$f'(x) = \cos(x)\sin\left(\frac{1}{x}\right)-\frac{\cos(\frac{1}{x})\sin(x)}{x^2}$

Explanation:

To find the derivative of this function we must use the Product Rule and the Chain Rule. If we have

$h(x) = \sin(x)$ and $g(x) = \sin\left(\frac{1}{x}\right)$ then

$h'(x) = \cos(x)$ and $g'(x) = \cos\left(\frac{1}{x}\right)\left(\frac{-1}{x^2}\right)$

Applying this to the product rule, we find

$f'(x) = \cos(x)\sin\left(\frac{1}{x}\right)+\cos\left(\frac{1}{x}\right)\sin(x)\left(\frac{-1}{x^2}\right) \rightarrow$

$f'(x) = \cos(x)\sin\left(\frac{1}{x}\right)-\frac{\cos(\frac{1}{x})\sin(x)}{x^2}$

Report an Error

Example Question #271 : Functions

Find the derivative of the function

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

Possible Answers:

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

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

None of these answers are correct.

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

Correct answer:

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

Explanation:

To find the derivative of this function, we must use the Product Rule, Quotient Rule, and the Chain Rule. To do this, we first find the derivative of each part.

$h(x) = x^4 \Rightarrow h'(x) = 4x^3$

$g(x) = \cos(3x) \Rightarrow g'(x) = -3\sin(3x)$

$j(x) = \sin(x) \Rightarrow j'(x) = \cos(x)$

Using the derivatives of each part we can find the derivative of the numerator using the Product Rule

$h(x)g(x) = x^4\cos(3x) \Rightarrow \\ h'(x)g(x) + h(x)g'(x) = 4x^3\cos(3x)+(-3x^4)\sin(3x)$

Finally, putting this into the Quotient Rule gives

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

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

Report an Error

Example Question #274 : Differential Functions

Differentiate the function

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

Possible Answers:

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

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

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

None of these answers are correct.

Correct answer:

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

Explanation:

To differentiate this function we must use the Quotient Rule

$f'(x) = \frac{h'(x)g(x)-h(x)g'(x)}{(g(x))^2}$

Using $h(x) = \sin(x)$ and g(x) = x^2 .

The derivative of the function is then

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

Report an Error

Example Question #275 : Differential Functions

Differentiate the following function

$f(x) = \sin^2(\cos(2x))$

Possible Answers:

$f'(x) = -4\sin(\cos(2x))\cos(\cos(2x))\sin(2x)$

$f'(x) = -4\cos(\cos(2x))\cos(\sin(2x))\sin(2x)$

$f'(x) = 2\sin(\cos(2x))\cos(\cos(2x))\sin(2x)$

None of these answers are correct.

Correct answer:

$f'(x) = -4\sin(\cos(2x))\cos(\cos(2x))\sin(2x)$

Explanation:

To differentiate this function we must use the Chain Rule. where

$f(x) = h(g(x)) \Rightarrow f'(x) = h'(g(x))\cdot g'(x)$

Therefore the derivative of the function is

$f(x) = 2\sin(\cos(2x))\cos(\cos(2x))(-\sin(2x))(2) \rightarrow$

$f'(x) = -4\sin(\cos(2x))\cos(\cos(2x))\sin(2x)$

Report an Error

Example Question #82 : Other Differential Functions

Find the derivative of the function

$f(x) = x^2\cos(x)$

Possible Answers:

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

None of these answers are correct.

$f'(x) = 2x\sin(x) + x^2\sin(x)$

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

Correct answer:

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

Explanation:

To find the derivative of the function, we must use the Product Rule,

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

Using h(x) = x^2 and $g(x) = \cos(x)$ .

Therefore

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

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

Report an Error

View Calculus Tutors

Ayan
Certified Tutor

Vellore Institute of Technology, Bachelor, Chemical Engineering. University of Calgary, Master's/Graduate, Chemical and Petro...

View Calculus Tutors

Shakeel
Certified Tutor

View Calculus Tutors

Sharon
Certified Tutor

Tabor College, Bachelor of Science, Mathematics Teacher Education. Tabor College, Masters in Education, Education.

All Calculus 1 Resources

10 Diagnostic Tests 438 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Algebra Tutors in Houston, MCAT Tutors in San Francisco-Bay Area, Computer Science Tutors in New York City, LSAT Tutors in Miami, Biology Tutors in Phoenix, GMAT Tutors in Phoenix, SAT Tutors in Miami, Chemistry Tutors in Washington DC, GRE Tutors in Chicago, Statistics Tutors in Chicago

Popular Courses & Classes

MCAT Courses & Classes in Los Angeles, MCAT Courses & Classes in Seattle, SAT Courses & Classes in New York City, GMAT Courses & Classes in Dallas Fort Worth, GRE Courses & Classes in Los Angeles, GMAT Courses & Classes in San Francisco-Bay Area, Spanish Courses & Classes in Philadelphia, MCAT Courses & Classes in San Francisco-Bay Area, GRE Courses & Classes in Atlanta, ISEE Courses & Classes in Washington DC

Popular Test Prep

MCAT Test Prep in New York City, LSAT Test Prep in Dallas Fort Worth, ACT Test Prep in Miami, SAT Test Prep in San Francisco-Bay Area, GMAT Test Prep in Seattle, GRE Test Prep in Phoenix, SAT Test Prep in San Diego, GRE Test Prep in Miami, ACT Test Prep in Phoenix, MCAT 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 1 : How to find differential functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #261 : Differential Functions

Example Question #81 : Other Differential Functions

Example Question #81 : Other Differential Functions

Example Question #81 : Other Differential Functions

Example Question #271 : Differential Functions

Example Question #272 : Differential Functions

Example Question #271 : Functions

Example Question #274 : Differential Functions

Example Question #275 : Differential Functions

Example Question #82 : Other Differential Functions

All Calculus 1 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: