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 #61 : How To Find Differential Functions

Differentiate the function:

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

Possible Answers:

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

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

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

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

f'(x)=1

Correct answer:

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

Explanation:

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

We evaluate this derivative using the quotient rule:

$if \ f(x)=\frac{g(x)}{h(x)}$ ,

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

Apply the above formula:

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

$f'(x)=\frac{x*cos(x) - sin(x)}{x^2}$ , which is our final answer.

Report an Error

Example Question #62 : How To Find Differential Functions

What is the slope of the line tangent to f(x) = x⁴ – 3x^–4 – 45 at x = 5?

Possible Answers:

500.00384

355.00384

422.125

400.096

355.096

Correct answer:

500.00384

Explanation:

First we must find the first derivative of f(x).

f'(x) = 4x³ + 12x^–5

To find the slope of the tangent line of f(x) at 5, we merely have to evaluate f'(x) at 5:

f'(5) = 4*5³ + 12* 5^–5 = 500 + 12/3125 = 500.00384

Report an Error

Example Question #63 : How To Find Differential Functions

Solve for g{}'(x) when

$g(x)=tan^{-1}(4x^3+7x^2)$

Possible Answers:

$g{}'(x)=\frac{4x^3+7x^2}{(12x^2+14x)^2}$

$g{}'(x)=sec^2(4x^3+7x^2)$

$g{}'(x)=\frac{12x^2+14x}{1+(4x^3+7x^2)^2}$

$g{}'(x)=sec^2(12x^2+14x)$

Correct answer:

$g{}'(x)=\frac{12x^2+14x}{1+(4x^3+7x^2)^2}$

Explanation:

$g{}'(x)=tan^{-1}(4x^3+7x^2)$

using the identity:

$\frac{\mathrm{d} }{\mathrm{d} x}\ tan^{-1}[(f(x)]=\frac{f{}'(x)}{1+[f(x)]^2}$

$g{}'(x)=\frac{12x^2+14x}{1+(4x^3+7x^2)^2}$

Report an Error

Example Question #251 : Functions

Differentiate $\small f(x)=\frac{x^2}{2x+7}$

Possible Answers:

$\small f'(x)=\frac{2x(3x+7)}{(2x+7)^2}$

$\small f'(x)=\frac{-2x(x+7)}{(2x+7)^2}$

$\small f'(x)=\frac{2x(x+7)}{(2x+7)}$

$\small f'(x)=\frac{2x(x+7)}{(2x+7)^2}$

Correct answer:

$\small f'(x)=\frac{2x(x+7)}{(2x+7)^2}$

Explanation:

The Quotient Rule applies when differentiating quotients of functions. Here, $\small f(x)$ equals the quotient of two functions, $\small x^2$ and $\small 2x+7$ . Let $\small l(x)=2x+7$ and $\small h(x)=x^2$ . (Think: $\small l(x)$ is the "low" function or denominator and $\small h(x)$ is the "high" function or numerator.) The Quotient Rule tells us to multiply the "low" function by the derivative of the "high" function, subtract the product of the "high" function and the derivative of the "low" function, and then divide the result by the square of the "low" function. In other words,

$\small \small f'(x)=\frac{l(x)h'(x)-h(x)l'(x)}{[l(x)]^2}$

Here, $\small l(x)=2x+7$ so $\small l'(x)=2$ . Similarly, $\small h(x)=x^2$ so $\small h'(x)=2x$ .

Then

$\small f'(x)=\frac{(2x+7)(2x)-x^2(2)}{(2x+7)^2}$

$\small f'(x)=\frac{(4x^2+14x)-2x^2}{(2x+7)^2}$

$\small f'(x)=\frac{2x^2+14x}{(2x+7)^2}$

Factoring out $\small 2x$ from the numerator gives

$\small \small f'(x)=\frac{2x(x+7)}{(2x+7)^2}$

$\small f'(x)=\frac{-2x(x+7)}{(2x+7)^2}$ inverts the order of the numerator, subtracting $\small l(x)h'(x)$ from $\small h(x)l'(x)$ .

$\small f'(x)=\frac{2x(3x+7)}{(2x+7)^2}$ adds the products in the numerator, rather than subtracting them.

$\small f'(x)=\frac{2x(x+7)}{(2x+7)}$ fails to square the denominator.

Report an Error

Example Question #64 : How To Find Differential Functions

Differentiate $\small f(x)=\frac{sec(x)}{x}$

Possible Answers:

$\small f'(x)=\frac{sec(x)(xtan(x)-1)}{x^2}$

$\small f'(x)=\frac{sec(x)(xtan(x)-1)}{x}$

$\small f'(x)=\frac{sec(x)(1-xtan(x))}{x^2}$

$\small f'(x)=\frac{sec(x)(xtan(x)+1)}{x^2}$

Correct answer:

$\small f'(x)=\frac{sec(x)(xtan(x)-1)}{x^2}$

Explanation:

The Quotient Rule applies when differentiating quotients of functions. Here, $\small f(x)$ equals the quotient of two functions, $\small sec(x)$ and $\small x$ . Let $\small l(x)=x$ and $\small h(x)=sec(x)$ . (Think: $\small l(x)$ is the "low" function or denominator and $\small h(x)$ is the "high" function or numerator.) The Quotient Rule tells us to multiply the "low" function by the derivative of the "high" function, subtract the product of the "high" function and the derivative of the "low" function, and then divide the result by the square of the "low" function. In other words,

$\small \small f'(x)=\frac{l(x)h'(x)-h(x)l'(x)}{[l(x)]^2}$

Here, $\small l(x)=x$ so $\small l'(x)=1$ . Similarly, $\small h(x)=sec(x)$ so $\small h'(x)=sec(x)tan(x)$ .

Then

$\small f'(x)=\frac{(x)[sec(x)tan(x)]-[sec(x)](1)}{x^2}$

$\small f'(x)=\frac{xsec(x)tan(x)-sec(x)}{x^2}$

Factoring out $\small sec(x)$ gives

$\small f'(x)=\frac{sec(x)(xtan(x)-1)}{x^2}$

$\small f'(x)=\frac{sec(x)(1-xtan(x))}{x^2}$ inverts the order of the numerator, subtracting $\small l(x)h'(x)$ from $\small h(x)l'(x)$ .

$\small f'(x)=\frac{sec(x)(xtan(x)+1)}{x^2}$ adds the products in the numerator, rather than subtracting them.

$\small f'(x)=\frac{sec(x)(xtan(x)-1)}{x}$ fails to square the denominator.

Report an Error

Example Question #65 : How To Find Differential Functions

Differentiate $\small f(x)=\frac{x-1}{x+5}$

Possible Answers:

$\small \small f'(x)=\frac{6}{(x+5)}$

$\small \small f'(x)=\frac{6}{(x+5)^2}$

$\small \small \small f'(x)=\frac{-6}{(x+5)^2}$

$\small \small \small f'(x)=\frac{2x+4}{(x+5)^2}$

Correct answer:

$\small \small f'(x)=\frac{6}{(x+5)^2}$

Explanation:

The Quotient Rule applies when differentiating quotients of functions. Here, $\small f(x)$ equals the quotient of two functions, $\small x-1$ and $\small x+5$ . Let $\small \small \small \small \small l(x)=x+5$ and $\small \small h(x)=x-1$ . (Think: $\small l(x)$ is the "low" function or denominator and $\small h(x)$ is the "high" function or numerator.) The Quotient Rule tells us to multiply the "low" function by the derivative of the "high" function, subtract the product of the "high" function and the derivative of the "low" function, and then divide the result by the square of the "low" function. In other words,

$\small \small f'(x)=\frac{l(x)h'(x)-h(x)l'(x)}{[l(x)]^2}$

Here, $\small l(x)=x+5$ so $\small l'(x)=1$ . Similarly, $\small h(x)=x-1$ so $\small h'(x)=1$ .

Then

$\small f'(x)=\frac{(x+5)(1)-(x-1)(1)}{(x+5)^2}$

$\small \small f'(x)=\frac{(x+5)-(x-1)}{(x+5)^2}$

$\small \small \small f'(x)=\frac{x-x+5+1}{(x+5)^2}$

$\small \small f'(x)=\frac{6}{(x+5)^2}$

$\small \small \small f'(x)=\frac{-6}{(x+5)^2}$ inverts the order of the numerator, subtracting $\small l(x)h'(x)$ from $\small h(x)l'(x)$ .

$\small \small \small f'(x)=\frac{2x+4}{(x+5)^2}$ adds the products in the numerator, rather than subtracting them.

$\small \small f'(x)=\frac{6}{(x+5)}$ fails to square the denominator.

Report an Error

Example Question #66 : How To Find Differential Functions

Differentiate $\small f(x)=\sqrt{x^2+1}$

Possible Answers:

$\small f'(x)=\frac{1}{2\sqrt{x^2+1}}+2x$

$\small f'(x)=\frac{x}{\sqrt{x^2+1}}$

$\small f'(x)=x^3+x$

$\small f'(x)=\frac{1}{2\sqrt{x^2+1}}$

Correct answer:

$\small f'(x)=\frac{x}{\sqrt{x^2+1}}$

Explanation:

The Chain Rule applies when differentiating compositions of functions. Here, $\small \small f(x)$ equals the composition of two functions, $\small \sqrt{x}$ and $\small x^2+1$ . Let $\small g(x)=\sqrt{x}$ and $\small h(x)=x^2+1$ . Then $\small \small f(x)=g(h(x))$ and the Chain Rule tells us to differentiate the outside function $\small \small g(x)$ and multiply the result by the derivative of the inside function $\small \small h(x)$ . In other words, $\small \small f'(x)=g'(h(x))h'(x)$ . Note that the inside function $\small \small h(x)$ is left untouched when the outside function $\small \small g(x)$ is differentiated. Here, $\small g'(h(x))=\frac{1}{2}(x^2+1)^{-\frac{1}{2}}$ and $\small h'(x)=2x$ . Remember, roots can (and should) be rewritten as fractional exponents, so $\small \sqrt{x^2+1}$ becomes $\small (x^2+1)^{\frac{1}{2}}$ which is then differentiated like any other exponent. So

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

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

$\small \small f'(x)=\frac{x}{\sqrt{x^2+1}}$

$\small f'(x)=\frac{1}{2\sqrt{x^2+1}}$ is an incomplete application of the Chain Rule which neglects to multiply the derivative of the outside function by the derivative of the inside function.

$\small f'(x)=\frac{1}{2\sqrt{x^2+1}}+2x$ is a misapplication of the Chain Rule which adds the derivative of the outside and inside functions rather than multiplying them.

$\small f'(x)=x^3+x$ is a misapplication of the Power Rule which fails to subtract 1 from the original exponent.

Report an Error

Example Question #67 : How To Find Differential Functions

Differentiate f(x)=cos^2(x)

Possible Answers:

f'(x)=2cos(x)

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

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

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

Correct answer:

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

Explanation:

The Chain Rule applies when differentiating compositions of functions. Here, $\small \small f(x)$ equals the composition of two functions, $\small \small x^2$ and $\small \small cos(x)$ . Let $\small \small \small g(x)=x^2$ and $\small \small \small h(x)=cos(x)$ . Then $\small \small f(x)=g(h(x))$ and the Chain Rule tells us to differentiate the outside function $\small \small g(x)$ and multiply the result by the derivative of the inside function $\small \small h(x)$ . In other words, $\small \small f'(x)=g'(h(x))h'(x)$ . Note that the inside function $\small \small h(x)$ is left untouched when the outside function $\small \small g(x)$ is differentiated. Here, $\small \small g'(h(x))=2cos(x)$ and $\small \small h'(x)=-sin(x)$ , so $\small f'(x)=(2cos(x))(-sin(x))$ which simplifies to $\small \small \small f'(x)=-2sin(x)cos(x)$ .

$\small f'(x)=2cos(x)$ is an incomplete application of the Chain Rule which neglects to multiply the derivative of the outside function by the derivative of the inside function.

$\small f'(x)=2cos(x)-sin(x)$ is a misapplication of the Chain Rule which adds the derivative of the outside and inside functions rather than multiplying them.

$\small f'(x)=-2sin(x)$ is a misapplication of the Chain Rule which substitutes the derivative of the inside function for the original inside function rather than multiplying the derivative of the outside function by the derivative of the inside function.

Report an Error

Example Question #68 : How To Find Differential Functions

Differentiate $\small f(x)=sin(6x)$

Possible Answers:

$\small f'(x)=cos(6x)$

$\small f'(x)=6cos(x)$

$\small f'(x)=cos(6x)+6$

$\small f'(x)=6cos(6x)$

Correct answer:

$\small f'(x)=6cos(6x)$

Explanation:

The Chain Rule applies when differentiating compositions of functions. Here, $\small \small f(x)$ equals the composition of two functions, $\small sin(x)$ and $\small 6x$ . Let $\small \small \small \small g(x)=sin(x)$ and $\small \small \small \small h(x)=6x$ . Then $\small \small f(x)=g(h(x))$ and the Chain Rule tells us to differentiate the outside function $\small \small g(x)$ and multiply the result by the derivative of the inside function $\small \small h(x)$ . In other words, $\small \small f'(x)=g'(h(x))h'(x)$ . Note that the inside function $\small \small h(x)$ is left untouched when the outside function $\small \small g(x)$ is differentiated. Here, $\small \small \small g'(h(x))=cos(6x)$ and $\small \small \small h'(x)=6$ , so $\small \small f'(x)=(cos(6x))(6)$ which simplifies to $\small \small \small \small f'(x)=6cos(6x)$ .

$\small \small f'(x)=cos(6x)$ is an incomplete application of the Chain Rule which neglects to multiply the derivative of the outside function by the derivative of the inside function.

$\small \small f'(x)=cos(6x)+6$ is a misapplication of the Chain Rule which adds the derivative of the outside and inside functions rather than multiplying them.

$\small \small f'(x)=6cos(x)$ is a misapplication of the Chain Rule which fails to preserve the original inside function when differentiating the outside function.

Report an Error

Example Question #69 : How To Find Differential Functions

Differentiate $\small f(x)=(3x+2)^5$

Possible Answers:

$\small f'(x)=5(3x+2)^4+3x$

$\small f'(x)=5(3x+2)^4+3$

$\small f'(x)=5(3x+2)^4$

$\small f'(x)=15(3x+2)^4$

Correct answer:

$\small f'(x)=15(3x+2)^4$

Explanation:

The Chain Rule applies when differentiating compositions of functions. Here, $\small \small f(x)$ equals the composition of two functions, $\small x^5$ and $\small 3x+2$ . Let $\small \small \small \small g(x)=x^5$ and $\small \small \small \small h(x)=3x+2$ . Then $\small \small f(x)=g(h(x))$ and the Chain Rule tells us to differentiate the outside function $\small \small g(x)$ and multiply the result by the derivative of the inside function $\small \small h(x)$ . In other words, $\small \small f'(x)=g'(h(x))h'(x)$ . Note that the inside function $\small \small h(x)$ is left untouched when the outside function $\small \small g(x)$ is differentiated. Here, $\small \small \small \small g'(h(x))=5(3x+2)^4$ and $\small \small \small h'(x)=3$ , so $\small \small \small \small f'(x)=(5(3x+2)^4)(3)$ which simplifies to $\small \small \small \small f'(x)=15(3x+2)^4$ .

$\small f'(x)=5(3x+2)^4$ is an incomplete application of the Chain Rule which neglects to multiply the derivative of the outside function by the derivative of the inside function.

$\small f'(x)=5(3x+2)^4+3$ is a misapplication of the Chain Rule which adds the derivative of the outside and inside functions rather than multiplying them.

$\small f'(x)=5(3x+2)^4+3x$ is a misapplication of the Chain Rule which adds the derivative of the outside function to an incorrect derivation of the inside function.

Report an Error

View Calculus Tutors

Muhammad
Certified Tutor

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

View Calculus Tutors

John
Certified Tutor

The University of Texas at Arlington, Bachelor of Science, Mathematics. Stanford University, Master of Science, Aerospace Eng...

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

Computer Science Tutors in New York City, SSAT Tutors in Atlanta, English Tutors in New York City, Math Tutors in Phoenix, GMAT Tutors in Miami, Spanish Tutors in Philadelphia, Computer Science Tutors in Chicago, MCAT Tutors in Los Angeles, GRE Tutors in Los Angeles, LSAT Tutors in Philadelphia

Popular Courses & Classes

SAT Courses & Classes in Denver, LSAT Courses & Classes in Phoenix, Spanish Courses & Classes in Denver, SSAT Courses & Classes in Seattle, SSAT Courses & Classes in Houston, LSAT Courses & Classes in Miami, GRE Courses & Classes in Phoenix, LSAT Courses & Classes in San Francisco-Bay Area, GMAT Courses & Classes in Chicago, ACT Courses & Classes in Los Angeles

Popular Test Prep

ISEE Test Prep in San Francisco-Bay Area, GMAT Test Prep in New York City, MCAT Test Prep in Atlanta, ISEE Test Prep in Houston, ACT Test Prep in Los Angeles, GMAT Test Prep in Denver, LSAT Test Prep in Seattle, SAT Test Prep in San Francisco-Bay Area, SSAT Test Prep in Los Angeles, GRE 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 #61 : How To Find Differential Functions

Example Question #62 : How To Find Differential Functions

Example Question #63 : How To Find Differential Functions

Example Question #251 : Functions

Example Question #64 : How To Find Differential Functions

Example Question #65 : How To Find Differential Functions

Example Question #66 : How To Find Differential Functions

Example Question #67 : How To Find Differential Functions

Example Question #68 : How To Find Differential Functions

Example Question #69 : How To Find Differential Functions

All Calculus 1 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: