We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

AP Calculus BC : Derivative Defined as Limit of Difference Quotient

Study concepts, example questions & explanations for AP Calculus BC

Create An Account

All AP Calculus BC Resources

2 Diagnostic Tests 86 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #63 : Derivatives

$g(x) = \sec \frac{\pi}{x}$

Evaluate g '(6) .

Possible Answers:

$- \frac{\pi}{54}$

$- \frac{\pi \sqrt{3}}{9}$

$- \frac{\pi}{9}$

$- \frac{\pi \sqrt{3}}{54}$

$- \frac{\pi \sqrt{3}}{27}$

Correct answer:

$- \frac{\pi}{54}$

Explanation:

To find g'(x) , substitute $u =\frac{ \pi }{x} = \pi x ^{-1}$ and use the chain rule:

$g(x) = \sec \left (\frac{\pi}{x} \right )$

$g'(x) = \frac{\mathrm{d} }{\mathrm{d} x} \sec\frac{\pi}{x}$

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

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

$= \frac{\mathrm{d} }{\mathrm{d} x} \pi x^{-1} \cdot \frac{\mathrm{d} } {\mathrm{d}u} \sec u$

$= \pi \cdot (-1) \cdot x^{-1-1} \cdot \tan u \sec u$

$= - \pi x^{-2} \cdot \tan \frac{\pi}{x} \sec \frac{\pi}{x}$

$= - \frac{\pi}{x^{2}} \tan \frac{\pi}{x} \sec \frac{\pi}{x}$

So $g '(x)= - \frac{\pi}{x^{2}}\tan \frac{\pi}{x} \sec \frac{\pi}{x}$

and

$g '(6)= - \frac{\pi}{6^{2}}\tan \frac{\pi}{6} \sec \frac{\pi}{6}$

$= - \frac{\pi}{36} \cdot \frac{\sqrt{3}}{3} \cdot \frac{2\sqrt{3}}{3}$

$= - \frac{\pi}{36} \cdot \frac{2\cdot 3}{9} = - \frac{\pi}{54}$

Report an Error

Example Question #43 : Derivatives

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

$f(x)= 1+\frac{4}{x}$

at the point (2, 3) ?

Possible Answers:

y = - x+5

y =2 x-1

y = x+1

y = 3

y =4 x-5

Correct answer:

y = - x+5

Explanation:

The slope of the line tangent to the graph of $f(x)= 1+\frac{4}{x}$ at (2, 3) is

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

$f(x)= 1+\frac{4}{x} = 1 + 4x^{-1}$

$f'(x) = (-1) 4x^{-1-1}$

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

$f'(2) = - \frac{4}{2^{2}}= - 1$

The equation of the line with slope through (2, 3) is:

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

y - 3 = -1 (x-2)

y - 3 = - x+2

y = - x+5

Report an Error

Example Question #2 : Derivative At A Point

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

$f(x) = 2 \sin \frac{\pi x}{6}$

at the point (7, -1) ?

Possible Answers:

$y = - \frac {\pi \sqrt{3 }}{6} x+ \frac {7 \pi \sqrt{3 }}{6} -1$

$y = \frac {\pi \sqrt{3 }}{6} x - \frac {7 \pi \sqrt{3 }}{6} -1$

$y = - \frac {\pi \sqrt{2 }}{4} x+ \frac {7 \pi \sqrt{2 }}{4} -1$

$y = \frac {\pi \sqrt{2 }}{4} x- \frac {7 \pi \sqrt{2 }}{4} -1$

$y = - \frac {\pi }{6} x+ \frac {7 \pi }{6} -1$

Correct answer:

$y = - \frac {\pi \sqrt{3 }}{6} x+ \frac {7 \pi \sqrt{3 }}{6} -1$

Explanation:

The slope of the line tangent to the graph of $f(x) = 2 \sin \frac{\pi x}{6}$ at (7, -1) is

f'(7) , which can be evaluated as follows:

$f(x) = 2 \sin \frac{\pi x}{6} = 2 \sin \frac{\pi }{6}x$

$f'(x) = 2 \cdot\frac {\pi }{6} \cos \frac{\pi }{6}x$

$f'(x) = \frac {\pi }{3} \cos \frac{\pi }{6}x$

$f'(7) = \frac {\pi }{3} \cos \frac{7 \pi }{6}$

$= \frac {\pi }{3} \left ( - \frac{\sqrt{3}}{2}\right ) = - \frac {\pi \sqrt{3 }}{6}$ , the slope of the line.

The equation of the line with slope $- \frac {\pi \sqrt{3 }}{6}$ through (7, -1) is:

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

$y - (-1) = - \frac {\pi \sqrt{3 }}{6} (x-7)$

$y +1 = - \frac {\pi \sqrt{3 }}{6} x+\left ( \frac {\pi \sqrt{3 }}{6} \right ) 7$

$y = - \frac {\pi \sqrt{3 }}{6} x+ \frac {7 \pi \sqrt{3 }}{6} -1$

Report an Error

Example Question #44 : Derivatives

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

$f(x)= \frac{8}{x}$

at (-2,-4) ?

Possible Answers:

y = 2 x

y = -2 x-8

$y = \frac{1}{2}x-3$

y = -8 x-20

$y = - \frac{1}{2}x-5$

Correct answer:

y = -2 x-8

Explanation:

The slope of the line tangent to the graph of $f(x)= \frac{8}{x}$ at (-2,-4) is

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

$f(x)= \frac{8}{x} = 8x^{-1}$

$f'(x)= 8 \cdot\left ( -1 x^{-2} \right )$

$f'(x)= -\frac{8}{x^{ 2}}$

$f'(-2)= -\frac{8}{(-2)^{ 2}} = -\frac{8}{4} = -2$ , the slope of the line.

The equation of the line with slope through (-2,-4) is:

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

y - (-4) = -2 [x- (-2)]

y +4 = -2 (x+2)

y +4 = -2 x-4

y = -2 x-8

Report an Error

Example Question #1194 : Calculus Ii

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

$f(x)= \log (x+3)$

at the point (7,1) ?

Possible Answers:

y = x-6

y = 2x - 13

$y = \frac{x + 3 }{10 }$

$y = \frac{x -7 + 10 \ln 10}{10 \ln 10}$

$y = \frac{x -7 + e \ln 10}{e \ln 10}$

Correct answer:

$y = \frac{x -7 + 10 \ln 10}{10 \ln 10}$

Explanation:

The slope of the line tangent to the graph of $f(x)= \log (x+3)$ at the point (7,1) is f'(7) , which can be evaluated as follows:

$f(x)= \log (x+3)$

$f(x)=\frac{ \ln (x+3) }{\ln 10}$

$f'(x)=\frac{\mathrm{d} }{\mathrm{d} x} \frac{ \ln (x+3) }{\ln 10}$

$f'(x)= \frac{1 }{\ln 10} \cdot \frac{\mathrm{d} }{\mathrm{d} x}\ln (x+3)$

$f'(x)= \frac{1 }{\ln 10} \cdot \frac{1}{x+3}$

$f'(7)= \frac{1 }{\ln 10} \cdot \frac{1}{7+3}$

$f'(7)= \frac{1 }{10 \ln 10}$

The line with this slope through (7,1) has equation:

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

$y - 1= \frac{1 }{10 \ln 10} (x-7)$

$y - 1= \frac{1 }{10 \ln 10} \cdot x- \frac{1 }{10 \ln 10} \cdot 7$ $y = \frac{x }{10 \ln 10} - \frac{7 }{10 \ln 10} +1$

$y = \frac{x -7 + 10 \ln 10}{10 \ln 10}$

Report an Error

Example Question #4 : Derivative At A Point

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 = 28 x- 71

y = 26 x- 65

y = 10x -17

y = 11x -20

y = 27 x- 68

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 #83 : Derivative Review

Given the function y=log_e(x)+5x , find the slope of the point (10,51) .

Possible Answers:

$\frac{51}{10}$

The slope cannot be determined.

Correct answer:

$\frac{51}{10}$

Explanation:

To find the slope at a point of a function, take the derivative of the function.

y=log_e(x)+5x

The derivative of log_a(x) is $\frac{1}{x\:ln(a)}$ .

Therefore the derivative becomes,

$\frac{1}{x\:ln(e)}+5=\frac{1}{x}+5$ since ln(e)=1 .

Now we substitute the given point to find the slope at that point.

$\frac{dy}{dx}=\frac{1}{x} +5 = \frac{1}{10}+5 =\frac{51}{10}$

Report an Error

Example Question #1 : Derivative At A Point

Find the value of the following derivative at the point $\left(2,\frac{1}{2}\right)$ :

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

Possible Answers:

$\frac{1}{16}$

$\frac{1}{4}$

$-\frac{1}{16}$

$\frac{1}{\sqrt{\frac{5}{2}}}$

${}-\frac{1}{4}$

Correct answer:

$-\frac{1}{16}$

Explanation:

To solve this problem, first we need to take the derivative of the function. It will be easier to rewrite the equation as $f(x)=(x+2)^{-\frac{1}{2}}$ from here we can take the derivative and simplify to get

$\\f'(x)=-\frac{1}{2}(x+2)^{-\frac{1}{2}-1}\cdot \frac{\mathrm{d} }{\mathrm{d} x}(x+2)\\ f'(x)=-\frac{1}{2}(x+2)^{-\frac{3}{2}}\cdot 1\\f'(x)=-\frac{1}{2}(x+2)^{-\frac{3}{2}}$

From here we need to evaluate at the given point $\left(2,\frac{1}{2}\right)$ . In this case, only the x value is important, so we evaluate our derivative at x=2 to get $f'(2)=-\frac{1}{2}(2+2)^{-\frac{3}{2}}=-\frac{1}{2}\cdot \frac{1}{8}=-\frac{1}{16}$ .

Report an Error

Example Question #2 : Derivative At A Point

Evaluate the value of the derivative of the given function at the point (-1,4) :

f(x)=3x^4+2x^2-1

Possible Answers:

Correct answer:

Explanation:

To solve this problem, first we need to take the derivative of the function.

$\\f'(x)=4\cdot 3x^{4-1}+2\cdot 2x^{2-1}\\f'(x)=12x^3+4x$

From here we need to evaluate at the given point (-1,4) . In this case, only the x value is important, so we evaluate our derivative at x=1 to get

$\\f'(-1)=12(-1)^3+4(-1)\\ f'(-1)=12(-1)+4(-1)\\ f'(-1)=-16$ .

Report an Error

Example Question #94 : Derivatives

Given $f(x)=5x^{2}+6x-12$ , find the value of f'(x) at the point (-1,-1) .

Possible Answers:

Correct answer:

Explanation:

Given the function $f(x)=5x^{2}+6x-12$ , we can use the Power Rule

$\frac{d}{dx}x^{n}=nx^{n-1}$ for all $n\neq0$ to find its derivative:

$f'(x)=(2)5x^{2-1}+(1)6x^{1-1}-(0)12=10x+6$ .

Plugging in the -value of the point (-1,-1) into f'(x) , we get

f'(-1)=10(-1)+6=-10+6=-4 .

Report an Error

View AP Calculus BC Tutors

Mohammed
Certified Tutor

University of Maryland-College Park, Bachelor of Science, Electrical Engineering. University of Maryland-College Park, Master...

View AP Calculus BC Tutors

Yoonsik
Certified Tutor

Seoul National University, Bachelor of Science, Physics. University of Pennsylvania, Doctor of Philosophy, Atomic and Molecul...

View AP Calculus BC Tutors

Nathan
Certified Tutor

California Polytechnic State University, San Luis Obispo, Bachelor, Electrical Engineering.

All AP Calculus BC Resources

2 Diagnostic Tests 86 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

ISEE Tutors in New York City, Statistics Tutors in Chicago, LSAT Tutors in Phoenix, SAT Tutors in Miami, Statistics Tutors in San Francisco-Bay Area, Math Tutors in Houston, Spanish Tutors in Chicago, Spanish Tutors in Seattle, Spanish Tutors in Miami, Algebra Tutors in Denver

Popular Courses & Classes

GMAT Courses & Classes in San Francisco-Bay Area, MCAT Courses & Classes in Houston, Spanish Courses & Classes in Denver, LSAT Courses & Classes in Boston, LSAT Courses & Classes in Seattle, MCAT Courses & Classes in Chicago, SSAT Courses & Classes in Chicago, MCAT Courses & Classes in Los Angeles, MCAT Courses & Classes in Denver, ISEE Courses & Classes in Washington DC

Popular Test Prep

ISEE Test Prep in Atlanta, GRE Test Prep in Denver, ACT Test Prep in Chicago, ISEE Test Prep in Los Angeles, LSAT Test Prep in New York City, LSAT Test Prep in Boston, MCAT Test Prep in Houston, GMAT Test Prep in Boston, ACT Test Prep in Denver, GMAT Test Prep in Houston

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

AP Calculus BC : Derivative Defined as Limit of Difference Quotient

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

Example Question #63 : Derivatives

Example Question #43 : Derivatives

Example Question #2 : Derivative At A Point

Example Question #44 : Derivatives

Example Question #1194 : Calculus Ii

Example Question #4 : Derivative At A Point

Example Question #83 : Derivative Review

Example Question #1 : Derivative At A Point

Example Question #2 : Derivative At A Point

Example Question #94 : Derivatives

All AP Calculus BC Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: