We are open Saturday and Sunday!

Call Now to Set Up Tutoring:

(888) 888-0446

AP Calculus BC : Derivative at a Point

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 #41 : Derivatives

Calculate the derivative of f(x)=4x^2-x+2 at the point x=1 .

Possible Answers:

Correct answer:

Explanation:

There are 2 steps to solving this problem.

First, take the derivative of f(x) .

Then, replace the value of x with the given point.

For example, if x=a , then we are looking for the value of f'(x=a) , or the derivative of f(x) at x=a .

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

Calculate f'(x)

Derivative rules that will be needed here:

Derivative of a constant is 0. For example, $\frac{\mathrm{d} }{\mathrm{d} t} 5 = 0$
Taking a derivative on a term, or using the power rule, can be done by doing the following: $\frac{\mathrm{d} }{\mathrm{d} t} t^n = n * t^{n-1}$

f'(x)=8x-1

Then, plug in the value of x and evaluate

f'(x=1)=8*1-1 = 7

Report an Error

Example Question #42 : Derivatives

Evaluate the first derivative if

f(x)=2e^x and x=0 .

Possible Answers:

f'(0)=2

f'(0)=e

f'(0)=0

f'(0)=1

Correct answer:

f'(0)=2

Explanation:

First we must find the first derivative of the function.

Because the derivative of the exponential function is the exponential function itelf, or

$\frac{\mathrm{d} }{\mathrm{d} x}[e^x]=e^x$

and taking the derivative is a linear operation,

we have that

$f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}[f(x)]$

$=\frac{\mathrm{d} }{\mathrm{d} x}[2e^x]$

$=2\frac{\mathrm{d} }{\mathrm{d} x}[e^x]$

=2e^x

Now setting x=0

f'(0)=2e^0=2*1=2

Thus

f'(0)=2

Report an Error

Example Question #3 : Derivative At A Point

Find the rate of change of f(x) when x=3.

f(x)=3e^x-x^3+12x^2

Possible Answers:

-27.2

105.26

72.41

123.23

Correct answer:

105.26

Explanation:

Find the rate of change of f(x) when x=3.

f(x)=3e^x-x^3+12x^2

To find a rate of change, we need to find the derivative.

First, recall the following rules:

$1) \frac{d}{dx}e^x=e^x$

$2)\frac{d}{dx} x^n =nx^{n-1}$

We can apply these two derivative rules to our function to get our first derivative. Then we need to plug in 3 for x and solve.

f(x)=3e^x-x^3+12x^2

f'(x)=3e^x-3x^2+24x

f'(3)=3e^3-3(3)^2+24(3)=105.26

So, our answer is 105.26

Report an Error

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

View AP Calculus BC Tutors

James
Certified Tutor

University of Georgia, Bachelor in Arts, Political Science and Government.

View AP Calculus BC Tutors

Bhupinder
Certified Tutor

Panjab University, Bachelor of Engineering, Biotechnology. Hamburg University of Technology, Master of Science, Chemical Engi...

View AP Calculus BC Tutors

Shakeel
Certified Tutor

All AP Calculus BC Resources

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

Popular Subjects

Computer Science Tutors in Houston, Algebra Tutors in Dallas Fort Worth, French Tutors in Atlanta, MCAT Tutors in Dallas Fort Worth, SAT Tutors in Chicago, Calculus Tutors in Atlanta, Calculus Tutors in Los Angeles, Spanish Tutors in Phoenix, Math Tutors in San Diego, ISEE Tutors in Los Angeles

Popular Courses & Classes

SSAT Courses & Classes in Seattle, LSAT Courses & Classes in Washington DC, Spanish Courses & Classes in Houston, Spanish Courses & Classes in Seattle, ACT Courses & Classes in New York City, GMAT Courses & Classes in New York City, SAT Courses & Classes in Atlanta, SSAT Courses & Classes in Philadelphia, SAT Courses & Classes in Denver, Spanish Courses & Classes in Los Angeles

Popular Test Prep

GMAT Test Prep in Boston, ACT Test Prep in Denver, ISEE Test Prep in Miami, GMAT Test Prep in New York City, MCAT Test Prep in Houston, ACT Test Prep in Boston, MCAT Test Prep in Washington DC, LSAT Test Prep in Los Angeles, MCAT Test Prep in Boston, ISEE Test Prep in Los Angeles

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 at a Point

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

Example Question #41 : Derivatives

Example Question #42 : Derivatives

Example Question #3 : Derivative At A Point

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

All AP Calculus BC Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: