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 #61 : Derivatives

Find the derivative of $f(x)=\frac{x}{x+1}$ using the definition of the derivative.

Possible Answers:

$f'(x)=\frac{1}{5x^2-3}$

f'(x)=1

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

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

Correct answer:

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

Explanation:

The definition of a derivative is $f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

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

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

Substituting these expressions into the definition of a derivative gives us

$f'(x)=\lim_{h\rightarrow 0}\frac{1}{h}*(f(x+h)-f(x))$

$f'(x)=\lim_{h\rightarrow 0}\frac{1}{h}*\left (\frac{x+h}{x+h+1}-\frac{x}{x+1} \right )$

$f'(x)=\lim_{h\rightarrow 0}\frac{1}{h}*\left (\frac{(x+h)(x+1)-x(x+h+1)}{(x+h+1)(x+1))} \right )$

$f'(x)=\lim_{h\rightarrow 0}\left (\frac{(x^2+hx+x+h) -(x^2+hx+x)}{h(x^2+hx+x+x+h+1)} \right )$

$f'(x)=\lim_{h\rightarrow 0}\left (\frac{h}{h(x^2+hx+2x+h+1)} \right )$

$f'(x)=\lim_{h\rightarrow 0}\left (\frac{1}{x^2+hx+2x+h+1} \right )$

$f'(x)=\left (\frac{1}{x^2+(0)x+2x+(0)+1} \right )$

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

Report an Error

Example Question #61 : Derivative Review

Possible Answers:

x^2

4x+C

Correct answer:

Explanation:

Report an Error

Example Question #62 : Derivative Review

What is the derivative of a term: ax^2 ?

Possible Answers:

3x^2

2ax

2x^3

Correct answer:

2ax

Explanation:

Step 1: Move the exponent to the coefficient..

We get, 2ax..

Step 2: The new exponent after moving the old one down is less than it was before:

$2ax^{2-1}=2ax$

The derivative of ax^2 is 2ax

Report an Error

Example Question #63 : Definition Of Derivative

Find the function represented by the following limit,

$\lim_{ h->0}\frac{\ln[(y+h)^2]-\ln(h^2)}{h}$

Possible Answers:

y^2 -1

$\frac{2}{y}$

$\frac{1}{y^2}+1$

Constant

$\ln(y+4)$

Correct answer:

$\frac{2}{y}$

Explanation:

$\lim_{ h->0}\frac{\ln[(y+h)^2]-\ln(h^2)}{h}$ (1)

Recall the first principles definition of a derivative,

$f'(y) = \lim_{h->0}\frac{f(y+h)-f(h)}{h}$ (2)

The limit in Equation (1) can be determined by inspection if we compare to the general definition of a derivative, Equation (2). It's apparent that equation (1) is the definition of the derivative for the function $\ln(y^2)$ . So the limit can be found by simply differentiating the function $\ln(y^2)$ with respect to .

$\frac{d}{dy}(\ln(y^2))=(2y)\frac{1}{y^2}=\frac{2}{y}$

$\lim_{ h->0}\frac{\ln[(y+h)^2]-\ln(h^2)}{h}=\frac{2}{y}$

Report an Error

Example Question #61 : Definition Of Derivative

Find the derivative of the following function:

$f(x)=\frac{cosx}{sinx}$ .

Possible Answers:

sec^2x

secxtanx

-csc^2x

csc^2x

-sec^2x

Correct answer:

-csc^2x

Explanation:

The function in the problem is simply cotx ! Therefore, by knowing the derivative of cotx , we know the solution is -csc^2x . You would obtain the same results if you completed this problem using the quotient rule and simplifying.

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 #63 : Derivative Review

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-8

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

y = 2 x

y = -8 x-20

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

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

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 Los Angeles, LSAT Tutors in Seattle, GRE Tutors in San Francisco-Bay Area, Chemistry Tutors in Boston, Spanish Tutors in Philadelphia, Calculus Tutors in Houston, Biology Tutors in San Diego, LSAT Tutors in Miami, SSAT Tutors in Denver, ACT Tutors in Atlanta

Popular Courses & Classes

MCAT Courses & Classes in Dallas Fort Worth, MCAT Courses & Classes in San Diego, SSAT Courses & Classes in Philadelphia, LSAT Courses & Classes in Atlanta, ISEE Courses & Classes in Phoenix, LSAT Courses & Classes in Los Angeles, ACT Courses & Classes in Chicago, MCAT Courses & Classes in San Francisco-Bay Area, MCAT Courses & Classes in Phoenix, SSAT Courses & Classes in Phoenix

Popular Test Prep

GMAT Test Prep in Dallas Fort Worth, ACT Test Prep in Chicago, ACT Test Prep in San Diego, ACT Test Prep in Denver, SSAT Test Prep in Washington DC, SAT Test Prep in Phoenix, GRE Test Prep in Washington DC, LSAT Test Prep in Los Angeles, GRE Test Prep in Phoenix, SSAT Test Prep in Philadelphia

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 #61 : Derivatives

Example Question #61 : Derivative Review

Example Question #62 : Derivative Review

Example Question #63 : Definition Of Derivative

Example Question #61 : Definition Of Derivative

Example Question #63 : Derivatives

Example Question #43 : Derivatives

Example Question #2 : Derivative At A Point

Example Question #63 : Derivative Review

Example Question #1194 : Calculus Ii

All Calculus 2 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: