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 Create Tests & Flashcards

All Calculus 2 Resources

9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

← Previous 1 2 … 5 6 7 8 9 10 11 12 13 … 51 52 Next →

Example Question #81 :Derivatives

Let the initial approximation of a solution of the equation

$x + 2 \ln x = 7$

be $x_{1} = 4$ .

使用一个牛顿法的迭代找到approximation to $x_{2}$ . Give your answer to the nearest thousandth.

Possible Answers:

3.952

4.152

4.102

4.002

4.052

Correct answer:

4.152

Explanation:

Rewrite the equation to be solved foras

$x + 2 \ln x - 7 = 0$ .

因此我们试图鳍d a zero of the function

$f(x) = x + 2 \ln x - 7$ .

$f'(x) = \frac{\mathrm{d} }{\mathrm{d} x}\left ( x + 2 \ln x - 7 \right ) = 1 + 2\cdot \frac{1}{x} = 1 + \frac{2}{x}$

Using Newton's method, we can find $x_{2}$ from the formula

$x_{2} = x_{1} - \frac{f(x_{1})}{f'(x_{1})}$ .

$f(x_{1}) = f (4) = 4 + 2 \ln4 - 7$

$= - 3 + \ln 16$

$\approx -3 + 2.7726 \approx -0.2274$

$f'(x_{1}) = f (4) = 1 + \frac{2}{4} = 1.5$

$x_{2} \approx 4 - \frac{-0.2274}{1.5} \approx 4 + 0.1516 \approx 4.152$

Report an Error

Example Question #82 :Derivatives

Let the initial approximation of a solution of the equation

$x = \cos 2x$

be $x_{1}= 0.6$ .

使用一个牛顿法的迭代找到approximation for $x_{2}$ . Give your answer to the nearest thousandth.

Possible Answers:

0.643

0.603

0.517

0.557

0.683

Correct answer:

0.517

Explanation:

Rewrite the equation to be solved foras

$x - \cos 2x = 0$ .

因此我们试图鳍d a zero of the function

$f(x) = x - \cos 2x$ .

$f'(x) =\frac{\mathrm{d} }{\mathrm{d} x}\left ( x - \cos 2x \right ) = 1 + 2 \sin 2x$ .

Using Newton's method, we can find $x_{2}$ from the formula

$x_{2} = x_{1} - \frac{f(x_{1})}{f'(x_{1})}$ .

$f(x_{1}) = f (0.6) =0.6 - \cos \left (2 \cdot 0.6 \right )$

$=0.6 - \cos 1.2$

$\approx 0.6 - 0.3624 \approx 0.2376$

$f'(x_{1}) = f'(0.6)= 1 + 2 \sin \left (2 \cdot 0.6 \right )$

$f'(x_{1}) = f'(0.6)= 1 + 2 \sin1.2$

$= 1 + 2 \cdot 0.9320$

$\approx 1 + 2 \cdot 0.9320 \approx 2.8641$

$x_{2} \approx 0.6 - \frac{0.2376}{2.8641} \approx 0.6 - 0.0830 \approx 0.517$

Report an Error

Example Question #1 :Derivative At A Point

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 $\frac{1}{x\:ln(a)}$ .

因此导数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 #2 :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:

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

$-\frac{1}{16}$

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

$\frac{1}{4}$

$\frac{1}{16}$

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 #3 :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 #4 :Derivative At A Point

Find the value of the derivate of the given function at the point (2,1) :

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

Possible Answers:

$\frac{3}{5}$

$\frac{1}2{}$

Correct answer:

$\frac{1}2{}$

Explanation:

To solve this problem, first we need to take the derivative of the function. To do this we need to use the quotient rule and simplified as follows:

$\\f'(x)=\frac{(x+4)\cdot \frac{\mathrm{d} }{\mathrm{d} x}(x^2+2)-(x^2+2)\cdot \frac{\mathrm{d} }{\mathrm{d} x}(x+4)}{(x+4)^2}\\ f'(x)=\frac{(x+4)\cdot 2x-(x^2+2)\cdot 1}{(x+4)^2}\\f'(x)=\frac{2x^2+8x-x^2-2}{(x+4)^2}\\f'(x)=\frac{x^2+8x-2}{(x+4)^2}$

From here we need to evaluate at the given point (2,1) . In this case, only the x value is important, so we evaluate our derivative at x=2 to get

$\\f'(2)=\frac{(2)^2+8(2)-2}{(2+4)^2}\\ f'(2)=\frac{4+16-2}{36}\\f'(2)=\frac{18}{36}=\frac{1}{2}$ .

Report an Error

Example Question #5 :Derivative At A Point

Find the slope at (1,0) for the function y=-ln(x) .

Possible Answers:

$\infty$

$-\infty$

Correct answer:

Explanation:

The derivative of y=-ln(x) is:

$y'=-\frac{1}{x}$

Substitute the point at x=1 .

$y'=-\frac{1}{x}=-\frac{1}{1}=-1$

Report an Error

Example Question #6 :Derivative At A Point

f(x)=ln(x)x^2

What is the slope of f(x) at x=1 ?

Possible Answers:

Correct answer:

Explanation:

In order to find the slope of a function at a certain point, plug in that point into the first derivative of the function. Our first step here is to take the first derivative.

Since we see that f(x) is composed of two different functions, we must use the product rule. Remember that the product rule goes as follows:

f(x)=g(x)h(x)

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

Following that procedure, we set ln(x) equal to g(x) and x^2 equal to h(x) .

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

which can be simplified to

f'(x)=x+2xln(x) = x(1+2ln(x)) .

Now plug in 1 to find the slope at x=1.

f'(1)= 1(1+2(0))=1

Remember that ln(1)=0 .

Report an Error

Example Question #7 :Derivative At A Point

Consider the function: 2y^3=3x^2 . What is the derivative at (3,2) ?

Possible Answers:

$\frac{3}{4}$

$\frac{3}{2}$

$\frac{1}{4}$

$\frac{1}{2}$

$\frac{2}{3}$

Correct answer:

$\frac{3}{4}$

Explanation:

To solve for the derivative of 2y^3=3x^2 , use implicit differentiation which means to take the derivative of each term with respect to the variable in that term.

$6y^2 \cdot \frac{dy}{dx}=6x$

$\frac{dy}{dx}=\frac{6x}{6y^2}$

Substitute the point into the derivative.

$\frac{dy}{dx}=\frac{6x}{6y^2}=\frac{6(3)}{6(2)^2}=\frac{3}{4}$

Report an Error

Example Question #8 :Derivative At A Point

Use implicit differentiation to find the slope of the tangent line to $\small y^3-lny=x^2$ at the point $\small (1,1)$ .

Possible Answers:

$\small 0$

$\small 2$

$\small -1$

$\small 1$

$\small -2$

Correct answer:

$\small 1$

Explanation:

We must take the derivative $\small \left(\small \frac{\mathrm{d} y}{\mathrm{d} x}\right)$ because that will give us the slope. On the left side we'll get

$\small 3y^2\frac{\mathrm{d}y }{\mathrm{d} x}-(1/y)\frac{\mathrm{d} y}{\mathrm{d} x}$ , and on the right side we'll get $\small 2x$ .

We include the $\small \frac{\mathrm{d} y}{\mathrm{d} x}$ on the left side because $\small y$ is a function of $\small x$ ,所以它的导数is unknown (hence we are trying to solve for it!).

Now we can factor out a $\small \frac{\mathrm{d} y}{\mathrm{d} x}$ on the left side to get

$\small \small \frac{\mathrm{d} y}{\mathrm{d} x}(3y^2-1/y)=2x$ and divide by $\small 3y^2-1/y$ in order to solve for $\small \frac{\mathrm{d} y}{\mathrm{d} x}$ .

Doing this gives you

$\small \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2x}{3y^2-1/y}$ .

We want to find the slope at $\small (1,1)$ , so we can sub in $\small 1$ for $\small x$ and $\small y$ .

$\small \small \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2(1)}{3(1)^2-1/(1)}=2/2=1$ .

Report an Error

← Previous 1 2 … 5 6 7 8 9 10 11 12 13 … 51 52 Next →

View Calculus 2 Tutors

Ping
Certified Tutor

The University of Texas at Dallas, Bachelor of Science, Computer Science. University of North Texas, Master of Arts, Education.

View Calculus 2 Tutors

Mohammed
Certified Tutor

South Carolina State University, Bachelor of Science, Mathematics. South Carolina State University, Masters in Education, Mat...

View Calculus 2 Tutors

Francisco
Certified Tutor

University of Texas at Brownsville, Bachelors, Mathematics. University of Texas at Brownsville, Masters, Mathematics.

All Calculus 2 Resources

9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

ISEE Tutors in San Diego,Physics Tutors in San Diego,Math Tutors in San Francisco-Bay Area,Math Tutors in Chicago,French Tutors in Houston,English Tutors in Boston,Calculus Tutors in Phoenix,Spanish Tutors in Atlanta,Biology Tutors in Chicago,Physics Tutors in Miami

Popular Courses & Classes

GRE Courses & Classes in Seattle,SSAT Courses & Classes in Boston,Spanish Courses & Classes in New York City,ACT Courses & Classes in Philadelphia,LSAT Courses & Classes in Denver,GRE Courses & Classes in Atlanta,MCAT Courses & Classes in Houston,Spanish Courses & Classes in Washington DC,GMAT Courses & Classes in Dallas Fort Worth,Spanish Courses & Classes in Miami

Popular Test Prep

SSAT Test Prep in Dallas Fort Worth,SAT Test Prep in Boston,ACT Test Prep in Seattle,GRE Test Prep in New York City,ACT Test Prep in Washington DC,LSAT Test Prep in San Francisco-Bay Area,GMAT Test Prep in Atlanta,SSAT Test Prep in Seattle,ACT Test Prep in Denver,MCAT Test Prep in Washington DC

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 #81 :Derivatives

Example Question #82 :Derivatives

Example Question #1 :Derivative At A Point

Example Question #2 :Derivative At A Point

Example Question #3 :Derivative At A Point

Example Question #4 :Derivative At A Point

Example Question #5 :Derivative At A Point

Example Question #6 :Derivative At A Point

Example Question #7 :Derivative At A Point

Example Question #8 :Derivative At A Point

All Calculus 2 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Find the Best Tutors

Email address:
Your name:
Feedback: