Call Now to Set Up Tutoring:

(888) 888-0446

Calculus 3 : Partial Derivatives

Study concepts, example questions & explanations for Calculus 3

Create An Account

All Calculus 3 Resources

6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #391 : Partial Derivatives

Find $\frac{\partial f}{dx}$ .

$f(x,y,z)=xyz^{10}+\frac{xy}{\ln(yz)}+10e^z+e^{xy}$

Possible Answers:

$\frac{\partial f}{dx}=yz^{10}+\frac{y}{\ln(yz)}$

$\frac{\partial f}{dx}=xyz^{10}+\frac{xy}{\ln(yz)}+ye^{xy}$

$\frac{\partial f}{dx}=yz^{10}+\frac{y}{\ln(yz)}+ye^{xy}$

$\frac{\partial f}{dx}=xz^{10}+\frac{1}{\ln(yz)}+xe^{xy}$

$\frac{\partial f}{dx}=\frac{y}{\ln(yz)}+ye^{xy}$

Correct answer:

$\frac{\partial f}{dx}=yz^{10}+\frac{y}{\ln(yz)}+ye^{xy}$

Explanation:

In order to find $\frac{\partial f}{dx}$ , we need to take the derivative of in respect to , and treat , and as constants. We also need to remember what the derivatives of natural log, exponential functions and power functions are for single variables.

Natural Log:

$f(x)=\ln(g(x))$

$f'(x)=\frac{g'(x)}{g(x)}$

Exponential Functions:

$f(x)=e^{g(x)}$

$f'(x)=g'(x)e^{g(x)}$

Power Functions:

f(x)=x^n

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

$\frac{\partial f}{dx}=yz^{10}+\frac{y}{\ln(yz)}+0+ye^{xy}$

$\frac{\partial f}{dx}=yz^{10}+\frac{y}{\ln(yz)}+ye^{xy}$

Report an Error

Example Question #392 : Partial Derivatives

True or False

$\frac{\partial}{\partial y}\Big(x^x+wz\ln(uvxw)\Big)=0$

Possible Answers:

False

True

Correct answer:

True

Explanation:

True:

Since there are no y's in the equation, the derivative of a constant is .

Report an Error

Example Question #2761 : Calculus 3

What is the partial derivative f_x(x,y) of the function

$f(x,y)=e^{xy}$ ?

Possible Answers:

$f_x(x,y)=xe^{xy}$

$f_x(x,y)=e^{xy}$

$f_x(x,y)=xye^{xy}$

$f_x(x,y)=ye^{xy}$

Correct answer:

$f_x(x,y)=ye^{xy}$

Explanation:

We can find f_x(x,y) given $f(x,y)=e^{xy}$ by differentiating the function while holding constant, i.e. we treat as a number. So we get

$f_x(x,y)=\frac{\partial}{\partial x}e^{xy}=e^{xy}\frac{\partial}{\partial x}xy=e^{xy}\cdot y=ye^{xy}$

Report an Error

Example Question #2762 : Calculus 3

Find the partial derivative f_y(x,y) for the function f(x,y)=2xy+2 .

Possible Answers:

f_y(x,y)=0

f_y(x,y)=2xy

f_y(x,y)=2y

f_y(x,y)=2x

Correct answer:

f_y(x,y)=2x

Explanation:

We can find f_y(x,y) from the function f(x,y)=2xy+2 by taking the derivative holding constant and letting be the variable. This means

$f_y(x,y)=\frac{\partial}{\partial y}(2xy+2)=2x$

Report an Error

Example Question #393 : Partial Derivatives

Given the function $f(x,y)=\sin x^2y$ , find the partial derivative f_x(x,y) .

Possible Answers:

$f_x(x,y)=x^2y\cos x^2y$

$f_x(x,y)=2xy\cos x^2y$

$f_x(x,y)=2x\cos x^2y$

$f_x(x,y)=y\cos x^2y$

Correct answer:

$f_x(x,y)=2xy\cos x^2y$

Explanation:

To find the partial derivative f_x(x,y) of the function $f(x,y)=\sin x^2y$ , we calculate the derivative of f(x,y) while treating as a constant. So we get

$f_x(x,y)=\frac{\partial}{\partial x}f(x,y)=\frac{\partial}{\partial x}\sin x^2y=(\cos x^2y)\cdot \frac{\partial}{\partial x}x^2y$

$=2xy\cos x^2y$

Report an Error

Example Question #391 : Partial Derivatives

Given the function $f(x,y)=\sin x^2y$ , find the partial derivative f_y(x,y) .

Possible Answers:

$f_y(x,y)=x^2\cos x^2y$

$f_y(x,y)=2x\cos x^2y$

$f_y(x,y)=2xy\cos x^2y$

$f_y(x,y)=y\cos x^2y$

$f_y(x,y)=2xy\cos x^2y$

$f_y(x,y)=y\cos x^2y$

Correct answer:

$f_y(x,y)=x^2\cos x^2y$

Explanation:

To find the partial derivative f_y(x,y) of the function $f(x,y)=\sin x^2y$ , we calculate the derivative of f(x,y) while treating as a constant. So we get

$f_y(x,y)=\frac{\partial}{\partial y}f(x,y)=\frac{\partial}{\partial y}\sin x^2y=(\cos x^2y)\cdot \frac{\partial}{\partial y}x^2y$

$=x^2\cos x^2y$

Report an Error

Example Question #2763 : Calculus 3

Given the function $f(x,y)=\ln (1+x^2+y^2)$ , find the partial derivative f_x(x,y) .

Possible Answers:

$f_x(x,y)=\frac{2x+2y}{1+x^2+y^2}$

f_x(x,y)=0

$f_x(x,y)=\frac{2y}{1+x^2+y^2}$

$f_x(x,y)=\frac{2x}{1+x^2+y^2}$

Correct answer:

$f_x(x,y)=\frac{2x}{1+x^2+y^2}$

Explanation:

To find the partial derivative f_x(x,y) of the function $f(x,y)=\ln (1+x^2+y^2)$ , we differentiate the function with respect to , which means we hold constant. Then we get (using the chain rule):

$f_x(x,y)=\frac{\partial}{\partial x}f(x,y)=\frac{\partial}{\partial x}\ln (1+x^2+y^2)=\frac{1}{1+x^2+y^2}\cdot \frac{\partial}{\partial x}(1+x^2+y^2)$

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

Report an Error

Example Question #2761 : Calculus 3

Given the function $f(x,y)=\ln (2+x^2+y^2)$ , find the partial derivative f_y(x,y) .

Possible Answers:

f_y(x,y)=0

$f_y(x,y)=\frac{2x}{2+x^2+y^2}$

$f_y(x,y)=\frac{2y}{1+x^2+y^2}$

$f_y(x,y)=\frac{2x+2y}{1+x^2+y^2}$

Correct answer:

$f_y(x,y)=\frac{2y}{1+x^2+y^2}$

Explanation:

To find the partial derivative f_y(x,y) of the function $f(x,y)=\ln (2+x^2+y^2)$ , we differentiate the function with respect to , which means we hold constant. Then we get (using the chain rule):

$f_y(x,y)=\frac{\partial}{\partial y}f(x,y)=\frac{\partial}{\partial y}\ln (2+x^2+y^2)=\frac{1}{2+x^2+y^2}\cdot \frac{\partial}{\partial y}(2+x^2+y^2)$

$=\frac{2y}{2+x^2+y^2}$

Report an Error

Example Question #2764 : Calculus 3

Given the function $f(x,y)=e^{24xy+25x^2}$ , find the partial derivative

f_x(x,y)

Possible Answers:

$f_x(x,y)=24xe^{24xy+25x^2}\,$

$f_x(x,y)=(24y+50x)e^{24xy+25x^2}$

$f_x(x,y)=e^{24y+50x}$

$f_x(x,y)=(24y+50x)e^{24y+50x}$

Correct answer:

$f_x(x,y)=(24y+50x)e^{24xy+25x^2}$

Explanation:

To find the partial derivative f_x of $f(x,y)=e^{24xy+25x^2}$ , we need to differentiate f(x,y) with respect to while holding constant. We can use the chain rule to get

$f_x(x,y)=\frac{\partial}{\partial x}e^{24xy+25x^2}=e^{24xy+25x^2}\cdot \frac{\partial}{\partial x}(24xy+25x^2)$

$=(24y+50x)e^{24xy+25x^2}$

Report an Error

Example Question #397 : Partial Derivatives

Given the function $f(x,y)=e^{24xy+25x^2}$ , find the partial derivative

f_y(x,y)

Possible Answers:

$f_y(x,y)=(24y+50x)e^{24y+50x}$

$f_y(x,y)=e^{24y+50x}$

$f_y(x,y)=(24y+50x)e^{24xy+25x^2}\,$

$f_y(x,y)=24xe^{24xy+25x^2}$

Correct answer:

$f_y(x,y)=24xe^{24xy+25x^2}$

Explanation:

To find the partial derivative f_y of $f(x,y)=e^{24xy+25x^2}$ , we need to differentiate f(x,y) with respect to while holding constant. We can use the chain rule to get

$f_y(x,y)=\frac{\partial}{\partial y}e^{24xy+25x^2}=e^{24xy+25x^2}\cdot \frac{\partial}{\partial y}(24xy+25x^2)$

$=24xe^{24xy+25x^2}$

Report an Error

View Calculus 3 Tutors

Jan
Certified Tutor

University of Connecticut, Bachelor of Science, Computer Science.

View Calculus 3 Tutors

Philipp
Certified Tutor

University of Nevada-Las Vegas, Bachelor of Science, Mechanical Engineering.

View Calculus 3 Tutors

Omar
Certified Tutor

Arizona State University, Bachelors, Civil Engineering.

All Calculus 3 Resources

6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Popular Subjects

Computer Science Tutors in Dallas Fort Worth, Statistics Tutors in Seattle, GRE Tutors in San Francisco-Bay Area, Reading Tutors in Los Angeles, Spanish Tutors in Atlanta, GRE Tutors in Seattle, Biology Tutors in Houston, LSAT Tutors in San Francisco-Bay Area, LSAT Tutors in Dallas Fort Worth, MCAT Tutors in Washington DC

Popular Courses & Classes

ACT Courses & Classes in Miami, SAT Courses & Classes in San Francisco-Bay Area, MCAT Courses & Classes in San Francisco-Bay Area, Spanish Courses & Classes in Dallas Fort Worth, SSAT Courses & Classes in San Francisco-Bay Area, GMAT Courses & Classes in Dallas Fort Worth, ISEE Courses & Classes in San Francisco-Bay Area, SAT Courses & Classes in Dallas Fort Worth, GRE Courses & Classes in New York City, SAT Courses & Classes in Boston

Popular Test Prep

MCAT Test Prep in Miami, MCAT Test Prep in San Francisco-Bay Area, GMAT Test Prep in New York City, SAT Test Prep in Washington DC, ISEE Test Prep in Phoenix, GMAT Test Prep in San Diego, GRE Test Prep in Boston, LSAT Test Prep in Boston, SAT Test Prep in Boston, GMAT Test Prep in Dallas Fort Worth

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 3 : Partial Derivatives

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #391 : Partial Derivatives

Example Question #392 : Partial Derivatives

Example Question #2761 : Calculus 3

Example Question #2762 : Calculus 3

Example Question #393 : Partial Derivatives

Example Question #391 : Partial Derivatives

Example Question #2763 : Calculus 3

Example Question #2761 : Calculus 3

Example Question #2764 : Calculus 3

Example Question #397 : Partial Derivatives

All Calculus 3 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: