We are open Saturday and Sunday!

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 #1061 : Partial Derivatives

Find $f_{xy}$ of the function $f(x,y,z)=3y\sin(x)+8x^2yz^3$

Possible Answers:

$5\cos(x)+16x^2z^4$

$3\cos(x)+16xz^3$

$3\sin(x)+16xz^3$

$3\cos(x)+14xz^3$

Correct answer:

$3\cos(x)+16xz^3$

Explanation:

To find $f_{xy}$ of the function, you take two consecutive partial derivatives:

$\frac{\partial }{\partial x}(3y\sin(x)+8x^2yz^3)=3y\cos(x)+16xyz^3$

$\frac{\partial }{\partial y}(3y\cos(x)+16xyz^3)=3\cos(x)+16xz^3$

Report an Error

Example Question #1062 : Partial Derivatives

Find $f_{xy}$ of the function $f(x,y,z)=e^{xy}+7xz^3$

Possible Answers:

$e^{xy}+xye^{xy}$

$e^{xy}+ye^{xy}$

$e^{y}+xye^{x}$

$e^{xy}+xe^{xy}$

Correct answer:

$e^{xy}+xye^{xy}$

Explanation:

To find $f_{xy}$ of the function, you take two consecutive partial derivatives:

$\frac{\partial }{\partial x}(e^{xy}+7xz^3)=ye^{xy}+7z^3$

$\frac{\partial }{\partial y}(ye^{xy}+7z^3)=e^{xy}+xye^{xy}$

Report an Error

Example Question #1063 : Partial Derivatives

Find the derivative of the function with respect to by taking the natural logarithm of both sides and then differentiating both sides with respect to .

$z=\frac{xy+2xy}{4y\sqrt{y^2+2x}}$

Possible Answers:

$\frac{\partial z}{\partial y}=\frac{xy+2xy}{4y\sqrt{2x}} \left(\frac{x^2}{x^3y+xy} +\frac{5}{5y+x} \right )$

$\frac{\partial z}{\partial y}=\frac{xy+2xy}{4y\sqrt{2x}} \left(\frac{x+2x}{xy+2xy} +\frac{4}{x}+\frac{4x+1}{2x^2+x} \right )$

$\frac{\partial z}{\partial y}=\frac{xy+2xy}{4y\sqrt{2x}} \left(\frac{3y}{xy+2xy} +\frac{4y+1}{2y^2+y} \right )$

$\frac{\partial z}{\partial y}=\frac{xy+2xy}{4y\sqrt{y^2+2x}} \left(\frac{x+2x}{xy+2xy} -\frac{4}{y}-\frac{2y}{2y^2+4x} \right )$

$\frac{\partial z}{\partial y}=\frac{xy+2xy}{4y\sqrt{2x}} \left(\frac{x}{x^3y+x} +\frac{5y}{5y^2+x} \right )$

Correct answer:

$\frac{\partial z}{\partial y}=\frac{xy+2xy}{4y\sqrt{y^2+2x}} \left(\frac{x+2x}{xy+2xy} -\frac{4}{y}-\frac{2y}{2y^2+4x} \right )$

Explanation:

$z=f(x,y)=\frac{xy+2xy}{4y\sqrt{y^2+2x}}$

Take the natural logarithm of both sides and expand the right-side using the properties of logarithms.

$\ln(z)=\ln\left(\frac{xy+2xy}{4y\sqrt{y^2+2x}}\right)$

Apply the rule for the logarithm of a quotient:

$\ln(z)=\ln\left(xy+2xy\right)-\ln\left(4y\sqrt{y^2+2x} \right)$

Apply the rule for the logarithm of a product:

$\ln(z)=\ln\left(xy+2xy\right)-\left[\ln(4y)+\ln\left(\sqrt{y^2+2x}\right) \right]$

Apply the rule for the logarithm of a quantity raised to a power:

$\ln(z)=\ln\left(xy+2xy\right)-\ln(4y)-\frac{1}{2}\ln\left(y^2+2x\right)$

Now differentiate both sides implicitly, remembering that is a multi-variable function of and .

$\frac{1}{z}\frac{\partial z}{\partial y}=\frac{\partial}{\partial y}\left[\ln\left(xy+2xy\right)-\ln(4y)-\frac{1}{2}\ln\left(y^2+2x\right)\right]$

Proceeding with the differentiation on the right-side with respect to .

$\frac{1}{z}\frac{\partial z}{\partial y}=\frac{x+2x}{xy+2xy} -\frac{4}{y}-\frac{2y}{2y^2+4x}$

$\frac{\partial z}{\partial y}=\frac{xy+2xy}{4y\sqrt{y^2+2x}} \left[\frac{x+2x}{xy+2xy} -\frac{4}{y}-\frac{2y}{2y^2+4x} \right ]$

Report an Error

Example Question #1064 : Partial Derivatives

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

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

Possible Answers:

$\frac{\partial f}{dy}=x+\frac{z}{\ln(yz)}-\frac{y}{\ln^2{(yz)}}+ye^{xy}$

$\frac{\partial f}{dy}=xz^{10}+\frac{x}{\ln(yz)}-\frac{x}{\ln^2{(yz)}}+xe^{xy}$

$\frac{\partial f}{dy}=\frac{x}{\ln(yz)}$

$\frac{\partial f}{dy}=xz^{10}$

$\frac{\partial f}{dy}=z^{10}+\frac{1}{\ln(yz)}-\frac{y}{\ln^2{(yz)}}+e^{xy}$

Correct answer:

$\frac{\partial f}{dy}=xz^{10}+\frac{x}{\ln(yz)}-\frac{x}{\ln^2{(yz)}}+xe^{xy}$

Explanation:

In order to find $\frac{\partial f}{dy}$ , 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

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

$\frac{\partial f}{dy}=xz^{10}+x(\ln(yz))^{-1}-xy(\ln(yz))^{-2}\cdot\frac{z}{yz}+0+xe^{xy}$

$\frac{\partial f}{dy}=xz^{10}+\frac{x}{\ln(yz)}-\frac{x}{\ln^2{(yz)}}+xe^{xy}$

Report an Error

Example Question #1065 : 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 #1061 : Partial Derivatives

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

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

Possible Answers:

$\frac{\partial f}{dy}=x+\frac{z}{\ln(yz)}-\frac{y}{\ln^2{(yz)}}+ye^{xy}$

$\frac{\partial f}{dy}=xz^{10}+\frac{x}{\ln(yz)}-\frac{x}{\ln^2{(yz)}}+xe^{xy}$

$\frac{\partial f}{dy}=z^{10}+\frac{1}{\ln(yz)}-\frac{y}{\ln^2{(yz)}}+e^{xy}$

$\frac{\partial f}{dy}=xz^{10}$

$\frac{\partial f}{dy}=\frac{x}{\ln(yz)}$

Correct answer:

$\frac{\partial f}{dy}=xz^{10}+\frac{x}{\ln(yz)}-\frac{x}{\ln^2{(yz)}}+xe^{xy}$

Explanation:

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

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

$\frac{\partial f}{dy}=xz^{10}+x(\ln(yz))^{-1}-xy(\ln(yz))^{-2}\cdot\frac{z}{yz}+0+xe^{xy}$

$\frac{\partial f}{dy}=xz^{10}+\frac{x}{\ln(yz)}-\frac{x}{\ln^2{(yz)}}+xe^{xy}$

Report an Error

Example Question #1062 : Partial Derivatives

True or False

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

Possible Answers:

True

False

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 #1 : Directional Derivatives

Calculate $D_{\vec{u}}f(x,y,z)$ , where $f(x,y,z)=x^2y+xz^{10}+xy^2z^5$ in the direction of $\vec{v}=(1,2,3)$ .

Possible Answers:

$\\ \\ D_{\vec{u}}f(x,y,z)= \\ \\ ((2xy+z^{10}+y^2z^5)+3(10xz^9+5xy^2z^4))$

$\\ \\ D_{\vec{u}}f(x,y,z)= \\ \\ \frac{1}{\sqrt{14}}((2xy+z^{10}+y^2z^5)+2(x^2+2xyz^5)+3(10xz^9+5xy^2z^4))$

$\\ \\ D_{\vec{u}}f(x,y,z)= \\ \\ ((2xy+z^{10}+y^2z^5)+2(x^2+2xyz^5)+3(10xz^9+5xy^2z^4))$

$\\ \\ D_{\vec{u}}f(x,y,z)= \\ \\ \frac{1}{\sqrt{14}}((2xy+z^{10}+y^2z^5)+3(10xz^9+5xy^2z^4))$

$\\ \\ D_{\vec{u}}f(x,y,z)= 2(x^2+2xyz^5)+3(10xz^9+5xy^2z^4)$

Correct answer:

$\\ \\ D_{\vec{u}}f(x,y,z)= \\ \\ \frac{1}{\sqrt{14}}((2xy+z^{10}+y^2z^5)+2(x^2+2xyz^5)+3(10xz^9+5xy^2z^4))$

Explanation:

The first thing to check is to see if the direction vector is a unit vector.

In order to see if it is a unit vector, we need to take the magnitude and see if it is equal to .

$||\vec{v}||=\sqrt{1^2+2^2+3^2}=\sqrt{1+4+9}=\sqrt{14}\neq1$

$\vec{u}=\frac{1}{\sqrt{14}}\vec{v}=\Big(\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}\Big)$

Now we are going to take partial derivatives in respect to , , and then , and then multiply each partial by the component of the unit vector that corresponds to it.

The formula is:

$D_{\vec{u}}f(x,y,z)=f_x(x,y,z)a+f_y(x,y,z)b+f_z(x,y,z)c$

$\\ \\ D_{\vec{u}}f(x,y,z)= \\ \\ \frac{1}{\sqrt{14}}(2xy+z^{10}+y^2z^5)+\frac{2}{\sqrt{14}}(x^2+0+2xyz^5)+\frac{3}{\sqrt{14}}(0+10xz^9+5xy^2z^4)$

$\\ \\ D_{\vec{u}}f(x,y,z)= \\ \\ \frac{1}{\sqrt{14}}(2xy+z^{10}+y^2z^5)+\frac{2}{\sqrt{14}}(x^2+2xyz^5)+\frac{3}{\sqrt{14}}(10xz^9+5xy^2z^4)$

$\\ \\ D_{\vec{u}}f(x,y,z)= \\ \\ \frac{1}{\sqrt{14}}((2xy+z^{10}+y^2z^5)+2(x^2+2xyz^5)+3(10xz^9+5xy^2z^4))$

Report an Error

Example Question #2 : Directional Derivatives

Calculate $D_{\vec{u}}f(x,y,z)$ , where $f(x,y,z)=z^2y\ln(x)+xy^2z^3$ in the direction of $\vec{v}=(0,1,1)$ .

Possible Answers:

$\\ \\ D_{\vec{u}}f(x,y,z)= \\ \\ \frac{1}{\sqrt{2}}(2zy\ln(x)+3xy^2z^2)$

$\\ \\ D_{\vec{u}}f(x,y,z)= \\ \\ \frac{1}{\sqrt{2}}((z^2\ln(x)+2xyz^3)+(2zy\ln(x)+3xy^2z^2))$

$\\ \\ D_{\vec{u}}f(x,y,z)= \\ \\ ((z^2\ln(x)+2xyz^3)+(2zy\ln(x)+3xy^2z^2))$

$\\ \\ D_{\vec{u}}f(x,y,z)= \\ \\ (2zy\ln(x)+3xy^2z^2)$

$\\ \\ D_{\vec{u}}f(x,y,z)= \\ \\ (z^2\ln(x)+2xyz^3)$

Correct answer:

$\\ \\ D_{\vec{u}}f(x,y,z)= \\ \\ \frac{1}{\sqrt{2}}((z^2\ln(x)+2xyz^3)+(2zy\ln(x)+3xy^2z^2))$

Explanation:

The first thing to check is to see if the direction vector is a unit vector.

In order to see if it is a unit vector, we need to take the magnitude and see if it is equal to .

$||\vec{v}||=\sqrt{0^2+1^2+1^2}=\sqrt{0+1+1}=\sqrt{2}\neq1$

$\vec{u}=\frac{1}{\sqrt{2}}\vec{v}=\Big(0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\Big)$

Now we are going to take partial derivatives in respect to , , and then , and then multiply each partial by the component of the unit vector that corresponds to it.

The formula is:

$D_{\vec{u}}f(x,y,z)=f_x(x,y,z)a+f_y(x,y,z)b+f_z(x,y,z)c$

$\\ \\ D_{\vec{u}}f(x,y,z)= \\ \\ 0(\frac{yz^2}{x}+y^2z^3)+\frac{1}{\sqrt{2}}(z^2\ln(x)+2xyz^3)+\frac{1}{\sqrt{2}}(2zy\ln(x)+3xy^2z^2)$

$\\ \\ D_{\vec{u}}f(x,y,z)= \\ \\ \frac{1}{\sqrt{2}}(z^2\ln(x)+2xyz^3)+\frac{1}{\sqrt{2}}(2zy\ln(x)+3xy^2z^2)$

$\\ \\ D_{\vec{u}}f(x,y,z)= \\ \\ \frac{1}{\sqrt{2}}((z^2\ln(x)+2xyz^3)+(2zy\ln(x)+3xy^2z^2))$

Report an Error

Example Question #3 : Directional Derivatives

$\begin{align*}&\text{Find }D_{\overrightarrow{u}}(-5,1,4)\text{ where }f(x,y,z)=y^3cos{(x^2)}sin{(z)}\\&\text{In the direction of }\overrightarrow{v}=(-2,20,-14).\end{align*}$

Possible Answers:

23.13

-1.55

18.92

-66.53

Correct answer:

-1.55

Explanation:

$\begin{align*}&\text{To find the directional derivative }D_{\overrightarrow{u}}(-5,1,4)\\&\text{a good first step is to make sure that }\overrightarrow{v}\\&\text{is in unit vector form. First, find the magnitude:}\\&|\overrightarrow{v}|=\sqrt{x^2+y^2+z^2}=\sqrt{(-2)^2+(20)^2+(-14)^2}=24.49\\&\text{With this, unit vector components can be found:}\\&u_x=\frac{v_x}{|\overrightarrow{v}|}=\frac{-2}{24.49}=-0.08\\&u_y=\frac{v_y}{|\overrightarrow{v}|}=\frac{20}{24.49}=0.82\\&u_z=\frac{v_z}{|\overrightarrow{v}|}=\frac{-14}{24.49}=-0.57\\&\text{The directional derivative takes the form: }D_{\overrightarrow{u}}(x,y,z)=u_x\frac{df}{dx}+u_y\frac{df}{dy}+u_z\frac{df}{dz}\\&\text{Therefore, for our values:}\\&D_{\overrightarrow{u}}(-5,1,4)=(-0.08)(-2xy^3sin{(x^2)}sin{(z)})+(0.82)(3y^2cos{(x^2)}sin{(z)})+(-0.57)(y^3cos{(x^2)}cos{(z)})=-1.55\end{align*}$

Report an Error

View Calculus 3 Tutors

Dalton
Certified Tutor

University of Massachusetts Amherst, Doctor of Philosophy, Mathematics.

View Calculus 3 Tutors

Fathima
Certified Tutor

BITS Pilani Goa, Doctorate (PhD), Mathematics.

View Calculus 3 Tutors

Wysheka
Certified Tutor

Clemson University, Doctorate (PhD), Engineering and Science Education.

All Calculus 3 Resources

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

Popular Subjects

Chemistry Tutors in Denver, MCAT Tutors in Atlanta, French Tutors in Denver, French Tutors in Seattle, LSAT Tutors in San Francisco-Bay Area, Biology Tutors in Chicago, Physics Tutors in San Diego, Biology Tutors in New York City, MCAT Tutors in Dallas Fort Worth, Math Tutors in Chicago

Popular Courses & Classes

ACT Courses & Classes in San Diego, ISEE Courses & Classes in Seattle, SAT Courses & Classes in Atlanta, SSAT Courses & Classes in New York City, GRE Courses & Classes in Boston, Spanish Courses & Classes in San Diego, LSAT Courses & Classes in Washington DC, Spanish Courses & Classes in New York City, GRE Courses & Classes in Chicago, ISEE Courses & Classes in Dallas Fort Worth

Popular Test Prep

SAT Test Prep in Seattle, ISEE Test Prep in Boston, LSAT Test Prep in Boston, MCAT Test Prep in Seattle, ISEE Test Prep in Philadelphia, SAT Test Prep in Chicago, LSAT Test Prep in Philadelphia, SSAT Test Prep in Philadelphia, MCAT Test Prep in New York City, GRE Test Prep in Chicago

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 #1061 : Partial Derivatives

Example Question #1062 : Partial Derivatives

Example Question #1063 : Partial Derivatives

Example Question #1064 : Partial Derivatives

Example Question #1065 : Partial Derivatives

Example Question #1061 : Partial Derivatives

Example Question #1062 : Partial Derivatives

Example Question #1 : Directional Derivatives

Example Question #2 : Directional Derivatives

Example Question #3 : Directional Derivatives

All Calculus 3 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: