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 #801 : Partial Derivatives

$\begin{align*}&\text{Perform the partial derivation, }f_{yx}\\&\text{Where }f(x,y,z)=\frac{(2x^{2}e^{(z^{2})})}{(9y^{2})}\end{align*}$

Possible Answers:

${-\frac{(16x^{3}e^{(2z^{2})})}{(81y^{5})}}$

${\frac{(32x^{3}e^{(2z^{2})})}{(81y^{6})}}$

${\frac{(4xe^{(z^{2})})}{(9y^{2})}-\frac{ (4x^{2}e^{(z^{2})})}{(9y^{3})}}$

${-\frac{(8xe^{(z^{2})})}{(9y^{3})}}$

Correct answer:

${-\frac{(8xe^{(z^{2})})}{(9y^{3})}}$

Explanation:

Report an Error

Example Question #802 : Partial Derivatives

$\begin{align*}&\text{Perform the partial derivation, }f_{zy}\\&\text{Where }f(x,y,z)=\frac{(ln(4y)e^{(4x)}tan(z^{2} - 12z))}{5}\end{align*}$

Possible Answers:

${\frac{(e^{(4x)}\cdot (2z - 12)\cdot (tan(z^{2} - 12z)^{2} + 1))}{(5y)}}$

${\frac{(e^{(4x)}tan(z^{2} - 12z))}{(5y)}+\frac{ (ln(4y)e^{(4x)}\cdot (2z - 12)\cdot (tan(z^{2} - 12z)^{2} + 1))}{5}}$

${\frac{(ln(4y)e^{(8x)}tan(z^{2} - 12z)\cdot (2z - 12)\cdot (tan(z^{2} - 12z)^{2} + 1))}{(25y)}}$

${\frac{(ln(4y)e^{(8x)}\cdot (2z - 12)^{2}\cdot (tan(z^{2} - 12z)^{2} + 1)^{2})}{(25y)}}$

Correct answer:

${\frac{(e^{(4x)}\cdot (2z - 12)\cdot (tan(z^{2} - 12z)^{2} + 1))}{(5y)}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&d[ln(u)]=\frac{du}{u}\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{(ln(4y)e^{(4x)}tan(z^{2} - 12z))}{5}\\&f_{z}=\frac{(ln(4y)e^{(4x)}\cdot (2z - 12)\cdot (tan(z^{2} - 12z)^{2} + 1))}{5}\\&f_{zy}=\frac{(e^{(4x)}\cdot (2z - 12)\cdot (tan(z^{2} - 12z)^{2} + 1))}{(5y)}\end{align*}$

Report an Error

Example Question #803 : Partial Derivatives

$\begin{align*}&\text{Perform the partial derivation, }f_{xy}\\&\text{Where }f(x,y,z)=\frac{(2\cdot 3^{(3z)}tan(y^{3})e^{(x^{2})})}{5}+ 4x^{4}ycos(z)\end{align*}$

Possible Answers:

${(4x^{4}cos(z) +\frac{ (6\cdot 3^{(3z)}y^{2}e^{(x^{2})}\cdot (tan(y^{3})^{2} + 1))}{5})\cdot (16x^{3}ycos(z) +\frac{ (4\cdot 3^{(3z)}xtan(y^{3})e^{(x^{2})})}{5})}$

${4x^{4}cos(z) + 16x^{3}ycos(z) +\frac{ (4\cdot 3^{(3z)}xtan(y^{3})e^{(x^{2})})}{5}+\frac{ (6\cdot 3^{(3z)}y^{2}e^{(x^{2})}\cdot (tan(y^{3})^{2} + 1))}{5}}$

${(16x^{3}ycos(z) +\frac{ (4\cdot 3^{(3z)}xtan(y^{3})e^{(x^{2})})}{5})\cdot (16x^{3}cos(z) +\frac{ (12\cdot 3^{(3z)}xy^{2}e^{(x^{2})}\cdot (tan(y^{3})^{2} + 1))}{5})}$

${16x^{3}cos(z) +\frac{ (12\cdot 3^{(3z)}xy^{2}e^{(x^{2})}\cdot (tan(y^{3})^{2} + 1))}{5}}$

Correct answer:

${16x^{3}cos(z) +\frac{ (12\cdot 3^{(3z)}xy^{2}e^{(x^{2})}\cdot (tan(y^{3})^{2} + 1))}{5}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=\frac{(2\cdot 3^{(3z)}tan(y^{3})e^{(x^{2})})}{5}+ 4x^{4}ycos(z)\\&f_{x}=16x^{3}ycos(z) +\frac{ (4\cdot 3^{(3z)}xtan(y^{3})e^{(x^{2})})}{5}\\&f_{xy}=16x^{3}cos(z) +\frac{ (12\cdot 3^{(3z)}xy^{2}e^{(x^{2})}\cdot (tan(y^{3})^{2} + 1))}{5}\end{align*}$

Report an Error

Example Question #803 : Partial Derivatives

$\begin{align*}&\text{Perform the partial derivation, }f_{yx}\\&\text{Where }f(x,y,z)=-\frac{ (2cos(x^{2})ln(4z)e^{(y)})}{5}-\frac{ (x^{2}sin(z))}{y^{2}}\end{align*}$

Possible Answers:

${\frac{(2x^{2}sin(z))}{y^{3}}-\frac{ (2cos(x^{2})ln(4z)e^{(y)})}{5}-\frac{ (2xsin(z))}{y^{2}}+\frac{ (4xln(4z)sin(x^{2})e^{(y)})}{5}}$

${-(\frac{(2cos(x^{2})ln(4z)e^{(y)})}{5}-\frac{ (2x^{2}sin(z))}{y^{3}})\cdot (\frac{(4xsin(z))}{y^{3}}+\frac{ (4xln(4z)sin(x^{2})e^{(y)})}{5})}$

${(\frac{(2cos(x^{2})ln(4z)e^{(y)})}{5}-\frac{ (2x^{2}sin(z))}{y^{3}})\cdot (\frac{(2xsin(z))}{y^{2}}-\frac{ (4xln(4z)sin(x^{2})e^{(y)})}{5})}$

${\frac{(4xsin(z))}{y^{3}}+\frac{ (4xln(4z)sin(x^{2})e^{(y)})}{5}}$

Correct answer:

${\frac{(4xsin(z))}{y^{3}}+\frac{ (4xln(4z)sin(x^{2})e^{(y)})}{5}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&d[cos(u)]=-sin(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=-\frac{ (2cos(x^{2})ln(4z)e^{(y)})}{5}-\frac{ (x^{2}sin(z))}{y^{2}}\\&f_{y}=\frac{(2x^{2}sin(z))}{y^{3}}-\frac{ (2cos(x^{2})ln(4z)e^{(y)})}{5}\\&f_{yx}=\frac{(4xsin(z))}{y^{3}}+\frac{ (4xln(4z)sin(x^{2})e^{(y)})}{5}\end{align*}$

Report an Error

Example Question #805 : Partial Derivatives

$\begin{align*}&\text{Perform the partial derivation, }f_{xy}\\&\text{Where }f(x,y,z)=-\frac{ 2}{(x^{3}z)}-\frac{ (2^{(4y)}ln(x^{2})cos(z))}{4}\end{align*}$

Possible Answers:

${-\frac{(2\cdot 2^{(4y)}ln(2)cos(z))}{x}}$

${\frac{6}{(x^{4}z)}-\frac{ (2^{(4y)}cos(z))}{(2x)}- 2^{(4y)}ln(x^{2})ln(2)cos(z)}$

${-\frac{(2\cdot 2^{(4y)}ln(2)cos(z)\cdot (\frac{6}{(x^{4}z)}-\frac{ (2^{(4y)}cos(z))}{(2x)}))}{x}}$

${-2^{(4y)}ln(x^{2})ln(2)cos(z)\cdot (\frac{6}{(x^{4}z)}-\frac{ (2^{(4y)}cos(z))}{(2x)})}$

Correct answer:

${-\frac{(2\cdot 2^{(4y)}ln(2)cos(z))}{x}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[ln(u)]=\frac{du}{u}\\&d[a^u]=a^uduln(a)\\&\text{Solve step-by-step:}\\&f(x,y,z)=-\frac{ 2}{(x^{3}z)}-\frac{ (2^{(4y)}ln(x^{2})cos(z))}{4}\\&f_{x}=\frac{6}{(x^{4}z)}-\frac{ (2^{(4y)}cos(z))}{(2x)}\\&f_{xy}=-\frac{(2\cdot 2^{(4y)}ln(2)cos(z))}{x}\end{align*}$

Report an Error

Example Question #806 : Partial Derivatives

$\begin{align*}&\text{Perform the partial derivation, }f_{zzz}\\&\text{Where }f(x,y,z)=-\frac{ (2^{(4y)}sin(x^{3} - 3x)sin(z))}{2}- 3x^{2}yz\end{align*}$

Possible Answers:

${- 9x^{2}y -\frac{ (3\cdot 2^{(4y)}sin(x^{3} - 3x)cos(z))}{2}}$

${-(3x^{2}y +\frac{ (2^{(4y)}sin(x^{3} - 3x)cos(z))}{2})^{3}}$

${\frac{(2^{(4y)}sin(x^{3} - 3x)cos(z))}{2}}$

${-\frac{(2^{(8y)}sin(x^{3} - 3x)^{2}cos(z)sin(z)\cdot (3x^{2}y +\frac{ (2^{(4y)}sin(x^{3} - 3x)cos(z))}{2}))}{4}}$

Correct answer:

${\frac{(2^{(4y)}sin(x^{3} - 3x)cos(z))}{2}}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[sin(u)]=cos(u)du\\&d[cos(u)]=-sin(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=-\frac{ (2^{(4y)}sin(x^{3} - 3x)sin(z))}{2}- 3x^{2}yz\\&f_{z}=- 3x^{2}y -\frac{ (2^{(4y)}sin(x^{3} - 3x)cos(z))}{2}\\&f_{zz}=\frac{(2^{(4y)}sin(x^{3} - 3x)sin(z))}{2}\\&f_{zzz}=\frac{(2^{(4y)}sin(x^{3} - 3x)cos(z))}{2}\end{align*}$

Report an Error

Example Question #807 : Partial Derivatives

$\begin{align*}&\text{Perform the partial derivation, }f_{yzy}\\&\text{Where }f(x,y,z)=3y^{2}sin(z^{3})sin(x) - 2\cdot 3^{(y^{2})}sin(x)sin(z)\end{align*}$

Possible Answers:

${-(6ysin(z^{3})sin(x) - 4\cdot 3^{(y^{2})}yln(3)sin(x)sin(z))^{2}\cdot (2\cdot 3^{(y^{2})}cos(z)sin(x) - 9y^{2}z^{2}cos(z^{3})sin(x))}$

${-(18yz^{2}cos(z^{3})sin(x) - 4\cdot 3^{(y^{2})}yln(3)cos(z)sin(x))\cdot (6ysin(z^{3})sin(x) - 4\cdot 3^{(y^{2})}yln(3)sin(x)sin(z))\cdot (4\cdot 3^{(y^{2})}ln(3)cos(z)sin(x) - 18z^{2}cos(z^{3})sin(x) + 8\cdot 3^{(y^{2})}y^{2}ln(3)^{2}cos(z)sin(x))}$

${12ysin(z^{3})sin(x) - 2\cdot 3^{(y^{2})}cos(z)sin(x) + 9y^{2}z^{2}cos(z^{3})sin(x) - 8\cdot 3^{(y^{2})}yln(3)sin(x)sin(z)}$

${18z^{2}cos(z^{3})sin(x) - 4\cdot 3^{(y^{2})}ln(3)cos(z)sin(x) - 8\cdot 3^{(y^{2})}y^{2}ln(3)^{2}cos(z)sin(x)}$

Correct answer:

${18z^{2}cos(z^{3})sin(x) - 4\cdot 3^{(y^{2})}ln(3)cos(z)sin(x) - 8\cdot 3^{(y^{2})}y^{2}ln(3)^{2}cos(z)sin(x)}$

Explanation:

$\begin{align*}&\text{When taking partial derivatives, order does not matter; you can}\\&\text{simply treat any variable that you're not deriving the equation for as constant.}\\&\text{Utilizing derivative rules:}\\&d[uv]=udv+vdu\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[a^u]=a^uduln(a)\\&d[sin(u)]=cos(u)du\\&\text{Solve step-by-step:}\\&f(x,y,z)=3y^{2}sin(z^{3})sin(x) - 2\cdot 3^{(y^{2})}sin(x)sin(z)\\&f_{y}=6ysin(z^{3})sin(x) - 4\cdot 3^{(y^{2})}yln(3)sin(x)sin(z)\\&f_{yz}=18yz^{2}cos(z^{3})sin(x) - 4\cdot 3^{(y^{2})}yln(3)cos(z)sin(x)\\&f_{yzy}=18z^{2}cos(z^{3})sin(x) - 4\cdot 3^{(y^{2})}ln(3)cos(z)sin(x) - 8\cdot 3^{(y^{2})}y^{2}ln(3)^{2}cos(z)sin(x)\end{align*}$

Report an Error

Example Question #808 : Partial Derivatives

Find $f_{xyz}$ of the following function:

$f(x, y, z)=z^3x^2y+\cos(z)xy^2+xyz$

Possible Answers:

$6xz^2-2y\sin(z)$

$2xz^3+2y\cos(z)+z$

$6xz^2+2y\sin(z)+1$

$6xz^2-2y\sin(z)+1$

Correct answer:

$6xz^2-2y\sin(z)+1$

Explanation:

To find the given partial derivative of the function, we must treat the other variable(s) as constants. For higher order partial derivatives, we work from left to right for the given variables.

To start, we must find the partial derivative of the function with respect to x:

$f_x=2xz^3y+\cos(z)y^2+yz$

The derivative was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

Next, we find the partial derivative of the above function with respect to y:

$f_{xy}=2xz^3+2y\cos(z)+z$

We used rules stated above.

Finally, we find the derivative of the above function with respect to z:

$f_{xyz}=6xz^2-2y\sin(z)+1$

We used all of the rules used previously along with

$\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$

Report an Error

Example Question #809 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(\frac{(e^{(4y)}e^{(x)}cos(z^{2} - 3z))}{2})\widehat{i}+(\frac{(4sin(y^{2}))}{z})\widehat{j}+(5e^{(x^{2})}e^{(z)}sin(y))\widehat{k}\\&\text{at the point }(0.98,1.11,2.39)\end{align*}$

Possible Answers:

-70.83

849.95

-283.32

141.66

Correct answer:

141.66

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(\frac{(e^{(4y)}e^{(x)}cos(z^{2} - 3z))}{2})\widehat{i}+(\frac{(4sin(y^{2}))}{z})\widehat{j}+(5e^{(x^{2})}e^{(z)}sin(y))\widehat{k}\\&\text{at the point }(0.98,1.11,2.39)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&d[e^u]=e^udu\\&d[sin(u)]=cos(u)du\\&divF(x,y,z)=(\frac{(e^{(4y)}e^{(x)}cos(z^{2} - 3z))}{2})+(\frac{(8ycos(y^{2}))}{z})+(5e^{(x^{2})}e^{(z)}sin(y))\\&divF(0.98,1.11,2.39)=(12.72)+(1.23)+(127.7)=141.66\end{align*}$

Report an Error

Example Question #810 : Partial Derivatives

$\begin{align*}&\text{Calculate the divergence of the formula:}\\&F(x,y,z)=(\frac{(5e^{(y^{2})}e^{(4z)})}{x})\widehat{i}+(\frac{(3ln(x^{2})sin(z^{2}))}{y^{3}})\widehat{j}+(2x^{3}ze^{(y^{4})})\widehat{k}\\&\text{at the point }(1.57,0.68,1.71)\end{align*}$

Possible Answers:

-3008.58

501.43

-1504.29

3008.58

Correct answer:

-3008.58

Explanation:

$\begin{align*}&\text{The divergence of a formula is determined in terms}\\&\text{of its components as follows:}\\&divF(x_1,x_2,...x_n)=\frac{\delta F_{x_1}}{\delta x_1}+\frac{\delta F_{x_2}}{\delta x_2}+...+\frac{\delta F_{x_n}}{\delta x_n}\\&\text{We're given the function:}\\&F(x,y,z)=(\frac{(5e^{(y^{2})}e^{(4z)})}{x})\widehat{i}+(\frac{(3ln(x^{2})sin(z^{2}))}{y^{3}})\widehat{j}+(2x^{3}ze^{(y^{4})})\widehat{k}\\&\text{at the point }(1.57,0.68,1.71)\\&\text{Utilizing derivative rules:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&divF(x,y,z)=(-\frac{(5e^{(y^{2})}e^{(4z)})}{x^{2}})+(-\frac{(9ln(x^{2})sin(z^{2}))}{y^{4}})+(2x^{3}e^{(y^{4})})\\&divF(1.57,0.68,1.71)=(-3009.97)+(-8.19)+(9.58)=-3008.58\end{align*}$

Report an Error

View Calculus 3 Tutors

Ruhhi
Certified Tutor

University of Delhi, Master's/Graduate, Physics.

View Calculus 3 Tutors

Dana
Certified Tutor

Texas A&M University - College Station, Bachelor, Chemical Engineering. Valencia College, Associate, Education.

View Calculus 3 Tutors

Arshia
Certified Tutor

IKIU, Bachelor, Mechanical Engineering. University of Victoria, Master's/Graduate, Biomedical Systems.

All Calculus 3 Resources

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

Popular Subjects

Math Tutors in San Diego, Spanish Tutors in Chicago, SAT Tutors in Chicago, Biology Tutors in Dallas Fort Worth, GRE Tutors in Denver, Biology Tutors in Miami, SSAT Tutors in Phoenix, ACT Tutors in Denver, SSAT Tutors in Atlanta, GRE Tutors in Atlanta

Popular Courses & Classes

Spanish Courses & Classes in Chicago, ISEE Courses & Classes in Philadelphia, GRE Courses & Classes in Philadelphia, ISEE Courses & Classes in Miami, SSAT Courses & Classes in Phoenix, SSAT Courses & Classes in San Diego, GMAT Courses & Classes in Atlanta, MCAT Courses & Classes in San Diego, GRE Courses & Classes in Boston, ACT Courses & Classes in San Francisco-Bay Area

Popular Test Prep

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

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 #801 : Partial Derivatives

Example Question #802 : Partial Derivatives

Example Question #803 : Partial Derivatives

Example Question #803 : Partial Derivatives

Example Question #805 : Partial Derivatives

Example Question #806 : Partial Derivatives

Example Question #807 : Partial Derivatives

Example Question #808 : Partial Derivatives

Example Question #809 : Partial Derivatives

Example Question #810 : Partial Derivatives

All Calculus 3 Resources

Report an issue with this question

DMCA Complaint

Contact Information

Address

Complaint Details

Email address:
Your name:
Feedback: