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Multivariable Calculus UAF MATH F253X

Dr. Latrice Bowman

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Section Partial Derivatives

Worksheet DC-1 Reading

All of the readings for this course come from the OpenStax Calculus Volume 3 textbook, which is available for free online at https://openstax.org/details/books/calculus-volume-3. You can read the textbook online or download a PDF version. The textbook is also available in print from various retailers if you prefer a physical copy.
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For this lesson, you should read sections 4.3 Partial Derivatives. You should read through the sections carefully, making sure to understand the definitions, examples, and key ideas. You should also work through the exercises at the end of each section to test your understanding and practice applying the concepts. Some of the exercises at the end of the section are assigned in the WP assignments.
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Worksheet DC-1 Videos

Partial Derivatives
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Worksheet DC-1 Written Practice

INFORMATION ABOUT ALL WRITTEN PRACTICE (WP) ASSIGNMENTS:
  • All of the WP problems come from our online textbook: https://openstax.org/details/books/calculus-volume-3
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  • Complete solutions to all WP assignments can be found in Canvas within the section Module.
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  • Links to the text and PDFs of the text can be found in the Canvas module under the Reading and Information.
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THE WP ASSIGNMENTS ARE DESIGNED TO PREPARE YOU FOR THE QUIZZES AND ASSESSMENTS
  • Work on every problem on every assignment.
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  • If you get stuck or do not understand a solution, ask questions.
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  • For each problem, try it once or twice before looking at solutions or asking for help.
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  • Mark the problems you did correctly on the first try.
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  • Highlight problems where you got help (using solutions, videos, a tutor, etc) -These are the problem types that you will need more practice on to be ready for the assessments.
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  • Understand that just copying the solutions instead of working through the problems will greatly reduce your chances of success in this class.
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WP5
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Partial Derivatives: Complete problems 119, 121, 125, 149, 157, 161
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Worksheet DC-1 Sample Outcomes

For the quiz on this outcome you would be given 30 minutes to complete 3 problems. Here are samples of the types of problems you will encounter:
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1.

Given the function \(g(x,y)=\frac{y}{2x+3y}\)
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(a)
Find the first-order partial derivatives of \(g(x,y)\text{.}\)
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Hint.
To find the partial derivatives, use the quotient rule.
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Answer.
\(\frac{\partial g}{\partial x} = \frac{-2y}{(2x+3y)^2}\)
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\(\frac{\partial g}{\partial y} = \frac{2x}{(2x+3y)^2}\)
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Solution.
To find the partial with respect to \(x\text{,}\) treat \(y\) as a constant and apply the quotient rule:
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\(\frac{\partial g}{\partial x} = \frac{(2x+3y)(0) - y(2)}{(2x+3y)^2} = \frac{-2y}{(2x+3y)^2}\)
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To find the partial with respect to \(y\text{,}\) treat \(x\) as a constant and apply the quotient rule:
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\(\frac{\partial g}{\partial y} = \frac{(2x+3y)(1) - y(3)}{(2x+3y)^2} = \frac{2x}{(2x+3y)^2}\)
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(b)
Find the second-order partial derivatives of \(g(x,y)\text{.}\)
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Hint.
To find the second-order partial derivatives, differentiate the first-order partial derivatives.
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Answer.
\(\frac{\partial^2 g}{\partial x^2} = \frac{8y^2}{(2x+3y)^3}\)
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\(\frac{\partial^2 g}{\partial y^2} = \frac{8x^2}{(2x+3y)^3}\)
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\(\frac{\partial^2 g}{\partial x \partial y} = \frac{\partial^2 g}{\partial y \partial x} = \frac{-4x - 6y}{(2x+3y)^3}\)
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Solution.
To find the second-order partial derivative with respect to \(x\text{,}\) differentiate the first-order partial derivative with respect to \(x\) with respect to \(x\text{:}\)
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\(\frac{\partial^2 g}{\partial x^2} = \frac{\partial}{\partial x}\left(\frac{-2y}{(2x+3y)^2}\right) = \frac{8y^2}{(2x+3y)^3}\)
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To find the second-order partial derivative with respect to \(y\text{,}\) differentiate the first-order partial derivative with respect to \(y\) with respect to \(y\text{:}\)
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\(\frac{\partial^2 g}{\partial y^2} = \frac{\partial}{\partial y}\left(\frac{2x}{(2x+3y)^2}\right) = \frac{8x^2}{(2x+3y)^3}\)
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To find the mixed partial derivatives, differentiate the first-order partial derivatives:
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\(\frac{\partial^2 g}{\partial x \partial y} = \frac{\partial}{\partial y}\left(\frac{-2y}{(2x+3y)^2}\right) = \frac{-4x - 6y}{(2x+3y)^3}\)
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\(\frac{\partial^2 g}{\partial y \partial x} = \frac{\∂}{\∂x}\left(\frac{2x}{(2x+3y)^2}\right) = \frac{-4x - 6y}{(2x+3y)^3}\)
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(c)
Evaluate \(g_{xyyx}(1,2)\text{.}\)
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Hint.
Use the second-order partial derivative found earlier to find the third and fourth-order partial derivatives.
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Answer.
\(g_{xyyx}(1,2) = -\frac{1}{32}\)
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Solution.
First we find the third and fourth-order partial derivatives:
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Given that \(g_{xy} = \frac{-4x - 6y}{(2x+3y)^3}\text{,}\) we can find the third-order partial derivative \(g_{xyy}\) by differentiating \(g_{xy}\) with respect to \(y\text{:}\)
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This will give \(g_{xyy} = \frac{\partial}{\partial y}\left(\frac{-4x - 6y}{(2x+3y)^3}\right)\text{.}\) Then we can find the fourth-order partial derivative \(g_{xyyx}\) by differentiating \(g_{xyy}\) with respect to \(x\text{:}\)
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\(g_{xyyx} = \frac{\partial}{\partial y}\left(\frac{\partial^2 g}{\partial x \partial y}\right) = \frac{\partial}{\partial y}\left(\frac{-4x - 6y}{(2x+3y)^3}\right)\)
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Substitute \(x = 1\) and \(y = 2\) into the expression for \(g_{xyyx}\text{:}\)
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\(g_{xyyx}(1,2) = \frac{-4(1) - 6(2)}{(2(1)+3(2))^3} = \frac{-16}{(8)^3} = \frac{-16}{512} = -\frac{1}{32}\)
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2.

Given the function \(f(x,y)=4x^3y^2-2x^2y+5x\)
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(a)
Find the first-order partial derivatives \(f_x\) and \(f_y\text{.}\)
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Hint.
Use the power rule and the chain rule to differentiate.
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Answer.
\(f_x = 12x^2y^2 - 4xy + 5\)
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\(f_y = 8x^3y - 2x^2\)
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Solution.
To find \(f_x\text{,}\) differentiate \(f(x,y)\) with respect to \(x\text{:}\)
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\(f_x = \frac{\partial}{\partial x}(4x^3y^2 - 2x^2y + 5x) = 12x^2y^2 - 4xy + 5\)
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To find \(f_y\text{,}\) differentiate \(f(x,y)\) with respect to \(y\text{:}\)
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\(f_y = \frac{\partial}{\partial y}(4x^3y^2 - 2x^2y + 5x) = 8x^3y - 2x^2\)
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(b)
Find the second-order partial derivatives \(f_{xx}\text{,}\) \(f_{yy}, f_{yx}\text{,}\) and \(f_{xy}\text{.}\)
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Hint.
Use the power rule and the chain rule to differentiate.
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Answer.
\(f_{xx} = 24x^2y^2 - 4y\)
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\(f_{yy} = 8x^3 - 4x^2\)
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\(f_{xy} =f_{yx}= 24x^2y - 4x\)
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Solution.
To find \(f_{xx}\text{,}\) differentiate \(f_x\) with respect to \(x\text{:}\)
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\(f_{xx} = \frac{\partial}{\partial x}(12x^2y^2 - 4xy + 5) = 24x^2y^2 - 4y\)
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To find \(f_{yy}\text{,}\) differentiate \(f_y\) with respect to \(y\text{:}\)
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\(f_{yy} = \frac{\partial}{\partial y}(8x^3y - 2x^2) = 8x^3 - 4x^2\)
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To find \(f_{xy}\text{,}\) differentiate \(f_x\) with respect to \(y\text{:}\)
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\(f_{xy} = \frac{\partial}{\partial y}(12x^2y^2 - 4xy + 5) = 24x^2y - 4x\)
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