Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 13 - Functions of Several Variables - 13.3 Exercises - Page 896: 24

Answer

$$\frac{\partial z}{\partial x}=\frac{y^3-x^2y}{(x^2+y^2)^2}$$ $$\frac{\partial z}{\partial y}=\frac{x^3-y^2x}{(x^2+y^2)^2}$$

Work Step by Step

To solve both partial derivatives we will use quotient rule. The partial derivative with respect to $x$ is: $$\frac{\partial z}{\partial x}=\frac{\partial }{\partial x}\left(\frac{xy}{x^2+y^2}\right)=\frac{(x^2+y^2)\frac{\partial }{\partial x}(xy)-xy\frac{\partial }{\partial x}(x^2+y^2)}{(x^2+y^2)^2}=\frac{(x^2+y^2)\cdot y-xy\cdot2x}{(x^2+y^2)^2}=\frac{x^2y+y^3-2x^2y}{(x^2+y^2)^2}=\frac{y^3-x^2y}{(x^2+y^2)^2}$$ because $y$ is treated as a constant. The partial derivative with respect to $y$ is: $$\frac{\partial z}{\partial y}=\frac{\partial }{\partial y}\left(\frac{xy}{x^2+y^2}\right)=\frac{(x^2+y^2)\frac{\partial}{\partial y}(xy)-xy\frac{\partial }{\partial y}(x^2+y^2)}{(x^2+y^2)^2}=\frac{(x^2+y^2)\cdot x-xy\cdot2y}{(x^2+y^2)^2}=\frac{x^3+y^2x-2y^2x}{(x^2+y^2)^2}=\frac{x^3-y^2x}{(x^2+y^2)^2}$$ because now $x$ is treated as a constant.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.