Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 9 - Multiveriable Calculus - Chapter Review - Review Exercises - Page 519: 60



Work Step by Step

We are given $z=\frac{x+5y}{x-2y}$ $f_{x}(x,y)=\frac{(x-2y)1-(x+5y)1}{(x-2y)^{2}}=\frac{7y}{(x-2y)^{2}}$ $f_{y}(x,y)=\frac{(x-2y)5-(x+5y)(-2)}{(x-2y)^{2}}=\frac{7x}{(x-2y)^{2}}$ By the definition $dz=f_{x}(x,y).dx+f_{y}(x,y).dy$ with $x=1; y=-2; dx=-0.04; dy=0.02$ we have $dz=(\frac{7y}{(x-2y)^{2}})(-0.04)+(\frac{7x}{(x-2y)^{2}})0.02$ $dz=\frac{7}{250}=0.028$
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.