Calculus with Applications (10th Edition)

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

Chapter 9 - Multiveriable Calculus - 9.5 Total Differentials and Approximations - 9.5 Exercises - Page 502: 4

Answer

$dz=\frac{-1}{130}$

Work Step by Step

We are given $z=f(x,y)=\ln(x^{2}+y^{2})$ $f_{x}(x,y)=\frac{2x}{x^{2}+y^{2}}$ $f_{y}(x,y)=\frac{2y}{x^{2}+y^{2}}$ By the definition $dz=f_{x}(x,y).dx+f_{y}(x,y)dy$ $dz=(\frac{2x}{x^{2}+y^{2}})dx+(\frac{2y}{x^{2}+y^{2}})dy$ With $x=2, y=3, dx=0.02, dy=-0.03$ $dz=\frac{-1}{130}$
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