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: 2



Work Step by Step

We are given $z=f(x,y)=5x^{3}+2xy^{2}-4y$ $f_{x}(x,y)=15x^{2}+2y^{2}$ $f_{y}(x,y)=4xy-4$ By the definition $dz=f_{x}(x,y).dx+f_{y}(x,y)dy$ $dz=(15x^{2}+2y^{2})dx+(4xy-4)dy$ $dz=[15\times(1)^{2}+2\times(3)^{2}]\times0.01+[4\times1\times3-4]\times0.02=0.49$
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