Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 14 - Section 14.3 - Partial Derivatives - 14.3 Exercise - Page 925: 71

Answer

$f_{xzy}=f_{yxz}=6yz^2$.

Work Step by Step

$f(x,y,z)=xy^2z^3+\arcsin{(x\sqrt z)}$ In order to find $f_{xzy}$ we can differentiate in whatever order we want because of Clairaut's Theorem. $f_y=2xyz^3\Rightarrow f_{yx}=2yz^3\Rightarrow f_{yxz}=6yz^2$
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