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 927: 101



Work Step by Step

Here, we have $f_{xyy}=(f_{xy})_{y}$ and $ f_{yxy}=(f_{yx})_{y}$ Further, $(f_{xy})_{y}=(f_{xy})_{y}=f_{xyy}$ This implies that $f_{yxy}=f_{xyy}$ Since, $ f_{yyx}=(f_{y})_{yx}$ and $(f_{y})_{yx}$=$(f_{y})_{xy}$ we have $f_{yyx}=(f_{y})_{xy}=f_{yxy}$ This gives: $f_{yxy}=f_{yxy},$ As per the transitive property of equality, we get $f_{xyy}=f_{yxy}=f_{yyx}$
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