Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 14 - Partial Derivatives - 14.3 Exercises - Page 939: 97

Answer

$f_{xyy}=(f_{xy})_{y}$ $ f_{yxy}=(f_{yx})_{y}\qquad$... by the theorem, $f_{yx}$=$f_{xy}$, so $(f_{xy})_{y}=(f_{xy})_{y}=f_{xyy}$ Thus, $f_{yxy}=f_{xyy}$ $ f_{yyx}=(f_{y})_{yx}\qquad$... by the theorem, $(f_{y})_{yx}$=$(f_{y})_{xy}$, so $f_{yyx}=(f_{y})_{xy}=f_{yxy}$ Thus, $f_{yxy}=f_{yxy},$ By the transitive property of equality, $f_{xyy}=f_{yxy}=f_{yyx}$

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