University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 13 - Section 13.3 - Partial Derivatives - Exercises - Page 702: 24

Answer

$y+z ; \\x+z ; \\y+x$

Work Step by Step

Our aim is to take the first partial derivatives of the given function $f(x,y)$ with respect to $x$, by keeping $y$ and $z$ as a constant, and vice versa: $\dfrac{\partial }{\partial x}f(x,y)=y+z ; \\\dfrac{\partial }{\partial y}f(x,y)=x+z ; \\\dfrac{\partial }{\partial z}f(x,y)=y+x$
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