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 924: 39

Answer

$u_{x_1}=\frac{x_1}{\sqrt{x_1^2+x_2^2+...x_n^2}}$, $u_{x_2}=\frac{x_2}{\sqrt{x_1^2+x_2^2+...x_n^2}}$,,, $u_{x_n}=\frac{x_n}{\sqrt{x_1^2+x_2^2+...x_n^2}}$.

Work Step by Step

$u=\sqrt{x_1^2+x_2^2+...x_n^2}$ In order to find $u_{x_1}$ we treat all other variables as constants and differentiate with respect to $x_1$. $u_{x_1}=\frac{x_1}{\sqrt{x_1^2+x_2^2+...x_n^2}}$ Analogously: $u_{x_2}=\frac{x_2}{\sqrt{x_1^2+x_2^2+...x_n^2}}$ . . . $u_{x_n}=\frac{x_n}{\sqrt{x_1^2+x_2^2+...x_n^2}}$
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