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: 36

Answer

$u_x=\frac{yx^{\frac{y}{z}-1}}{z}$, $f_y=\frac{ln{(x)}x^{\frac{y}{z}}}{z}$, $f_z=\frac{-yln{(x)}x^{\frac{y}{z}}}{z^2}$.

Work Step by Step

$u=x^{\frac{y}{z}}$ In order to find $u_x$ we treat $y$ and $z$ as constants and differentiate with respect to $x$. $u_x=\frac{yx^{\frac{y}{z}-1}}{z}$ Analogously: $f_y=\frac{ln{(x)}x^{\frac{y}{z}}}{z}$ $f_z=\frac{-yln{(x)}x^{\frac{y}{z}}}{z^2}$
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