Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 14 - Review - Exercises - Page 983: 16

Answer

$G_{x}=ze^{xz}sin(\frac{y}{z})$, $G_{y}=\frac{e^{xz} cos(\frac{y}{z})}{z}$ and $G_{z}=xe^{xz}sin(\frac{y}{z})-\frac{ye^{xz}cos(\frac{y}{z})}{z^{2}}$

Work Step by Step

Given: $G(x,y,z)=e^{xz}sin(\frac{y}{z})$ We need to find the first partial derivatives $G_{x}$,$G_{y}$ and $G_{z}$ Differentiate the function with respect to $x$ keeping $y$ and $z$ constant. $G_{x}=ze^{xz}sin(\frac{y}{z})$ Differentiate the function with respect to $y$ keeping $x$ and $z$ constant. $G_{y}=e^{xz}\times cos(\frac{y}{z}) \times \frac{1}{z}=\frac{e^{xz} cos(\frac{y}{z})}{z}$ Differentiate the function with respect to $z$ keeping $x$ and $y$ constant. Apply product rule. $G_{z}=xe^{xz}sin(\frac{y}{z})+e^{xz}\times cos(\frac{y}{z})\times \frac{-y}{z^{2}}$ $G_{z}=xe^{xz}sin(\frac{y}{z})-\frac{ye^{xz}cos(\frac{y}{z})}{z^{2}}$ Hence, $G_{x}=ze^{xz}sin(\frac{y}{z})$, $G_{y}=\frac{e^{xz} cos(\frac{y}{z})}{z}$ and $G_{z}=xe^{xz}sin(\frac{y}{z})-\frac{ye^{xz}cos(\frac{y}{z})}{z^{2}}$
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