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: 60

Answer

$u_{xy}=e^{xy}siny(xy+1)+ye^{xy}cosy$ and $u_{yx}=e^{xy}siny(xy+1)+ye^{xy}cosy$ Hence, $u_{xy}=u_{yx}$

Work Step by Step

Consider the function $u=e^{xy}siny$ Need to prove the conclusion of Clairaut’s Theorem holds, that is, $u_{xy}=u_{yx}$ In order to find this differentiate the function with respect to $x$ keeping $y$ constant. $u_{x}=ye^{xy}siny$ Differentiate the function with respect to $x$ keeping $y$ constant. $u_{y}=xe^{xy}siny+ye^{xy}cosy$ Differentiate $u_{x}$ with respect to $y$ keeping $x$ constant. $u_{xy}=\frac{∂}{∂y}[ye^{xy}siny]$ (apply product rule) $=e^{xy}siny\frac{∂y}{∂y}+ysiny\frac{∂e^{xy}}{∂y}+ye^{xy}\frac{∂(siny)}{∂y}$ $=e^{xy}siny+ysiny(xe^{xy})+ye^{xy}cosy$ Thus, $u_{xy}=e^{xy}siny(xy+1)+ye^{xy}cosy$ Differentiate $u_{y}$ with respect to $x$ keeping $y$ constant. $u_{yx}=\frac{∂}{∂x}[xe^{xy}siny+ye^{xy}cosy]$ $=\frac{∂}{∂x}[xe^{xy}siny]+\frac{∂}{∂x}[ye^{xy}cosy]$ $=e^{xy}siny\frac{∂}{∂x}[x]+x\times\frac{∂}{∂x}[e^{xy}siny]+ye^{xy}cosy$ $=e^{xy}siny+xysiny(e^{xy})+ye^{xy}cosy$ Thus, $u_{yx}=e^{xy}siny(xy+1)+ye^{xy}cosy$ Therefore, $u_{xy}=e^{xy}siny(xy+1)+ye^{xy}cosy$ and $u_{yx}=e^{xy}siny(xy+1)+ye^{xy}cosy$ Hence, $u_{xy}=u_{yx}$
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