## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 14 - Section 14.3 - Partial Derivatives - 14.3 Exercise: 62

#### Answer

$u_{xy}=-\frac{2}{(x+2y)^{2}}$ and $u_{yx}=-\frac{2}{(x+2y)^{2}}$ Hence, $u_{xy}=u_{yx}$

#### Work Step by Step

Consider the function $u=ln(x+2y)$ 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. $u_{x}=\frac{1}{(x+2y)}$ Differentiate the function with respect to $x$ keeping $y$ constant. $u_{y}=\frac{2}{(x+2y)}$ Differentiate $u_{x}$ with respect to $y$ keeping $x$ constant. $u_{xy}=\frac{∂}{∂y}[\frac{1}{(x+2y)}]$ $=-\frac{2}{(x+2y)^{2}}$ Thus, $u_{xy}=-\frac{2}{(x+2y)^{2}}$ Differentiate $u_{y}$ with respect to $x$ keeping $y$ constant. $u_{yx}=\frac{∂}{∂x}[\frac{2}{(x+2y)}]$ $=-\frac{2}{(x+2y)^{2}}$ Therefore, $u_{xy}=-\frac{2}{(x+2y)^{2}}$ and $u_{yx}=-\frac{2}{(x+2y)^{2}}$ Hence, $u_{xy}=u_{yx}$

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