Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 14 - Partial Derivatives - 14.4 Exercises - Page 947: 26

Answer

$du=\displaystyle \frac{x}{\sqrt{x^{2}+3y^{2}}}dx+\frac{3y}{\sqrt{x^{2}+3y^{2}}}dy$

Work Step by Step

$du=\displaystyle \frac{\partial u}{\partial x}dx+\frac{\partial u}{\partial y}dy$ $u=\sqrt{x^{2}+3y^{2}}=(x^{2}+3y^{2})^{1/2}$ $\displaystyle \frac{\partial u}{\partial x}dx=\frac{1}{2}(x^{2}+3y^{2})^{-1/2}(2x)dx=\frac{x}{\sqrt{x^{2}+3y^{2}}}dx$ $\displaystyle \frac{\partial u}{\partial y}dy=\frac{1}{2}(x^{2}+3y^{2})^{-1/2}(6y)dy=\frac{3y}{\sqrt{x^{2}+3y^{2}}}dy$ $du=\displaystyle \frac{x}{\sqrt{x^{2}+3y^{2}}}dx+\frac{3y}{\sqrt{x^{2}+3y^{2}}}dy$
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