Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 14 - Partial Derivatives - Review - Exercises: 14

Answer

$g_{u}=\frac{-u^{2}+v^{2}-4vu}{(u^{2}+v^{2})^{2}}$ and $g_{v}=\frac{2u^{2}-2v^{2}-2vu}{(u^{2}+v^{2})^{2}}$

Work Step by Step

Given: $g(u,v)=\frac{u+2v}{u^{2}+v^{2}}$ Need to find first partial derivatives $g_{u}$ and $g_{v}$. Differentiate the function with respect to $u$ keeping $v$ constant. $g_{u}=\frac{1(u^{2}+v^{2})-2u(u+2v)}{(u^{2}+v^{2})^{2}}$ $=\frac{-u^{2}+v^{2}-4vu}{(u^{2}+v^{2})^{2}}$ Differentiate the function with respect to $v$ keeping $u$ constant. $g_{v}=\frac{2(u^{2}+v^{2})-2v(u+2v)}{(u^{2}+v^{2})^{2}}$ $=\frac{2u^{2}-2v^{2}-2vu}{(u^{2}+v^{2})^{2}}$ Hence, $g_{u}=\frac{-u^{2}+v^{2}-4vu}{(u^{2}+v^{2})^{2}}$ and $g_{v}=\frac{2u^{2}-2v^{2}-2vu}{(u^{2}+v^{2})^{2}}$
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