Calculus 8th Edition

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

Chapter 14 - Partial Derivatives - 14.3 Partial Derivatives - 14.3 Exercises: 67

Answer

$\frac{∂^{3}W}{∂u^{2}∂v}=\frac{3}{4}v (u+v^{2})^{-5/2}$

Work Step by Step

Consider the function $W=\sqrt {u+v^{2}}$ Let us start differentiating the function with respect to $v$ keeping $u$ constant. $\frac{∂W}{∂v}=\frac{1}{2\sqrt {u+v^{2}}}\times2v $ $=v( {u+v^{2}})^{-1/2}$ Differentiate with respect to $u$ keeping $v$ constant. $\frac{∂^{2}W}{∂u∂v}=-\frac{1}{2}v (u+v^{2})^{-3/2}$ Apply power rule. Differentiate with respect to $u$ keeping $v$ constant. $\frac{∂^{3}W}{∂u^{2}∂v}=-\frac{1}{2}(\frac{-3}{2})v (u+v^{2})^{-5/2}$ $=\frac{3}{4}v (u+v^{2})^{-5/2}$ Hence, $\frac{∂^{3}W}{∂u^{2}∂v}=\frac{3}{4}v (u+v^{2})^{-5/2}$
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