Calculus 8th Edition

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

Chapter 14 - Partial Derivatives - 14.5 The Chain Rule - 14.5 Exercises - Page 984: 25

Answer

$\dfrac{\partial N}{\partial u} =\dfrac{5}{144} \\ \dfrac{\partial N}{\partial v} =\dfrac{-5}{96} \\ \dfrac{\partial N}{\partial w} =\dfrac{5}{144}$

Work Step by Step

We have $p=2+3(4)=14; q=3+(2)(4)=11;r=4+(2)(3)=10$ Now, $\dfrac{\partial N}{\partial p} =\dfrac{(p+r)-(p+q)}{(p+r)^2}$ and $\dfrac{\partial N}{\partial q} =\dfrac{(p+r)-0}{(p+r)^2}=\dfrac{1}{(p+r)}$ and $\dfrac{\partial N}{\partial r} =\dfrac{0-(p+q)}{(p+r)^2}$ For $u=2;v=3; w=4$: $\dfrac{\partial N}{\partial p} =\dfrac{-1}{(24)^2}$; $\dfrac{\partial N}{\partial q} =\dfrac{(14+10)-0}{(14+10)^2}=\dfrac{1}{24}$ $\dfrac{\partial N}{\partial r} =\dfrac{0-(14+11)}{(14+10)^2}=\dfrac{-25}{(24)^2}$
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