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: 21

Answer

$\dfrac{\partial z}{\partial s} =1582$; $\dfrac{\partial z}{\partial t} = 3164$ ; $\dfrac{\partial z}{\partial u} = - 700$

Work Step by Step

We have: $\dfrac{\partial z}{\partial x}= 4x^3 + 2xy$ and $\dfrac{\partial z}{\partial y} = x^2$ and $\dfrac{\partial x}{\partial s} = 1$ and $\dfrac{\partial x}{\partial t} =2$ and $\dfrac{\partial x}{\partial u} = -1$ and $\dfrac{\partial y}{\partial s} = tu^2$ $\dfrac{\partial y}{\partial t} =su^2$ and $\dfrac{\partial y}{\partial u} = 2stu $ Next, we will compute the value for each partial derivative. when $s = 4;t = 2; u = 1$ $x = s + 2t - u = (4) + 2(2) - (1) = 7$ and $ y = stu^2 = 4 \cdot 2 \cdot (1)^2 = 8$ $\dfrac{\partial z}{\partial s} =[4(7)^3 + 2 + (7)^2) \cdot (2)(1)^2) = 1582$; $\dfrac{\partial z}{\partial t} = 3164 \\ \dfrac{\partial z}{\partial u} =[(4(7)^3) + 2(7)(8)] \cdot (-1) + (7)^2)(2(4)(2)(1)] = - 700$
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