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: 7

Answer

$$\frac{\partial z}{\partial s}=5t^5s^4(s-t)^4(2s-t)$$ $$\frac{\partial z}{\partial t}=5s^5t^4(s-t)^4(s-2t)$$

Work Step by Step

First we will evaluate the partial derivative with respect to $s$: $$\frac{\partial z}{\partial s}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial s}= \frac{\partial}{\partial x}((x-y)^5)\frac{\partial }{\partial s}(s^2t)+\frac{\partial}{\partial y}((x-y)^5)\frac{\partial}{\partial s}(st^2)= 5(x-y)^4\frac{\partial}{\partial x}(x-y)\cdot2st+5(x-y)^4\frac{\partial }{\partial y}(x-y)\cdot t^2= 10st(x-y)^4\cdot1+5t^2(x-y)^4\cdot(-1)=5t(x-y)^4(2s-t)$$ Now we have to express solution in terms of $s$and $t$: $$\frac{\partial z}{\partial s}=5t(x-y)^4(2s-t)= 5t(s^2t-st^2)^4(2s-t)= 5t\cdot (st)^4(s-t)^4\cdot(2s-t)= 5t^5s^4(s-t)^4(2s-t)$$ Now we will evaluate the partial derivative with respect to $t$: $$\frac{\partial z}{\partial t}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial t}= \frac{\partial}{\partial x}((x-y)^5)\frac{\partial }{\partial t}(s^2t)+\frac{\partial}{\partial y}((x-y)^5)\frac{\partial}{\partial t}(st^2)= 5(x-y)^4\frac{\partial }{\partial x}(x-y)\cdot s^2+5(x-y)^4\frac{\partial}{\partial y}(x-y)\cdot 2st= 5s^2(x-y)^4\cdot1+10st(x-y)^4\cdot(-1)= 5s(x-y)^4(s-2t)$$ Now we will express solution in terms of $s$ and $t$: $$\frac{\partial z}{\partial t}=5s(x-y)^4(s-2t)= 5s(s^2t-st^2)^4(s-2t)= 5s^5t^4(s-t)^4(s-2t)$$
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