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: 12

Answer

$$\frac{\partial z}{\partial s}=-\frac{13t\sec^2(\frac{2s+3t}{3s-2t})}{(3s-2t)^2}$$ $$\frac{\partial z}{\partial t}=\frac{13s\sec^2(\frac{2s+3t}{3s-2t})}{(3s-2t)^2}$$

Work Step by Step

The partial derivative with respect to $s$ is: $$\frac{\partial z}{\partial s}=\frac{\partial z}{\partial u}\frac{\partial u}{\partial s}+\frac{\partial z}{\partial v}\frac{\partial v}{\partial s}= \frac{\partial }{\partial u}(\tan(\frac{u}{v}))\frac{\partial}{\partial s}(2s+3t)+\frac{\partial}{\partial v}(\tan(\frac{u}{v}))\frac{\partial}{\partial s}(3s-2t)= \frac{1}{\cos^2(\frac{u}{v})}\frac{\partial }{\partial u}(\frac{u}{v})\cdot2+\frac{1}{\cos^2(\frac{u}{v})}\frac{\partial}{\partial v}(\frac{u}{v})\cdot3= \frac{1}{\cos^2(\frac{u}{v})}(\frac{1}{v}\cdot2+(-\frac{u}{v^2})\cdot3)= \frac{1}{\cos(\frac{u}{v})}\frac{2v-3u}{v^2}$$ Expressing this in terms of $s$ and $t$ we get: $$\frac{\partial z}{\partial s}=\frac{1}{\cos^2(\frac{u}{v})}\frac{2v-3u}{v^2}= \frac{1}{\cos^2(\frac{2s+3t}{3s-2t})}\frac{2(3s-2t)-3(2s+3t)}{(3s-2t)^2}=\frac{1}{\cos^2(\frac{2s+3t}{3s-2t})}\frac{-13t}{(3s-2t)^2}=-\frac{13t\sec^2(\frac{2s+3t}{3s-2t})}{(3s-2t)^2}$$ The partial derivative with respect to $t$ is: $$\frac{\partial z}{\partial t}=\frac{\partial z}{\partial u}\frac{\partial u}{\partial t}+\frac{\partial z}{\partial v}\frac{\partial v}{\partial t}= \frac{\partial }{\partial u}(\tan(\frac{u}{v}))\frac{\partial}{\partial t}(2s+3t)+\frac{\partial}{\partial v}(\tan(\frac{u}{v}))\frac{\partial}{\partial t}(3s-2t)= \frac{1}{\cos^2(\frac{u}{v})}\frac{\partial}{\partial u}(\frac{u}{v})\cdot3+\frac{1}{\cos^2(\frac{u}{v})}\frac{\partial}{\partial v}(\frac{u}{v})\cdot(-2)= \frac{1}{\cos^2(\frac{u}{v})}(\frac{1}{u}\cdot3-(-\frac{u}{v^2}\cdot2))= \frac{1}{\cos^2(\frac{u}{v})}\frac{3v+2u}{v^2}$$ Expressing this in terms of $s$ and $t$ we get: $$\frac{\partial z}{\partial t}=\frac{1}{\cos^2(\frac{u}{v})}\frac{3v+2u}{v^2}=\sec^2(\frac{2s+3t}{3s-2t})\frac{3(3s-2t)+2(2s+3t)}{(3s-2t)^2}= \frac{13s\sec^2(\frac{2s+3t}{3s-2t})}{(3s-2t)^2}$$
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