Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 14 - Section 14.5 - The Chain Rule. - 14.5 Exercise - Page 944: 9


$$\frac{\partial z}{\partial s}=\frac{3\sin t-2t\sin s}{3s\sin t+2t\cos s}$$ $$\frac{\partial z}{\partial t}=\frac{3s\cos t+2\cos s}{3s\sin t+2t\cos s}$$

Work Step by Step

The partial derivative with respect to $s$ is: $$\frac{\partial z}{\partial s}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial s}= \frac{\partial }{\partial x}(\ln(3x+2y))\frac{\partial}{\partial s}(s\sin t)+\frac{\partial}{\partial y}(\ln(3x+2y))\frac{\partial}{\partial s}(t\cos s)= \frac{1}{3x+2y}\frac{\partial}{\partial x}(3x+2y)\cdot \sin t+\frac{1}{3x+2y}\frac{\partial}{\partial y}(3x+2y)\cdot t\cdot(-\sin s)= \frac{3\sin t-2t\sin s}{3x+2y}$$ Now we will express solution in terms of $s$ and $t$: $$\frac{\partial z}{\partial s}=\frac{3\sin t-2t\sin s}{3x+2y}= \frac{3\sin t-2t\sin s}{3s\sin t+2t\cos s}$$ The partial derivative with respect to $t$ is: $$\frac{\partial z}{\partial t}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial t}= \frac{\partial }{\partial x}(\ln(3x+2y))\frac{\partial}{\partial t}(s\sin t)+\frac{\partial}{\partial y}(\ln(3x+2y))\frac{\partial}{\partial t}(t\cos s)= \frac{1}{3x+2y}\frac{\partial}{\partial x}(3x+2y)\cdot s\cos t+\frac{1}{3x+2y}\frac{\partial}{\partial y}(3x+2y)\cdot\cos s= \frac{3s\cos t+2\cos s}{3x+2y}$$ Expressing solution in terms of $s$ and $t$ we get: $$\frac{\partial z}{\partial t}=\frac{3s\cos t+2\cos s}{3x+2y}= \frac{3s\cos t+2\cos s}{3s\sin t+2t\cos s}$$
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