Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 13 - Functions of Several Variables - 13.3 Exercises - Page 896: 31

Answer

$$\frac{\partial z}{\partial x}=\frac{2}{cos^2(2x-y)}$$ $$\frac{\partial z}{\partial y}=-\frac{1}{\cos^2(2x-y)}$$

Work Step by Step

We will use chain rule to find both partial derivatives. The partial derivative with respect to $x$ is: $$\frac{\partial z}{\partial x}=\frac{\partial }{\partial x}(\tan(2x-y))=\frac{1}{\cos^2(2x-y)}\frac{\partial }{\partial x}(2x-y)=\frac{1}{\cos^2(2x-y)}\cdot2=\frac{2}{\cos^2(2x-y)}$$ because $y$ is treated as a constant. $$\frac{\partial z}{\partial y}=\frac{\partial }{\partial y}(\tan(2x-y))=\frac{1}{\cos^2(2x-y)}\frac{\partial }{\partial y}(2x-y)=\frac{1}{\cos^2(2x-y)}\cdot(-1)=-\frac{1}{\cos^2(2x-y)}$$ because now $x$ is treated as a constant.
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