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: 34

Answer

$$\frac{\partial z}{\partial x}=-2x\sin(x^2+y^2)$$ $$\frac{\partial z}{\partial y}--2y\sin(x^2+y^2)$$

Work Step by Step

We will use chain rule to find both partial derivatives. The partial derivative with respect to $x$ is: $$\frac{\partial z}{\partial x}=\frac{\partial }{\partial x}(\cos(x^2+y^2))=-\sin(x^2+y^2)\frac{\partial }{\partial x}(x^2+y^2)=-\sin(x^2+y^2)\cdot2x=-2x\sin(x^2+y^2)$$ We treated $y$ as a constant here. The partial derivative with respect to $y$ is: $$\frac{\partial z}{\partial y}=\frac{\partial }{\partial y}(\cos(x^2+y^2))=-\sin(x^2+y^2)\frac{\partial }{\partial y}(x^2+y^2)=-\sin (x^2+y^2)\cdot2y=-2y\sin(x^2+y^2)$$ because now $x$ is treated as a constant.
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