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: 27

Answer

$$\frac{\partial f(x,y)}{\partial x}=\frac{x}{\sqrt{x^2+y^2}}$$ $$\frac{\partial f(x,y)}{\partial y}=\frac{y}{\sqrt{x^2+y^2}}$$

Work Step by Step

To solve both partial derivatives we will use chain rule. The partial derivative with respect to $x$ is: $$\frac{\partial f(x,y)}{\partial x}=\frac{\partial }{\partial x}(\sqrt{x^2+y^2})=\frac{1}{2\sqrt{x^2+y^2}}\frac{\partial }{\partial x}(x^2+y^2)=\frac{1}{2\sqrt{x^2+y^2}}\cdot2x=\frac{x}{\sqrt{x^2+y^2}}$$ because $y$ is treated as a constant. The partial derivative with respect to $y$ is: $$\frac{\partial f(x,y)}{\partial y}=\frac{\partial }{\partial y}(\sqrt{x^2+ y^2})=\frac{1}{2\sqrt{x^2+y^2}}\frac{\partial }{\partial y}(x^2+y^2)=\frac{1}{2\sqrt{x^2+y^2}}\cdot2y=\frac{y}{\sqrt{x^2+y^2}}$$ because now $x$ is treated as a constant.
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