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: 28

Answer

$$\frac{\partial f(x,y)}{\partial x}=\frac{1}{\sqrt{2x+y^3}}$$ $$\frac{\partial f(x,y)}{\partial y}=\frac{3y^2}{2\sqrt{2x+y^3}}$$

Work Step by Step

We will use chain rule to find both partial derivatives. The partial derivative with respect to $x$ is: $$\frac{\partial f(x,y)}{\partial x}=\frac{\partial }{\partial x}(\sqrt{2x+y^3})=\frac{1}{2\sqrt{2x+y^3}}\frac{\partial }{\partial x}(2x+y^3)=\frac{1}{2\sqrt{2x+y^3}}\cdot2=\frac{1}{\sqrt{2x+y^3}}$$ because $y$ is treated as a constant. The partial derivative with respect to $y$ is: $$\frac{\partial f(x,y)}{\partial y}=\frac{\partial }{\partial y}(\sqrt{2x+y^3})=\frac{1}{2\sqrt{2x+y^3}}\frac{\partial }{\partial y}(2x+y^3)=\frac{1}{2\sqrt{2x+y^3}}\cdot3y^2=\frac{3y^2}{2\sqrt{2x+y^3}}$$ because now $x$ is treated as a constant.
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