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: 16

Answer

$$\frac{\partial z}{\partial x}=\frac{e^{x/y}}{y}$$ $$\frac{\partial z}{\partial y}=-\frac{xe^{x/y}}{y^2}$$

Work Step by Step

We will use chain rule to solve both partial derivatives. The partial derivative with respect to $x$ is: $$\frac{\partial z}{\partial x}=\frac{\partial }{\partial x}(e^{x/y})=e^{x/y}\frac{\partial }{\partial x}(\frac{x}{y})=e^{x/y}\cdot\frac{1}{y}=\frac{e^{x/y}}{y}$$ because $y$ is treated as a constant. The partial derivative with respect to $y$ is: $$\frac{\partial z}{\partial y}=\frac{\partial }{\partial y}(e^{x/y})=e^{x/y}\frac{\partial }{\partial y}(\frac{x}{y})=e^{x/y}\cdot(-\frac{x}{y^2})=-\frac{xe^{x/y}}{y^2}$$ because here $x$ is treated as a constant.
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