Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 14 - Partial Derivatives - 14.3 Partial Derivatives - 14.3 Exercises - Page 964: 21

Answer

$$\frac{\partial}{\partial x}f(x,y)=\frac{1}{y};$$ $$\frac{\partial}{\partial y}f(x,y)=-\frac{x}{y^2}.$$

Work Step by Step

1) Regarding $y$ as constant and differentiating with respect to $x$ we get $$\frac{\partial}{\partial x}f(x,y)=\frac{\partial}{\partial x}\left(\frac{x}{y}\right)=\frac{1}{y}\frac{\partial}{\partial x}(x)=\frac{1}{y}\cdot 1=\frac{1}{y}.$$ 2) Regarding $x$ as constant and differentiating with respect to $y$ we get $$\frac{\partial}{\partial y}f(x,y)=\frac{\partial}{\partial y}\left(\frac{x}{y}\right)=x\frac{\partial}{\partial y}\left(\frac{1}{y}\right)=x\cdot\left(-\frac{1}{y^2}\right)=-\frac{x}{y^2}.$$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.