Calculus 8th Edition

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

Chapter 14 - Partial Derivatives - 14.2 Limits and Continuity - 14.2 Exercises - Page 951: 26

Answer

$h(x,y)=\dfrac{(1-xy)}{(1+x^2y^2)}+ln[\dfrac{(1-xy)}{(1+x^2y^2)}]$; set of points: {$(x,y) | 1\gt xy$}

Work Step by Step

As per the given values; $h(x,y)=\dfrac{(1-xy)}{(1+x^2y^2)}+ln[\dfrac{(1-xy)}{(1+x^2y^2)}]$ The function is continuous everywhere and the term $\dfrac{(1-xy)}{(1+x^2y^2)}$ is greater than $0$ such as $t \gt 0$ these shows the number of values of $t$ for which $\ln t$ exist for such values. Thus, $1-xy\gt 0$ or, $1\gt xy$ Hence, $h(x,y)=\dfrac{(1-xy)}{(1+x^2y^2)}+ln[\dfrac{(1-xy)}{(1+x^2y^2)}]$; set of points: {$(x,y) | 1\gt xy$}
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.