Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 14: Partial Derivatives - Section 14.6 - Tangent Planes and Differentials - Exercises 14.6 - Page 835: 38

Answer

$0.0832$

Work Step by Step

$$f_x=\dfrac{1}{x} \\ f_x(1,1)=1^{-1}=1 \\ f_y(x,y) =y^{-1} \\ f_{y}(1,1) =1^{-1}=1 \\f_{xx}(x,y)=-x^{-2} \\ f_{yy}(x,y)=-y^{-2} \\f_{xy}(x,y) =0$$ Error: $|E(x,y)| \leq \dfrac{1}{2} \times (1.04) [ |x-1| +|y-1|)^2$ or, $$ E \leq \dfrac{1.04}{2} \times (0.2+0.2)^2 =0.0832$$
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.