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 14.3 - Partial Derivatives - Exercises 14.3 - Page 807: 11


$f_x=\frac{-1-y^2}{(xy-1)^2}$ $f_y=\frac{-1-x^2}{(xy-1)^2}$

Work Step by Step

Take the first partial derivatives of the given function. When taking a partial derivative with respect to x, treat y as a constant, and vice versa: $f_x=\frac{(xy-1)(1)-(x+y)(y)}{(xy-1)^2}=\frac{-1-y^2}{(xy-1)^2}$ $f_y=\frac{(xy-1)(1)-(x+y)(x)}{(xy-1)^2}=\frac{-1-x^2}{(xy-1)^2}$
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