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: 25

Answer

$f_x=1$ $f_y=-1/2(y^2+z^2)^{-1/2}\times 2y$ $f_z=-1/2(y^2+z^2)^{-1/2}\times 2z$

Work Step by Step

Take the first partial derivatives of the given function. When taking a partial derivative with respect to x, treat y and z as constants, with respect to y, treat x and z as constants, and with respect to z, treat x and y as constants: $f_x=1$ $f_y=-1/2(y^2+z^2)^{-1/2}\times 2y$ $f_z=-1/2(y^2+z^2)^{-1/2}\times 2z$
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