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: 33


$sech^2 (x+2y+3z)$; $2sech^2 (x+2y+3z)$; $3 \space sech^2 (x+2y+3z)$

Work Step by Step

We need to take the first partial derivatives of the given function. In order to find the partial derivative, we will differentiate with respect to $x$, by keeping $y$ and $z$ as a constant, and vice versa: $f_x=sech^2 (x+2y+3z) \times \dfrac{\partial (x+2y-3z)}{\partial x} =sech^2 (x+2y+3z)$ $f_y=sech^2 (x+2y+3z) \times \dfrac{\partial (x+2y-3z)}{\partial y} =2sech^2 (x+2y+3z)$ $f_z= sech^2 (x+2y+3z) \times \dfrac{\partial (x+2y-3z)}{\partial z} =3 \space sech^2 (x+2y+3z)$
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