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: 34


$y \cosh (xy-z^2)$ ; $x \cosh (xy-z^2)$; $-2z \cosh (xy-z^2)$

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: $f_x=\cosh (xy-z^2) \times (y) =y \cosh (xy-z^2)$ $f_y=\cosh (xy-z^2) \times (x) =x \cosh (xy-z^2)$ $f_z= \cosh (xy-z^2) \times (-2z) =-2z \cosh (xy-z^2)$
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