Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 15 - Differentiation in Several Variables - 15.3 Partial Derivatives - Exercises - Page 781: 35


$$ z_x= 2xy\cosh (x^2y),\quad z_y= x^2\cosh (x^2y). $$

Work Step by Step

Recall that $(\sinh x)'=\cosh x$ Recall that $(x^n)'=nx^{n-1}$ Since $ z=\sinh (x^2y)$, by using the chain rule, we have $$ z_x= \cosh(x^2y) (2xy)=2xy\cosh (x^2y),\\ z_y= \cosh(x^2y) (x^2)=x^2\cosh (x^2y). $$
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