University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 13 - Section 13.3 - Partial Derivatives - Exercises - Page 702: 42

Answer

$$\dfrac{\partial^2 f}{\partial x^2}=-y^2 \sin (xy) \\\dfrac{\partial^2 f}{\partial y^2}=-x^2 \sin (xy) \\ \dfrac{\partial^2 f}{\partial x\partial y}=\dfrac{\partial^2 f}{\partial y \partial x }=-xy \sin xy+\cos xy$$

Work Step by Step

Our aim is to take the first partial derivative of the given function $f(x,y)$ with respect to one of its variables, by treating all others as a constant. $f_x= y \cos (xy) $ We repeat the process for the other partial derivative. $f_y=x \cos (xy) $ Now, compute the second partial derivatives: $\dfrac{\partial^2 f}{\partial x^2}=\dfrac{\partial }{\partial x} (y \cos (xy)=-y^2 \sin xy$ $\dfrac{\partial^2 f}{\partial y^2}=\dfrac{\partial }{\partial y} [x \cos (xy)]=-x^2 \sin xy$ Next, $\dfrac{\partial^2 f}{\partial x\partial y}=\dfrac{\partial^2 f}{\partial y \partial x }=\dfrac{\partial }{\partial y } [y \cos (xy)])=-x \space y \sin x \space y+\cos x \space y$
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