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: 42


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

Work Step by Step

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= \dfrac{\partial (\sin xy)}{\partial x}=y \cos (xy) $ $f_y=\dfrac{\partial (\sin xy)}{\partial y}=x \cos (xy) $ $\dfrac{\partial^2 f}{\partial x^2}=\dfrac{\partial (y \cos (xy)}{\partial x}=-y^2 \sin (xy)$ $\dfrac{\partial^2 f}{\partial y^2}=\dfrac{\partial [x \cos (xy)]}{\partial y}=-x^2 \sin (xy)$ Now, $\dfrac{\partial^2 f}{\partial x\partial y}=\dfrac{\partial^2 f}{\partial y \partial x }=\dfrac{\partial [y \cos (xy)]) }{\partial y } =-xy \sin xy+\cos xy$
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