Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 13 - Partial Derivatives - 13.3 Partial Derivatives - Exercises Set 13.3 - Page 939: 87

Answer

$${f_{xy}}\left( {x,y} \right) = {f_{yx}}\left( {x,y} \right) = - {e^x}\sin y$$

Work Step by Step

$$\eqalign{ & f\left( {x,y} \right) = {e^x}\cos y \cr & \cr & {\text{Find the first derivatives }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right) \cr & \cr & {\text{Calculate }}{f_x}\left( {x,y} \right){\text{ differentiating with respect to }}x;{\text{ treat }}y{\text{ as a constant}} \cr & {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left( {{e^x}\cos y} \right) \cr & {f_x}\left( {x,y} \right) = \cos y\frac{\partial }{{\partial x}}\left( {{e^x}} \right) \cr & {f_x}\left( {x,y} \right) = {e^x}\cos y \cr & \cr & {\text{Calculate }}{f_y}\left( {x,y} \right){\text{ differentiating with respect to }}y;{\text{ treat }}x{\text{ as a constant}} \cr & {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left( {{e^x}\cos y} \right) \cr & {f_y}\left( {x,y} \right) = {e^x}\frac{\partial }{{\partial y}}\left( {\cos y} \right) \cr & {f_y}\left( {x,y} \right) = - {e^x}\sin y \cr & \cr & {\text{Find the mixed second - partial derivatives }}{f_{xy}}\left( {x,y} \right){\text{ and }}{f_{yx}}\left( {x,y} \right){\text{ and}} \cr & {\text{confirm that they are the same}} \cr & {\text{Calculate }}{f_{xy}}\left( {x,y} \right){\text{ differentiating }}{f_x}\left( {x,y} \right){\text{with respect to }}y. \cr & {f_{xy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left( {{e^x}\cos y} \right) \cr & {f_{xy}}\left( {x,y} \right) = {e^x}\left( { - \sin y} \right) \cr & {f_{xy}}\left( {x,y} \right) = - {e^x}\sin y \cr & \cr & {\text{Calculate }}{f_{yx}}\left( {x,y} \right){\text{ differentiating }}{f_y}\left( {x,y} \right){\text{with respect to }}x. \cr & {f_{yx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left( { - {e^x}\sin y} \right) \cr & {f_{yx}}\left( {x,y} \right) = - \sin y\frac{\partial }{{\partial x}}\left( {{e^x}} \right) \cr & {f_{yx}}\left( {x,y} \right) = - {e^x}\sin y \cr & \cr & {\text{Then}}{\text{,}} \cr & {f_{xy}}\left( {x,y} \right) = {f_{yx}}\left( {x,y} \right) = - {e^x}\sin y \cr} $$
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