Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 9 - Multiveriable Calculus - 9.2 Partial Derivatives - 9.2 Exercises - Page 478: 22

Answer

$${g_{xx}}\left( {x,y} \right) = 60{x^2}{y^2},\,\,\,\,{g_{yy}}\left( {x,y} \right) = 40{x^4} + 72y,\,\,\,\,{g_{xy}}\left( {x,y} \right) = \,{g_{yx}}\left( {x,y} \right) = 140{x^3}y$$

Work Step by Step

$$\eqalign{ & g\left( {x,y} \right) = 5{x^4}{y^2} + 12{y^3} - 9x \cr & {\text{Find }}{g_x}\left( {x,y} \right){\text{ and }}{g_y}\left( {x,y} \right) \cr & {g_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {5{x^4}{y^2} + 12{y^3} - 9x} \right] \cr & {\text{treat }}y{\text{ as a constant and }}x{\text{ as a variable}}{\text{. then}} \cr & {g_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {5{x^4}{y^2}} \right] + \frac{\partial }{{\partial x}}\left[ {12{y^3}} \right] - \frac{\partial }{{\partial x}}\left[ {9x} \right] \cr & {g_x}\left( {x,y} \right) = 20{x^3}{y^2} - 9 \cr & \cr & {g_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {5{x^4}{y^2} + 12{y^3} - 9x} \right] \cr & {\text{treat }}x{\text{ as a constant and }}y{\text{ as a variable}}{\text{. then}} \cr & {g_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {5{x^4}{y^2}} \right] + \frac{\partial }{{\partial y}}\left[ {12{y^3}} \right] - \frac{\partial }{{\partial y}}\left[ {9x} \right] \cr & {g_y}\left( {x,y} \right) = 10{x^4}y + 36{y^2} \cr & \cr & {\text{find the second - order partial derivatives}} \cr & {g_{xy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {20{x^3}{y^2} - 9} \right] \cr & {g_{xy}}\left( {x,y} \right) = 40{x^3}y \cr & \cr & {g_{yx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {10{x^4}y + 36{y^2}} \right] \cr & {g_{yx}}\left( {x,y} \right) = 40{x^3}y \cr & \cr & {g_{xx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {20{x^3}{y^2} - 9} \right] \cr & {g_{xx}}\left( {x,y} \right) = 60{x^2}{y^2} \cr & {\text{and}} \cr & {g_{yy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {10{x^4}y + 36{y^2}} \right] \cr & {g_{yy}}\left( {x,y} \right) = 40{x^4} + 72y \cr} $$
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