Chapter 9 - Multiveriable Calculus - 9.2 Partial Derivatives - 9.2 Exercises - Page 478: 23

$$\eqalign{ & {R_{xx}}\left( {x,y} \right) = 8 + 24{y^2},\,\,\,\,{R_{yy}}\left( {x,y} \right) = - 30xy + 24{x^2},\,\,\,\, \cr & {R_{xy}}\left( {x,y} \right) = {R_{yx}}\left( {x,y} \right) = - 15{y^2} + 48xy \cr} $$

$$\eqalign{ & R\left( {x,y} \right) = 4{x^2} - 5x{y^3} + 12{x^2}{y^2} \cr & {\text{Find }}{R_x}\left( {x,y} \right){\text{ and }}{R_y}\left( {x,y} \right) \cr & {R_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {4{x^2} - 5x{y^3} + 12{x^2}{y^2}} \right] \cr & {\text{treat }}y{\text{ as a constant and }}x{\text{ as a variable}}{\text{. then}} \cr & {R_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {4{x^2}} \right] - \frac{\partial }{{\partial x}}\left[ {5x{y^3}} \right] + \frac{\partial }{{\partial x}}\left[ {12{x^2}{y^2}} \right] \cr & {R_x}\left( {x,y} \right) = 8x - 5{y^3} + 24x{y^2} \cr & \cr & {R_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {4{x^2} - 5x{y^3} + 12{x^2}{y^2}} \right] \cr & {\text{treat }}x{\text{ as a constant and }}y{\text{ as a variable}}{\text{. then}} \cr & {R_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {4{x^2}} \right] - \frac{\partial }{{\partial y}}\left[ {5x{y^3}} \right] + \frac{\partial }{{\partial y}}\left[ {12{x^2}{y^2}} \right] \cr & {R_y}\left( {x,y} \right) = - 15x{y^2} + 24{x^2}y \cr & \cr & {\text{find the second - order partial derivatives}} \cr & {R_{xy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {8x - 5{y^3} + 24x{y^2}} \right] \cr & {R_{xy}}\left( {x,y} \right) = - 15{y^2} + 48xy \cr & \cr & {R_{yx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ { - 15x{y^2} + 24{x^2}y} \right] \cr & {R_{yx}}\left( {x,y} \right) = - 15{y^2} + 48xy \cr & \cr & {R_{xx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {8x - 5{y^3} + 24x{y^2}} \right] \cr & {R_{xx}}\left( {x,y} \right) = 8 + 24{y^2} \cr & {\text{and}} \cr & {R_{yy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ { - 15x{y^2} + 24{x^2}y} \right] \cr & {R_{yy}}\left( {x,y} \right) = - 30xy + 24{x^2} \cr} $$