Chapter 9 - Multiveriable Calculus - 9.2 Partial Derivatives - 9.2 Exercises - Page 478: 21

$${f_{xx}}\left( {x,y} \right) = 8{y^2} - 32,\,\,\,\,{f_{yy}}\left( {x,y} \right) = 8{x^2},\,\,\,\,{f_{xy}}\left( {x,y} \right) = 16xy,\,\,\,{f_{yx}}\left( {x,y} \right) = 16xy$$

$$\eqalign{ & f\left( {x,y} \right) = 4{x^2}{y^2} - 16{x^2} + 4y \cr & \cr & {\text{Find }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right) \cr & {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {4{x^2}{y^2} - 16{x^2} + 4y} \right] \cr & {\text{treat }}y{\text{ as a constant and }}x{\text{ as a variable}}{\text{. then}} \cr & {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {4{x^2}{y^2}} \right] - \frac{\partial }{{\partial x}}\left[ {16{x^2}} \right] + \frac{\partial }{{\partial x}}\left[ {4y} \right] \cr & {f_x}\left( {x,y} \right) = 8x{y^2} - 32x \cr & \cr & {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {4{x^2}{y^2} - 16{x^2} + 4y} \right] \cr & {\text{treat }}x{\text{ as a constant and }}y{\text{ as a variable}}{\text{. then}} \cr & {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {4{x^2}{y^2}} \right] - \frac{\partial }{{\partial y}}\left[ {16{x^2}} \right] + \frac{\partial }{{\partial y}}\left[ {4y} \right] \cr & {f_y}\left( {x,y} \right) = 8{x^2}y + 4 \cr & \cr & {\text{find the second - order partial derivatives}} \cr & {f_{xy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {8x{y^2} - 32x} \right] \cr & {f_{xy}}\left( {x,y} \right) = 16xy \cr & \cr & {f_{yx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {8{x^2}y + 4} \right] \cr & {f_{yx}}\left( {x,y} \right) = 16xy \cr & \cr & {f_{xx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {8x{y^2} - 32x} \right] \cr & {f_{xx}}\left( {x,y} \right) = 8{y^2} - 32 \cr & {\text{and}} \cr & {f_{yy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {8{x^2}y + 4} \right] \cr & {f_{yy}}\left( {x,y} \right) = 8{x^2} \cr} $$