Chapter 9 - Multiveriable Calculus - 9.2 Partial Derivatives - 9.2 Exercises - Page 478: 27

$$\eqalign{ & {z_{xx}}\left( {x,y} \right) = 9{e^x}y,\,\,\,\,{z_{yy}}\left( {x,y} \right) = 0 \cr & {z_{xy}}\left( {x,y} \right) = {z_{yx}}\left( {x,y} \right) = 9{e^x} \cr} $$

$$\eqalign{ & z = 9y{e^x} \cr & {\text{Find }}{z_x}\left( {x,y} \right){\text{ and }}{z_y}\left( {x,y} \right) \cr & {z_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {9y{e^x}} \right] \cr & {\text{treat }}y{\text{ as a constant and }}x{\text{ as a variable}}{\text{. then}} \cr & {z_x}\left( {x,y} \right) = 9y\frac{\partial }{{\partial x}}\left[ {{e^x}} \right] \cr & {z_x}\left( {x,y} \right) = 9{e^x}y \cr & \cr & {z_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {9y{e^x}} \right] \cr & {\text{treat }}x{\text{ as a constant and }}y{\text{ as a variable}}{\text{. then}} \cr & {z_y}\left( {x,y} \right) = 9{e^x}\frac{\partial }{{\partial y}}\left[ y \right] \cr & {z_y}\left( {x,y} \right) = 9{e^x} \cr & \cr & {\text{find the second - order partial derivatives}} \cr & {z_{xy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {9{e^x}y} \right] \cr & {z_{xy}}\left( {x,y} \right) = 9{e^x} \cr & \cr & {z_{yx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {9{e^x}} \right] \cr & {z_{yx}}\left( {x,y} \right) = 9{e^x} \cr & \cr & {z_{xx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {9{e^x}y} \right] \cr & {z_{xx}}\left( {x,y} \right) = 9{e^x}y \cr & {\text{and}} \cr & {z_{yy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {9{e^x}} \right] \cr & {z_{yy}}\left( {x,y} \right) = 0 \cr} $$