## Calculus with Applications (10th Edition)

$${\text{ }}{f_{xx}}\left( {x,y} \right) = 4{x^2}y{e^{{x^2}}} + 2y{e^{{x^2}}}{\text{ and }}{f_{xy}}\left( {x,y} \right) = 2x{e^{{x^2}}}$$
\eqalign{ & f\left( {x,y} \right) = y{e^{{x^2}}} \cr & {\text{find }}{f_x}\left( {x,y} \right) \cr & {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {y{e^{{x^2}}}} \right] \cr & {\text{treat y as a constant and }}x{\text{ as a variable}}{\text{. then}} \cr & {f_x}\left( {x,y} \right) = y\frac{\partial }{{\partial x}}\left[ {{e^{{x^2}}}} \right] \cr & {\text{solving derivatives}} \cr & {f_x}\left( {x,y} \right) = y{e^{{x^2}}}\frac{\partial }{{\partial x}}\left[ {{x^2}} \right] \cr & {f_x}\left( {x,y} \right) = 2xy{e^{{x^2}}} \cr & \cr & {\text{find }}{f_{xx}}\left( {x,y} \right) \cr & {\text{ }}{f_{xx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {2xy{e^{{x^2}}}} \right] \cr & {\text{by using the product rule}} \cr & {\text{ }}{f_{xx}}\left( {x,y} \right) = 2xy\frac{\partial }{{\partial x}}\left[ {{e^{{x^2}}}} \right] + 2y{e^{{x^2}}}\frac{\partial }{{\partial x}}\left[ x \right] \cr & {\text{ }}{f_{xx}}\left( {x,y} \right) = 2xy\left( {2x{e^{{x^2}}}} \right) + 2y{e^{{x^2}}} \cr & {\text{ }}{f_{xx}}\left( {x,y} \right) = 4{x^2}y{e^{{x^2}}} + 2y{e^{{x^2}}} \cr & \cr & {\text{find }}{f_{xy}}\left( {x,y} \right) \cr & {\text{differentiate }}{f_x}\left( {x,y} \right){\text{ with respect to }}y \cr & {f_{xy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {2xy{e^{{x^2}}}} \right] \cr & {\text{treat }}x{\text{ as a constant and }}y{\text{ as a variable}}{\text{. then}} \cr & {f_{xy}}\left( {x,y} \right) = 2x{e^{{x^2}}}\frac{\partial }{{\partial y}}\left[ y \right] \cr & {f_{xy}}\left( {x,y} \right) = 2x{e^{{x^2}}}\left( 1 \right) \cr & {f_{xy}}\left( {x,y} \right) = 2x{e^{{x^2}}} \cr}