Chapter 9 - Multiveriable Calculus - 9.2 Partial Derivatives - 9.2 Exercises - Page 478: 15

$${f_x}\left( {x,y} \right) = 2{x^2}y{e^{{x^2}y}} + {e^{{x^2}y}},\,\,\,\,{f_y}\left( {x,y} \right) = {x^3}{e^{{x^2}y}},\,\,\,\,\,{f_x}\left( {2, - 1} \right) = - 7{e^{ - 4}}{\text{ and }}{f_y}\left( { - 4,3} \right) = - 64{e^{48}}$$

$$\eqalign{ & f\left( {x,y} \right) = x{e^{{x^2}y}} \cr & {\text{Find }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right) \cr & \cr & {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {x{e^{{x^2}y}}} \right] \cr & {\text{treating y as a constant and }}x{\text{ as a variable use }}\frac{d}{{dx}}\left[ {{e^u}} \right] = {e^u}\frac{{du}}{{dx}}{\text{ and product rule}}{\text{. then}} \cr & {f_x}\left( {x,y} \right) = x\frac{\partial }{{\partial x}}\left[ {{e^{{x^2}y}}} \right] + {e^{{x^2}y}}\frac{\partial }{{\partial x}}\left[ x \right] \cr & {f_x}\left( {x,y} \right) = x{e^{{x^2}y}}\frac{\partial }{{\partial x}}\left[ {{x^2}y} \right] + {e^{{x^2}y}}\frac{\partial }{{\partial x}}\left[ x \right] \cr & {\text{solve derivatives}} \cr & {f_x}\left( {x,y} \right) = x{e^{{x^2}y}}\left( {2xy} \right) + {e^{{x^2}y}}\left( 1 \right) \cr & {f_x}\left( {x,y} \right) = 2{x^2}y{e^{{x^2}y}} + {e^{{x^2}y}} \cr & {\text{evaluate }}{f_x}\left( {2, - 1} \right) \cr & {f_x}\left( {2, - 1} \right) = 2{\left( 2 \right)^2}\left( { - 1} \right){e^{{{\left( 2 \right)}^2}\left( { - 1} \right)}} + {e^{{{\left( 2 \right)}^2}\left( { - 1} \right)}} \cr & {f_x}\left( {2, - 1} \right) = - 8{e^{ - 4}} + {e^{ - 4}} \cr & {f_x}\left( {2, - 1} \right) = - 7{e^{ - 4}} \cr & \cr & {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {x{e^{{x^2}y}}} \right] \cr & {\text{treat x as a constant and y as a variable}}{\text{. then}} \cr & {f_y}\left( {x,y} \right) = x{e^{{x^2}y}}\frac{\partial }{{\partial y}}\left[ {{x^2}y} \right] \cr & {f_y}\left( {x,y} \right) = x{e^{{x^2}y}}\left( {{x^2}} \right) \cr & {f_y}\left( {x,y} \right) = {x^3}{e^{{x^2}y}} \cr & {\text{evaluate }}{f_y}\left( { - 4,3} \right) \cr & {f_y}\left( { - 4,3} \right) = {\left( { - 4} \right)^3}{e^{{{\left( { - 4} \right)}^2}\left( 3 \right)}} \cr & {f_y}\left( { - 4,3} \right) = - 64{e^{48}} \cr} $$