Chapter 9 - Multiveriable Calculus - 9.2 Partial Derivatives - 9.2 Exercises - Page 478: 16

$$\eqalign{ & {f_x}\left( {x,y} \right) = {y^2}{e^{x + 3y}} + {e^{x + 3y}},\,\,\,\,{f_y}\left( {x,y} \right) = 3{e^{x + 3y}}{y^2} + 2y{e^{x + 3y}},\,\,\,\,\, \cr & {f_x}\left( {2, - 1} \right) = 2{e^{ - 1}}{\text{ and }}{f_y}\left( { - 4,3} \right) = 33{e^5} \cr} $$

$$\eqalign{ & f\left( {x,y} \right) = {y^2}{e^{x + 3y}} \cr & {\text{Find }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right) \cr & \cr & {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {{y^2}{e^{x + 3y}}} \right] \cr & {\text{treating y as a constant and }}x{\text{ as a variable use }}\frac{d}{{dx}}\left[ {{e^u}} \right] = {e^u}\frac{{du}}{{dx}}{\text{ and product rule}}{\text{. then}} \cr & {f_x}\left( {x,y} \right) = {y^2}\frac{\partial }{{\partial x}}\left[ {{e^{x + 3y}}} \right]{e^{x + 3y}}\frac{\partial }{{\partial x}}\left[ {x + 3y} \right] \cr & {f_x}\left( {x,y} \right) = {y^2}{e^{x + 3y}}\frac{\partial }{{\partial x}}\left[ {x + 3y} \right] + {e^{x + 3y}}\frac{\partial }{{\partial x}}\left[ {x + 3y} \right] \cr & {\text{solve derivatives}} \cr & {f_x}\left( {x,y} \right) = {y^2}{e^{x + 3y}} + {e^{x + 3y}} \cr & {\text{evaluate }}{f_x}\left( {2, - 1} \right) \cr & {f_x}\left( {2, - 1} \right) = {\left( { - 1} \right)^2}{e^{2 + 3\left( { - 1} \right)}} + {e^{2 + 3\left( { - 1} \right)}} \cr & {f_x}\left( {2, - 1} \right) = 2{e^{ - 1}} \cr & \cr & {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {{y^2}{e^{x + 3y}}} \right] \cr & {\text{treating }}x{\text{ as a constant and }}y{\text{ as a variable use }}\frac{d}{{dx}}\left[ {{e^u}} \right] = {e^u}\frac{{du}}{{dx}}{\text{ and product rule}}{\text{. then}} \cr & {f_y}\left( {x,y} \right) = {y^2}\frac{\partial }{{\partial y}}\left[ {{e^{x + 3y}}} \right] + {e^{x + 3y}}\frac{\partial }{{\partial y}}\left[ {{y^2}} \right] \cr & {f_y}\left( {x,y} \right) = 3{e^{x + 3y}}{y^2} + 2y{e^{x + 3y}} \cr & {\text{evaluate }}{f_y}\left( { - 4,3} \right) \cr & {f_y}\left( { - 4,3} \right) = 3{e^{ - 4 + 3\left( 3 \right)}}{\left( 3 \right)^2} + 2\left( 3 \right){e^{ - 4 + 3\left( 3 \right)}} \cr & {f_y}\left( { - 4,3} \right) = 27{e^5} + 6{e^5} \cr & {f_y}\left( { - 4,3} \right) = 33{e^5} \cr} $$