Chapter 9 - Multiveriable Calculus - 9.2 Partial Derivatives - 9.2 Exercises - Page 478: 5

$${f_x}\left( {x,y} \right) = 10x{y^3},\,\,\,\,{f_y}\left( {x,y} \right) = 15{x^2}{y^2},\,\,\,\,\,{f_x}\left( {2, - 1} \right) = - 20{\text{ and }}{f_y}\left( { - 4,3} \right) = 2160$$

$$\eqalign{ & f\left( {x,y} \right) = 5{x^2}{y^3} \cr & {\text{Find }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right) \cr & \cr & {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {5{x^2}{y^3}} \right] \cr & {\text{treat y as a constant and }}x{\text{ as a variable}}{\text{. then}} \cr & {f_x}\left( {x,y} \right) = 5{y^3}\frac{\partial }{{\partial x}}\left[ {{x^2}} \right] \cr & {f_x}\left( {x,y} \right) = 5{y^3}\left( {2x} \right) \cr & {f_x}\left( {x,y} \right) = 10x{y^3} \cr & {\text{evaluate }}{f_x}\left( {2, - 1} \right) \cr & {f_x}\left( {2, - 1} \right) = 10\left( 2 \right){\left( { - 1} \right)^3} \cr & {f_x}\left( {2, - 1} \right) = - 20 \cr & \cr & {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {5{x^2}{y^3}} \right] \cr & {\text{treat x as a constant and y as a variable}}{\text{. then}} \cr & {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {5{x^2}{y^3}} \right] \cr & {f_y}\left( {x,y} \right) = 5{x^2}\frac{\partial }{{\partial y}}\left[ {{y^3}} \right] \cr & {f_y}\left( {x,y} \right) = 5{x^2}\left( {3{y^2}} \right) \cr & {f_y}\left( {x,y} \right) = 15{x^2}{y^2} \cr & {\text{evaluate }}{f_y}\left( { - 4,3} \right) \cr & {f_y}\left( { - 4,3} \right) = 15{\left( { - 4} \right)^2}{\left( 3 \right)^2} \cr & {f_y}\left( { - 4,3} \right) = 15\left( {16} \right)\left( 9 \right) \cr & {f_y}\left( { - 4,3} \right) = 2160 \cr} $$