Chapter 14 - Section 14.3 - Partial Derivatives - 14.3 Exercise - Page 970: 15

${f_x}\left( {x,y} \right) = {y^2}{e^{xy}}{\text{ and }}{f_y}\left( {x,y} \right) = \left( {xy + 1} \right){e^{xy}}$

$$\eqalign{ & f\left( {x,y} \right) = y{e^{xy}} \cr & {\text{Differentiate with respect to }}x \cr & {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {y{e^{xy}}} \right] \cr & {\text{Considering }}y{\text{ as a constant}} \cr & {f_x}\left( {x,y} \right) = y\frac{\partial }{{\partial x}}\left[ {{e^{xy}}} \right] \cr & {\text{Using the Chain Rule}} \cr & {f_x}\left( {x,y} \right) = y{e^{xy}}\frac{\partial }{{\partial x}}\left[ {xy} \right] \cr & {f_x}\left( {x,y} \right) = {y^2}{e^{xy}}\frac{\partial }{{\partial x}}\left[ x \right] \cr & {f_x}\left( {x,y} \right) = {y^2}{e^{xy}}\left( 1 \right) \cr & {f_x}\left( {x,y} \right) = {y^2}{e^{xy}} \cr & \cr & {\text{Differentiate with respect to }}y \cr & {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {y{e^{xy}}} \right] \cr & {\text{Considering }}x{\text{ as a constant and apply the product rule}} \cr & {f_y}\left( {x,y} \right) = y\frac{\partial }{{\partial y}}\left[ {{e^{xy}}} \right] + {e^{xy}}\frac{\partial }{{\partial y}}\left[ y \right] \cr & {\text{Using the Chain Rule}} \cr & {f_y}\left( {x,y} \right) = y{e^{xy}}\frac{\partial }{{\partial y}}\left[ {xy} \right] + {e^{xy}}\frac{\partial }{{\partial y}}\left[ y \right] \cr & {f_y}\left( {x,y} \right) = y{e^{xy}}\left( x \right) + {e^{xy}}\left( 1 \right) \cr & {\text{Simplify}} \cr & {f_y}\left( {x,y} \right) = xy{e^{xy}} + {e^{xy}} \cr & {f_y}\left( {x,y} \right) = \left( {xy + 1} \right){e^{xy}} \cr} $$