Chapter 13 - Partial Derivatives - 13.3 Partial Derivatives - Exercises Set 13.3 - Page 939: 88

$${f_{xy}}\left( {x,y} \right) = {f_{yx}}\left( {x,y} \right) = - 2y{e^{x - {y^2}}}$$

$$\eqalign{ & f\left( {x,y} \right) = {e^{x - {y^2}}} \cr & {\text{Find the first derivatives }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right) \cr & \cr & {\text{Calculate }}{f_x}\left( {x,y} \right){\text{ differentiating with respect to }}x;{\text{ treat }}y{\text{ as a constant}} \cr & {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left( {{e^{x - {y^2}}}} \right) \cr & {f_x}\left( {x,y} \right) = {e^{x - {y^2}}}\frac{\partial }{{\partial x}}\left( {x - {y^2}} \right) \cr & {f_x}\left( {x,y} \right) = {e^{x - {y^2}}}\left( 1 \right) \cr & {f_x}\left( {x,y} \right) = {e^{x - {y^2}}} \cr & \cr & {\text{Calculate }}{f_y}\left( {x,y} \right){\text{ differentiating with respect to }}y;{\text{ treat }}x{\text{ as a constant}} \cr & {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left( {{e^{x - {y^2}}}} \right) \cr & {f_y}\left( {x,y} \right) = {e^{x - {y^2}}}\frac{\partial }{{\partial y}}\left( {x - {y^2}} \right) \cr & {f_y}\left( {x,y} \right) = - 2y{e^{x - {y^2}}} \cr & \cr & {\text{Find the mixed second - partial derivatives }}{f_{xy}}\left( {x,y} \right){\text{ and }}{f_{yx}}\left( {x,y} \right){\text{ and}} \cr & {\text{confirm that they are the same}} \cr & {\text{Calculate }}{f_{xy}}\left( {x,y} \right){\text{ differentiating }}{f_x}\left( {x,y} \right){\text{with respect to }}y. \cr & {f_{xy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left( {{e^{x - {y^2}}}} \right) \cr & {f_{xy}}\left( {x,y} \right) = {e^{x - {y^2}}}\left( { - 2y} \right) \cr & {f_{xy}}\left( {x,y} \right) = - 2y{e^{x - {y^2}}} \cr & \cr & {\text{Calculate }}{f_{yx}}\left( {x,y} \right){\text{ differentiating }}{f_y}\left( {x,y} \right){\text{with respect to }}x. \cr & {f_{yx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left( { - 2y{e^{x - {y^2}}}} \right) \cr & {f_{yx}}\left( {x,y} \right) = - 2y{e^{x - {y^2}}}\left( 1 \right) \cr & {f_{yx}}\left( {x,y} \right) = - 2y{e^{x - {y^2}}} \cr & \cr & {\text{Then}}{\text{,}} \cr & {f_{xy}}\left( {x,y} \right) = {f_{yx}}\left( {x,y} \right) = - 2y{e^{x - {y^2}}} \cr} $$