Chapter 13 - Partial Derivatives - 13.3 Partial Derivatives - Exercises Set 13.3 - Page 939: 92

$${f_{xy}}\left( {x,y} \right) = {f_{yx}}\left( {x,y} \right) = \frac{{8{x^3}y - 8x{y^3}}}{{{{\left( {{x^2} + {y^2}} \right)}^3}}}$$

$$\eqalign{ & f\left( {x,y} \right) = \left( {{x^2} - {y^2}} \right)/\left( {{x^2} + {y^2}} \right) \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 & {\text{use the quotient rule}} \cr & {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left( {\frac{{{x^2} - {y^2}}}{{{x^2} + {y^2}}}} \right) \cr & {f_x}\left( {x,y} \right) = \frac{{\left( {{x^2} + {y^2}} \right)\left( {2x} \right) - \left( {{x^2} - {y^2}} \right)\left( {2x} \right)}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}} \cr & {f_x}\left( {x,y} \right) = \frac{{2{x^3} + 2x{y^2} - 2{x^3} + 2x{y^2}}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}} \cr & {f_x}\left( {x,y} \right) = \frac{{4x{y^2}}}{{{{\left( {{x^2} + {y^2}} \right)}^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( {\frac{{{x^2} - {y^2}}}{{{x^2} + {y^2}}}} \right) \cr & {f_y}\left( {x,y} \right) = \frac{{\left( {{x^2} + {y^2}} \right)\left( { - 2y} \right) - \left( {{x^2} - {y^2}} \right)\left( {2y} \right)}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}} \cr & {f_y}\left( {x,y} \right) = \frac{{ - 2{x^2}y - 2{y^3} - 2{x^2}y + 2{y^3}}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}} \cr & {f_y}\left( {x,y} \right) = \frac{{ - 4{x^2}y}}{{{{\left( {{x^2} + {y^2}} \right)}^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( {\frac{{4x{y^2}}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}}} \right) \cr & {f_{xy}}\left( {x,y} \right) = \frac{{8xy{{\left( {{x^2} + {y^2}} \right)}^2} - 4x{y^2}\left( 2 \right)\left( {{x^2} + {y^2}} \right)\left( {2y} \right)}}{{{{\left( {{x^2} + {y^2}} \right)}^4}}} \cr & {f_{xy}}\left( {x,y} \right) = \frac{{8xy\left( {{x^2} + {y^2}} \right) - 16x{y^3}}}{{{{\left( {{x^2} + {y^2}} \right)}^3}}} \cr & {f_{xy}}\left( {x,y} \right) = \frac{{8{x^3}y + 8x{y^3} - 16x{y^3}}}{{{{\left( {{x^2} + {y^2}} \right)}^3}}} \cr & {f_{xy}}\left( {x,y} \right) = \frac{{8{x^3}y - 8x{y^3}}}{{{{\left( {{x^2} + {y^2}} \right)}^3}}} \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 y}}\left( {\frac{{ - 4{x^2}y}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}}} \right) \cr & {f_{yx}}\left( {x,y} \right) = - \frac{{8xy{{\left( {{x^2} + {y^2}} \right)}^2} - 4{x^2}y\left( 2 \right)\left( {{x^2} + {y^2}} \right)\left( {2x} \right)}}{{{{\left( {{x^2} + {y^2}} \right)}^4}}} \cr & {f_{yx}}\left( {x,y} \right) = - \frac{{8xy\left( {{x^2} + {y^2}} \right) - 16{x^3}y}}{{{{\left( {{x^2} + {y^2}} \right)}^3}}} \cr & {f_{yx}}\left( {x,y} \right) = - \frac{{8{x^3}y + 8x{y^3} - 16{x^3}y}}{{{{\left( {{x^2} + {y^2}} \right)}^3}}} \cr & {f_{yx}}\left( {x,y} \right) = - \frac{{ - 8{x^3}y + 8x{y^3}}}{{{{\left( {{x^2} + {y^2}} \right)}^3}}} \cr & {f_{yx}}\left( {x,y} \right) = \frac{{8{x^3}y - 8x{y^3}}}{{{{\left( {{x^2} + {y^2}} \right)}^3}}} \cr & \cr & {\text{Then}}{\text{,}} \cr & {f_{xy}}\left( {x,y} \right) = {f_{yx}}\left( {x,y} \right) = \frac{{8{x^3}y - 8x{y^3}}}{{{{\left( {{x^2} + {y^2}} \right)}^3}}} \cr} $$