Chapter 13 - Partial Derivatives - 13.3 Partial Derivatives - Exercises Set 13.3 - Page 939: 91

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

$$\eqalign{ & f\left( {x,y} \right) = \left( {x - y} \right)/\left( {x + y} \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 - y}}{{x + y}}} \right) \cr & {f_x}\left( {x,y} \right) = \frac{{\left( {x + y} \right)\left( 1 \right) - \left( {x - y} \right)\left( 1 \right)}}{{{{\left( {x + y} \right)}^2}}} \cr & {f_x}\left( {x,y} \right) = \frac{{x + y - x + y}}{{{{\left( {x + y} \right)}^2}}} \cr & {f_x}\left( {x,y} \right) = \frac{{2y}}{{{{\left( {x + y} \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 - y}}{{x + y}}} \right) \cr & {f_y}\left( {x,y} \right) = \frac{{\left( {x + y} \right)\left( { - 1} \right) - \left( {x - y} \right)\left( 1 \right)}}{{{{\left( {x + y} \right)}^2}}} \cr & {f_y}\left( {x,y} \right) = \frac{{ - x - y - x + y}}{{{{\left( {x + y} \right)}^2}}} \cr & {f_y}\left( {x,y} \right) = - \frac{{2x}}{{{{\left( {x + y} \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{{2y}}{{{{\left( {x + y} \right)}^2}}}} \right) \cr & {f_{xy}}\left( {x,y} \right) = \frac{{2{{\left( {x + y} \right)}^2} - 2y\left( 2 \right)\left( {x + y} \right)}}{{{{\left( {x + y} \right)}^4}}} \cr & {f_{xy}}\left( {x,y} \right) = \frac{{2{{\left( {x + y} \right)}^2} - 4y\left( {x + y} \right)}}{{{{\left( {x + y} \right)}^4}}} \cr & {f_{xy}}\left( {x,y} \right) = \frac{{2\left( {x + y} \right) - 4y}}{{{{\left( {x + y} \right)}^3}}} \cr & {f_{xy}}\left( {x,y} \right) = \frac{{2x + 2y - 4y}}{{{{\left( {x + y} \right)}^3}}} \cr & {f_{xy}}\left( {x,y} \right) = \frac{{2x - 2y}}{{{{\left( {x + y} \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{{2x}}{{{{\left( {x + y} \right)}^2}}}} \right) \cr & {f_{yx}}\left( {x,y} \right) = - \frac{{2{{\left( {x + y} \right)}^2} - 2x\left( 2 \right)\left( {x + y} \right)}}{{{{\left( {x + y} \right)}^4}}} \cr & {f_{yx}}\left( {x,y} \right) = - \frac{{2\left( {x + y} \right) - 4x}}{{{{\left( {x + y} \right)}^3}}} \cr & {f_{yx}}\left( {x,y} \right) = - \frac{{2x + 2y - 4x}}{{{{\left( {x + y} \right)}^3}}} \cr & {f_{yx}}\left( {x,y} \right) = \frac{{2x - 2y}}{{{{\left( {x + y} \right)}^3}}} \cr & \cr & {\text{Then}}{\text{,}} \cr & {f_{xy}}\left( {x,y} \right) = {f_{yx}}\left( {x,y} \right) = \frac{{2x - 2y}}{{{{\left( {x + y} \right)}^3}}} \cr} $$