Chapter 13 - Partial Derivatives - 13.3 Partial Derivatives - Exercises Set 13.3 - Page 939: 90

$${f_{xy}}\left( {x,y} \right) = {f_{yx}}\left( {x,y} \right) =- \frac{{4xy}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}}$$

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