Chapter 13 - Partial Derivatives - 13.3 Partial Derivatives - Exercises Set 13.3 - Page 939: 86

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

$$\eqalign{ & f\left( {x,y} \right) = \sqrt {{x^2} + {y^2}} \cr & \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( {\sqrt {{x^2} + {y^2}} } \right) \cr & {\text{Recall that }}\frac{d}{{dx}}\left[ {\sqrt u } \right] = \frac{{u'}}{{2\sqrt u }} \cr & {f_x}\left( {x,y} \right) = \frac{{2x}}{{2\sqrt {{x^2} + {y^2}} }} \cr & {f_x}\left( {x,y} \right) = \frac{x}{{\sqrt {{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( {\sqrt {{x^2} + {y^2}} } \right) \cr & {f_y}\left( {x,y} \right) = \frac{{2y}}{{2\sqrt {{x^2} + {y^2}} }} \cr & {f_y}\left( {x,y} \right) = \frac{y}{{\sqrt {{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{x}{{\sqrt {{x^2} + {y^2}} }}} \right) \cr & {f_{xy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left( {x{{\left( {{x^2} + {y^2}} \right)}^{ - 1/1}}} \right) \cr & {\text{By using the chain rule}} \cr & {f_{xy}}\left( {x,y} \right) = x\frac{\partial }{{\partial y}}\left( {{{\left( {{x^2} + {y^2}} \right)}^{ - 1/2}}} \right) \cr & {f_{xy}}\left( {x,y} \right) = - \frac{1}{2}x{\left( {{x^2} + {y^2}} \right)^{ - 3/2}}\left( {2y} \right) \cr & {\text{Simplifying}} \cr & {f_{xy}}\left( {x,y} \right) = - \frac{{xy}}{{{{\left( {{x^2} + {y^2}} \right)}^{3/2}}}} \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( {\frac{y}{{\sqrt {{x^2} + {y^2}} }}} \right) \cr & {\text{By using the chain rule}} \cr & {f_{yx}}\left( {x,y} \right) = y\frac{\partial }{{\partial x}}\left( {{{\left( {{x^2} + {y^2}} \right)}^{ - 1/2}}} \right) \cr & {f_{yx}}\left( {x,y} \right) = - \frac{1}{2}y{\left( {{x^2} + {y^2}} \right)^{ - 3/2}}\left( {2x} \right) \cr & {\text{Simplifying}} \cr & {f_{yx}}\left( {x,y} \right) = - \frac{{xy}}{{{{\left( {{x^2} + {y^2}} \right)}^{3/2}}}} \cr & \cr & {\text{Then}}{\text{,}} \cr & {f_{xy}}\left( {x,y} \right) = {f_{yx}}\left( {x,y} \right) = - \frac{{xy}}{{{{\left( {{x^2} + {y^2}} \right)}^{3/2}}}} \cr} $$