Chapter 13 - Partial Derivatives - 13.3 Partial Derivatives - Exercises Set 13.3 - Page 937: 29

$$\frac{{\partial z}}{{\partial x}} = - \frac{{y\left( {{x^2} - {y^2}} \right)}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}}{\text{ and }}\frac{{\partial z}}{{\partial y}} = \frac{{x\left( {{x^2} - {y^2}} \right)}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}}$$

$$\eqalign{ & z = \frac{{xy}}{{{x^2} + {y^2}}} \cr & \cr & {\text{Calculate the partial derivative }}\frac{{\partial z}}{{\partial x}}{\text{ }} \cr & {\text{differentiating both sides with respect to }}x \cr & \frac{{\partial z}}{{\partial x}} = \frac{\partial }{{\partial x}}\left[ {\frac{{xy}}{{{x^2} + {y^2}}}} \right] \cr & {\text{Applying the quotient rule for derivatives}} \cr & \frac{{\partial z}}{{\partial x}} = \frac{{\left( {{x^2} + {y^2}} \right)\frac{\partial }{{\partial x}}\left[ {xy} \right] - xy\frac{\partial }{{\partial x}}\left[ {{x^2} + {y^2}} \right]}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}} \cr & {\text{Solving the derivatives}} \cr & \frac{{\partial z}}{{\partial x}} = \frac{{\left( {{x^2} + {y^2}} \right)\left( y \right) - xy\left( {2x} \right)}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}} \cr & {\text{Simplifying the resultant derivative}} \cr & \frac{{\partial z}}{{\partial x}} = \frac{{{x^2}y + {y^3} - 2{x^2}y}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}} \cr & \frac{{\partial z}}{{\partial x}} = \frac{{{y^3} - {x^2}y}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}} \cr & \frac{{\partial z}}{{\partial x}} = - \frac{{y\left( {{x^2} - {y^2}} \right)}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}} \cr & \cr & and \cr & {\text{Calculate the partial derivative }}\frac{{\partial z}}{{\partial y}}{\text{ }} \cr & {\text{differentiating both sides with respect to }}y \cr & \frac{{\partial z}}{{\partial y}} = \frac{\partial }{{\partial y}}\left[ {\frac{{xy}}{{{x^2} + {y^2}}}} \right] \cr & {\text{Applying the quotient rule for derivatives}} \cr & \frac{{\partial z}}{{\partial y}} = \frac{{\left( {{x^2} + {y^2}} \right)\frac{\partial }{{\partial y}}\left[ {xy} \right] - xy\frac{\partial }{{\partial y}}\left[ {{x^2} + {y^2}} \right]}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}} \cr & {\text{Solving the derivative}} \cr & \frac{{\partial z}}{{\partial y}} = \frac{{\left( {{x^2} + {y^2}} \right)\left( x \right) - xy\left( {2y} \right)}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}} \cr & {\text{Simplifying the resultant derivative}} \cr & \frac{{\partial z}}{{\partial y}} = \frac{{{x^3} + x{y^2} - 2x{y^2}}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}} \cr & \frac{{\partial z}}{{\partial y}} = \frac{{{x^3} - x{y^2}}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}} \cr & \frac{{\partial z}}{{\partial y}} = \frac{{x\left( {{x^2} - {y^2}} \right)}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}} \cr & \cr & {\text{The partial derivatives are:}} \cr & \frac{{\partial z}}{{\partial x}} = - \frac{{y\left( {{x^2} - {y^2}} \right)}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}}{\text{ and }}\frac{{\partial z}}{{\partial y}} = \frac{{x\left( {{x^2} - {y^2}} \right)}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}} \cr} $$