Chapter 13 - Partial Derivatives - 13.3 Partial Derivatives - Exercises Set 13.3 - Page 937: 30

$$\frac{{\partial z}}{{\partial x}} = \frac{{x{y^3}\left( {3x + 4y} \right)}}{{2{{\left( {x + y} \right)}^{3/2}}}}{\text{ and }}\frac{{\partial z}}{{\partial y}} = \frac{{6{x^3}{y^2} + 5{x^2}{y^3}}}{{2{{\left( {x + y} \right)}^{3/2}}}}$$

$$\eqalign{ & z = \frac{{{x^2}{y^3}}}{{\sqrt {x + y} }} \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{{{x^2}{y^3}}}{{\sqrt {x + y} }}} \right] \cr & {\text{Applying the quotient rule for derivatives}} \cr & \frac{{\partial z}}{{\partial x}} = \frac{{\sqrt {x + y} \frac{\partial }{{\partial x}}\left[ {{x^2}{y^3}} \right] - {x^2}{y^3}\frac{\partial }{{\partial x}}\left[ {\sqrt {x + y} } \right]}}{{{{\left( {\sqrt {x + y} } \right)}^2}}} \cr & {\text{Solving the derivatives}} \cr & \frac{{\partial z}}{{\partial x}} = \frac{{\sqrt {x + y} \left( {2x{y^3}} \right) - {x^2}{y^3}\left( {\frac{1}{{2\sqrt {x + y} }}} \right)}}{{{{\left( {\sqrt {x + y} } \right)}^2}}} \cr & {\text{Simplifying the resultant derivative}} \cr & \frac{{\partial z}}{{\partial x}} = \frac{{2x{y^3}\sqrt {x + y} - \frac{{{x^2}{y^3}}}{{2\sqrt {x + y} }}}}{{x + y}} \cr & \frac{{\partial z}}{{\partial x}} = \frac{{2\sqrt {x + y} \left( {2x{y^3}\sqrt {x + y} - \frac{{{x^2}{y^3}}}{{2\sqrt {x + y} }}} \right)}}{{2\sqrt {x + y} \left( {x + y} \right)}} \cr & \frac{{\partial z}}{{\partial x}} = \frac{{4x{y^3}\left( {x + y} \right) - {x^2}{y^3}}}{{2{{\left( {x + y} \right)}^{3/2}}}} \cr & \frac{{\partial z}}{{\partial x}} = \frac{{4{x^2}{y^3} + 4x{y^4} - {x^2}{y^3}}}{{2{{\left( {x + y} \right)}^{3/2}}}} \cr & \frac{{\partial z}}{{\partial x}} = \frac{{3{x^2}{y^3} + 4x{y^4}}}{{2{{\left( {x + y} \right)}^{3/2}}}} \cr & {\text{factoring}} \cr & \frac{{\partial z}}{{\partial x}} = \frac{{x{y^3}\left( {3x + 4y} \right)}}{{2{{\left( {x + y} \right)}^{3/2}}}} \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{{{x^2}{y^3}}}{{\sqrt {x + y} }}} \right] \cr & {\text{Applying the quotient rule for derivatives}} \cr & \frac{{\partial z}}{{\partial y}} = \frac{{\sqrt {x + y} \frac{\partial }{{\partial y}}\left[ {{x^2}{y^3}} \right] - {x^2}{y^3}\frac{\partial }{{\partial y}}\left[ {\sqrt {x + y} } \right]}}{{{{\left( {\sqrt {x + y} } \right)}^2}}} \cr & {\text{Solving the derivatives}} \cr & \frac{{\partial z}}{{\partial y}} = \frac{{\sqrt {x + y} \left( {3{x^2}{y^2}} \right) - {x^2}{y^3}\left( {\frac{1}{{2\sqrt {x + y} }}} \right)}}{{{{\left( {\sqrt {x + y} } \right)}^2}}} \cr & {\text{Simplifying the resultant derivative}} \cr & \frac{{\partial z}}{{\partial y}} = \frac{{3{x^2}{y^2}\sqrt {x + y} - \frac{{{x^2}{y^3}}}{{2\sqrt {x + y} }}}}{{x + y}} \cr & \frac{{\partial z}}{{\partial y}} = \frac{{2\sqrt {x + y} \left( {3{x^2}{y^2}\sqrt {x + y} - \frac{{{x^2}{y^3}}}{{2\sqrt {x + y} }}} \right)}}{{2\sqrt {x + y} \left( {x + y} \right)}} \cr & \frac{{\partial z}}{{\partial y}} = \frac{{6{x^2}{y^2}\left( {x + y} \right) - {x^2}{y^3}}}{{2{{\left( {x + y} \right)}^{3/2}}}} \cr & \frac{{\partial z}}{{\partial y}} = \frac{{6{x^3}{y^2} + 6{x^2}{y^3} - {x^2}{y^3}}}{{2{{\left( {x + y} \right)}^{3/2}}}} \cr & \frac{{\partial z}}{{\partial y}} = \frac{{6{x^3}{y^2} + 5{x^2}{y^3}}}{{2{{\left( {x + y} \right)}^{3/2}}}} \cr & {\text{factoring}} \cr & \frac{{\partial z}}{{\partial y}} = \frac{{{x^2}{y^2}\left( {6x + 5y} \right)}}{{2{{\left( {x + y} \right)}^{3/2}}}} \cr & \cr & {\text{The partial derivatives are:}} \cr & \frac{{\partial z}}{{\partial x}} = \frac{{x{y^3}\left( {3x + 4y} \right)}}{{2{{\left( {x + y} \right)}^{3/2}}}}{\text{ and }}\frac{{\partial z}}{{\partial y}} = \frac{{6{x^3}{y^2} + 5{x^2}{y^3}}}{{2{{\left( {x + y} \right)}^{3/2}}}} \cr} $$