Chapter 9 - Multiveriable Calculus - 9.2 Partial Derivatives - 9.2 Exercises - Page 478: 40

$$\eqalign{ & {f_x}\left( {x,y,z} \right) = \frac{{4x + y}}{{yz - 2}}{\text{,}}\,\,\,\,{f_y}\left( {x,y,z} \right) = - \frac{{2x + 2{x^2}z}}{{{{\left( {yz - 2} \right)}^2}}} \cr & {f_z}\left( {x,y,z} \right) = - \frac{{y\left( {2{x^2} + xy} \right)}}{{{{\left( {yz - 2} \right)}^2}}},\,\,\,\,{\text{ }}{f_{yz}}\left( {x,y,z} \right) = \frac{{2\left( {2{x^2} + 2xy + {x^2}yz} \right)}}{{{{\left( {yz - 2} \right)}^3}}} \cr} $$

$$\eqalign{ & f\left( {x,y,z} \right) = \frac{{2{x^2} + xy}}{{yz - 2}} \cr & {\text{Find }}{f_x}\left( {x,y,z} \right){\text{ treat }}y{\text{ and }}z{\text{ as constants}} \cr & {f_x}\left( {x,y,z} \right) = \frac{\partial }{{\partial x}}\left[ {\frac{{2{x^2} + xy}}{{yz - 2}}} \right] \cr & {f_x}\left( {x,y,z} \right) = \frac{1}{{yz - 2}}\frac{\partial }{{\partial x}}\left[ {2{x^2} + xy} \right] \cr & {\text{then}} \cr & {f_x}\left( {x,y,z} \right) = \frac{1}{{yz - 2}}\left( {4x + y} \right) \cr & {f_x}\left( {x,y,z} \right) = \frac{{4x + y}}{{yz - 2}} \cr & \cr & {\text{Find }}{f_y}\left( {x,y,z} \right){\text{ treat }}x{\text{ and }}z{\text{ as constants}} \cr & {f_y}\left( {x,y,z} \right) = \frac{\partial }{{\partial y}}\left[ {\frac{{2{x^2} + xy}}{{yz - 2}}} \right] \cr & {\text{by using the quotient rule}} \cr & {f_y}\left( {x,y,z} \right) = \frac{{\left( {yz - 2} \right)\left( x \right) - \left( {2{x^2} + xy} \right)\left( z \right)}}{{{{\left( {yz - 2} \right)}^2}}} \cr & {\text{then}} \cr & {f_y}\left( {x,y,z} \right) = \frac{{xyz - 2x - 2{x^2}z - xyz}}{{{{\left( {yz - 2} \right)}^2}}} \cr & {f_y}\left( {x,y,z} \right) = - \frac{{2x + 2{x^2}z}}{{{{\left( {yz - 2} \right)}^2}}} \cr & \cr & {\text{Find }}{f_z}\left( {x,y,z} \right){\text{ treat }}x{\text{ and }}y{\text{ as constants}} \cr & {f_z}\left( {x,y,z} \right) = \frac{\partial }{{\partial z}}\left[ {\frac{{2{x^2} + xy}}{{yz - 2}}} \right] \cr & {f_z}\left( {x,y,z} \right) = \left( {2{x^2} + xy} \right)\frac{\partial }{{\partial z}}\left[ {\frac{1}{{yz - 2}}} \right] \cr & {\text{then}} \cr & {f_z}\left( {x,y,z} \right) = \left( {2{x^2} + xy} \right)\left( {\frac{{ - y}}{{{{\left( {yz - 2} \right)}^2}}}} \right) \cr & {f_z}\left( {x,y,z} \right) = - \frac{{y\left( {2{x^2} + xy} \right)}}{{{{\left( {yz - 2} \right)}^2}}} \cr & \cr & {\text{Find }}{f_{yz}}\left( {x,y,z} \right){\text{ }} \cr & {\text{ }}{f_{yz}}\left( {x,y,z} \right) = \frac{\partial }{{\partial z}}\left[ { - \frac{{2x + 2{x^2}z}}{{{{\left( {yz - 2} \right)}^2}}}} \right] \cr & {\text{then by using the quotient rule}} \cr & {\text{ }}{f_{yz}}\left( {x,y,z} \right) = - \frac{{{{\left( {yz - 2} \right)}^2}\left( {2{x^2}} \right) - \left( {2x + 2{x^2}z} \right)\left( 2 \right)\left( {yz - 2} \right)\left( y \right)}}{{{{\left( {yz - 2} \right)}^4}}} \cr & {\text{simplifying}} \cr & {\text{ }}{f_{yz}}\left( {x,y,z} \right) = - \frac{{\left( {yz - 2} \right)\left( {2{x^2}} \right) - \left( {2x + 2{x^2}z} \right)\left( 2 \right)\left( y \right)}}{{{{\left( {yz - 2} \right)}^3}}} \cr & {\text{ }}{f_{yz}}\left( {x,y,z} \right) = - \frac{{2\left[ {\left( {yz - 2} \right)\left( {{x^2}} \right) - \left( {2x + 2{x^2}z} \right)\left( y \right)} \right]}}{{{{\left( {yz - 2} \right)}^3}}} \cr & {\text{ }}{f_{yz}}\left( {x,y,z} \right) = - \frac{{2\left( {{x^2}yz - 2{x^2} - 2xy - 2{x^2}yz} \right)}}{{{{\left( {yz - 2} \right)}^3}}} \cr & {\text{ }}{f_{yz}}\left( {x,y,z} \right) = - \frac{{2\left( { - 2{x^2} - 2xy - {x^2}yz} \right)}}{{{{\left( {yz - 2} \right)}^3}}} \cr & {\text{ }}{f_{yz}}\left( {x,y,z} \right) = \frac{{2\left( {2{x^2} + 2xy + {x^2}yz} \right)}}{{{{\left( {yz - 2} \right)}^3}}} \cr} $$