Chapter 9 - Multiveriable Calculus - 9.2 Partial Derivatives - 9.2 Exercises - Page 478: 18

$$\eqalign{ & {f_x}\left( {x,y} \right) = \frac{{14x + 18{y^2}}}{{3{{\left( {7{x^2} + 18x{y^2} + {y^3}} \right)}^{2/3}}}}\,\,\,\,\,\,\,{f_y}\left( {x,y} \right) = \frac{{36xy + 3{y^2}}}{{3{{\left( {7{x^2} + 18x{y^2} + {y^3}} \right)}^{2/3}}}} \cr & {f_x}\left( {2, - 1} \right) = \frac{{46}}{{3{{\left( {63} \right)}^{2/3}}}}{\text{ and }}{f_y}\left( { - 4,3} \right) = - \frac{{135}}{{{{\left( {509} \right)}^{2/3}}}} \cr} $$

$$\eqalign{ & f\left( {x,y} \right) = {\left( {7{x^2} + 18x{y^2} + {y^3}} \right)^{1/3}} \cr & {\text{Find }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right) \cr & {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}{\left( {7{x^2} + 18x{y^2} + {y^3}} \right)^{1/3}} \cr & {\text{treating y as a constant and }}x{\text{ as a variable use chain rule}} \cr & {f_x}\left( {x,y} \right) = \frac{1}{3}{\left( {7{x^2} + 18x{y^2} + {y^3}} \right)^{ - 2/3}}\frac{\partial }{{\partial x}}\left[ {7{x^2} + 18x{y^2} + {y^3}} \right] \cr & {\text{then}} \cr & {f_x}\left( {x,y} \right) = \frac{1}{3}{\left( {7{x^2} + 18x{y^2} + {y^3}} \right)^{ - 2/3}}\left( {14x + 18{y^2}} \right) \cr & {f_x}\left( {x,y} \right) = \frac{{14x + 18{y^2}}}{{3{{\left( {7{x^2} + 18x{y^2} + {y^3}} \right)}^{2/3}}}} \cr & \cr & {\text{evaluate }}{f_x}\left( {2, - 1} \right) \cr & {f_x}\left( {2, - 1} \right) = \frac{{14\left( 2 \right) + 18{{\left( { - 1} \right)}^2}}}{{3{{\left( {7{{\left( 2 \right)}^2} + 18\left( 2 \right){{\left( { - 1} \right)}^2} + {{\left( { - 1} \right)}^3}} \right)}^{2/3}}}} \cr & {f_x}\left( {2, - 1} \right) = \frac{{28 + 18}}{{3{{\left( {63} \right)}^{2/3}}}} \cr & {f_x}\left( {2, - 1} \right) = \frac{{46}}{{3{{\left( {63} \right)}^{2/3}}}} \cr & \cr & {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}{\left( {7{x^2} + 18x{y^2} + {y^3}} \right)^{1/3}} \cr & {\text{treating x as a constant and }}y{\text{ as a variable use chain rule}} \cr & {f_y}\left( {x,y} \right) = \frac{1}{3}{\left( {7{x^2} + 18x{y^2} + {y^3}} \right)^{ - 2/3}}\frac{\partial }{{\partial y}}\left[ {7{x^2} + 18x{y^2} + {y^3}} \right] \cr & {\text{then}} \cr & {f_y}\left( {x,y} \right) = \frac{1}{3}{\left( {7{x^2} + 18x{y^2} + {y^3}} \right)^{ - 2/3}}\left( {36xy + 3{y^2}} \right) \cr & {f_y}\left( {x,y} \right) = \frac{{36xy + 3{y^2}}}{{3{{\left( {7{x^2} + 18x{y^2} + {y^3}} \right)}^{2/3}}}} \cr & \cr & {\text{evaluate }}{f_y}\left( { - 4,3} \right) \cr & {f_y}\left( { - 4,3} \right) = \frac{{36\left( { - 4} \right)\left( 3 \right) + 3{{\left( 3 \right)}^2}}}{{3{{\left( {7{{\left( { - 4} \right)}^2} + 18\left( { - 4} \right){{\left( 3 \right)}^2} + {{\left( 3 \right)}^3}} \right)}^{2/3}}}} \cr & {f_y}\left( { - 4,3} \right) = \frac{{ - 432 + 27}}{{3{{\left( { - 509} \right)}^{2/3}}}} \cr & {f_y}\left( { - 4,3} \right) = \frac{{ - 405}}{{3{{\left( {509} \right)}^{2/3}}}} \cr & {f_y}\left( { - 4,3} \right) = - \frac{{135}}{{{{\left( {509} \right)}^{2/3}}}} \cr} $$