Chapter 9 - Multiveriable Calculus - 9.2 Partial Derivatives - 9.2 Exercises - Page 478: 17

$$\eqalign{ & {f_x}\left( {x,y} \right) = \frac{{4{x^3} + 3y}}{{2\sqrt {{x^4} + 3xy + {y^4} + 10} }}\,\,\,\,\,\,\,{f_y}\left( {x,y} \right) = \frac{{3x + 4{y^3}}}{{2\sqrt {{x^4} + 3xy + {y^4} + 10} }} \cr & {f_x}\left( {2, - 1} \right) = \frac{{29}}{{2\sqrt {21} }}{\text{ and }}{f_y}\left( { - 4,3} \right) = \frac{{48}}{{\sqrt {311} }} \cr} $$

$$\eqalign{ & f\left( {x,y} \right) = \sqrt {{x^4} + 3xy + {y^4} + 10} \cr & f\left( {x,y} \right) = {\left( {{x^4} + 3xy + {y^4} + 10} \right)^{1/2}} \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( {{x^4} + 3xy + {y^4} + 10} \right)^{1/2}} \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}{2}{\left( {{x^4} + 3xy + {y^4} + 10} \right)^{ - 1/2}}\frac{\partial }{{\partial x}}\left[ {{x^4} + 3xy + {y^4} + 10} \right] \cr & {\text{then}} \cr & {f_x}\left( {x,y} \right) = \frac{1}{2}{\left( {{x^4} + 3xy + {y^4} + 10} \right)^{ - 1/2}}\left( {4{x^3} + 3y} \right) \cr & {f_x}\left( {x,y} \right) = \frac{{4{x^3} + 3y}}{{2{{\left( {{x^4} + 3xy + {y^4} + 10} \right)}^{1/2}}}} \cr & {f_x}\left( {x,y} \right) = \frac{{4{x^3} + 3y}}{{2\sqrt {{x^4} + 3xy + {y^4} + 10} }} \cr & \cr & {\text{evaluate }}{f_x}\left( {2, - 1} \right) \cr & {f_x}\left( {2, - 1} \right) = \frac{{4{{\left( 2 \right)}^3} + 3\left( { - 1} \right)}}{{2\sqrt {{{\left( 2 \right)}^4} + 3\left( { - 2} \right)\left( { - 1} \right) + {{\left( { - 1} \right)}^4} + 10} }} \cr & {f_x}\left( {2, - 1} \right) = \frac{{32 - 3}}{{2\sqrt {21} }} \cr & {f_x}\left( {2, - 1} \right) = \frac{{29}}{{2\sqrt {21} }} \cr & \cr & {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}{\left( {{x^4} + 3xy + {y^4} + 10} \right)^{1/2}} \cr & {\text{treating y as a constant and }}x{\text{ as a variable use chain rule}} \cr & {f_y}\left( {x,y} \right) = \frac{1}{2}{\left( {{x^4} + 3xy + {y^4} + 10} \right)^{ - 1/2}}\frac{\partial }{{\partial y}}\left[ {{x^4} + 3xy + {y^4} + 10} \right] \cr & {\text{then}} \cr & {f_y}\left( {x,y} \right) = \frac{1}{2}{\left( {{x^4} + 3xy + {y^4} + 10} \right)^{ - 1/2}}\left( {3x + 4{y^3}} \right) \cr & {f_y}\left( {x,y} \right) = \frac{{3x + 4{y^3}}}{{2{{\left( {{x^4} + 3xy + {y^4} + 10} \right)}^{1/2}}}} \cr & {f_y}\left( {x,y} \right) = \frac{{3x + 4{y^3}}}{{2\sqrt {{x^4} + 3xy + {y^4} + 10} }} \cr & \cr & {\text{evaluate }}{f_y}\left( { - 4,3} \right) \cr & {f_y}\left( { - 4,3} \right) = \frac{{3\left( { - 4} \right) + 4{{\left( 3 \right)}^3}}}{{2\sqrt {{{\left( { - 4} \right)}^4} + 3\left( { - 4} \right)\left( 3 \right) + {{\left( 3 \right)}^4} + 10} }} \cr & {f_y}\left( { - 4,3} \right) = \frac{{ - 12 + 108}}{{2\sqrt {256 - 36 + 81 + 10} }} \cr & {f_y}\left( { - 4,3} \right) = \frac{{96}}{{2\sqrt {311} }} \cr & {f_y}\left( { - 4,3} \right) = \frac{{48}}{{\sqrt {311} }} \cr} $$