Chapter 9 - Multiveriable Calculus - 9.2 Partial Derivatives - 9.2 Exercises - Page 478: 12

$$\eqalign{ & {f_x}\left( {x,y} \right) = \frac{{6x{y^5}}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}},\,\,\,\,{f_y}\left( {x,y} \right) = \frac{{9{x^4}{y^2} + 3{x^2}{y^4}}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}},\,\,\,\,\,{f_x}\left( {2, - 1} \right) = - \frac{{12}}{{25}} \cr & {\text{ }}{f_y}\left( { - 4,3} \right)\frac{{24624}}{{625}} \cr} $$

$$\eqalign{ & f\left( {x,y} \right) = \frac{{3{x^2}{y^3}}}{{{x^2} + {y^2}}} \cr & {\text{Find }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right) \cr & \cr & {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {\frac{{3{x^2}{y^3}}}{{{x^2} + {y^2}}}} \right] \cr & {\text{treat y as a constant and }}x{\text{ as a variable and use the quotient rule}} \cr & {f_x}\left( {x,y} \right) = \frac{{\left( {{x^2} + {y^2}} \right)\frac{\partial }{{\partial x}}\left[ {3{x^2}{y^3}} \right] - \left( {3{x^2}{y^3}} \right)\frac{\partial }{{\partial x}}\left[ {{x^2} + {y^2}} \right]}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}} \cr & {\text{solving derivatives}} \cr & {f_x}\left( {x,y} \right) = \frac{{3{y^3}\left( {{x^2} + {y^2}} \right)\frac{\partial }{{\partial x}}\left[ {{x^2}} \right] - \left( {3{x^2}{y^3}} \right)\frac{\partial }{{\partial x}}\left[ {{x^2} + {y^2}} \right]}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}} \cr & {f_x}\left( {x,y} \right) = \frac{{6x{y^3}\left( {{x^2} + {y^2}} \right) - \left( {3{x^2}{y^3}} \right)\left( {2x} \right)}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}} \cr & {\text{multiply and simplify}} \cr & {f_x}\left( {x,y} \right) = \frac{{6{x^3}{y^3} + 6x{y^5} - 6{x^3}{y^3}}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}} \cr & {f_x}\left( {x,y} \right) = \frac{{6x{y^5}}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}} \cr & {\text{evaluate }}{f_x}\left( {2, - 1} \right) \cr & {f_x}\left( {2, - 1} \right) = \frac{{6\left( 2 \right){{\left( { - 1} \right)}^5}}}{{{{\left( {{2^2} + {{\left( { - 1} \right)}^2}} \right)}^2}}} \cr & {f_x}\left( {2, - 1} \right) = - \frac{{12}}{{25}} \cr & \cr & {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {\frac{{3{x^2}{y^3}}}{{{x^2} + {y^2}}}} \right] \cr & {\text{treat }}x{\text{ as a constant and }}y{\text{ as a variable and use the quotient rule}} \cr & {f_y}\left( {x,y} \right) = \frac{{\left( {{x^2} + {y^2}} \right)\frac{\partial }{{\partial y}}\left[ {3{x^2}{y^3}} \right] - \left( {3{x^2}{y^3}} \right)\frac{\partial }{{\partial y}}\left[ {{x^2} + {y^2}} \right]}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}} \cr & {\text{solving derivatives}} \cr & {f_y}\left( {x,y} \right) = \frac{{3{x^2}\left( {{x^2} + {y^2}} \right)\frac{\partial }{{\partial y}}\left[ {{y^3}} \right] - \left( {3{x^2}{y^3}} \right)\frac{\partial }{{\partial y}}\left[ {{x^2} + {y^2}} \right]}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}} \cr & {f_y}\left( {x,y} \right) = \frac{{3{x^2}\left( {{x^2} + {y^2}} \right)\left( {3{y^2}} \right) - \left( {3{x^2}{y^3}} \right)\left( {2y} \right)}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}} \cr & {\text{multiply and simplify}} \cr & {f_y}\left( {x,y} \right) = \frac{{9{x^4}{y^2} + 9{x^2}{y^4} - 6{x^2}{y^4}}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}} \cr & {f_y}\left( {x,y} \right) = \frac{{9{x^4}{y^2} + 3{x^2}{y^4}}}{{{{\left( {{x^2} + {y^2}} \right)}^2}}} \cr & {\text{evaluate }}{f_x}\left( { - 4,3} \right) \cr & {f_y}\left( { - 4,3} \right) = \frac{{9{{\left( { - 4} \right)}^4}{{\left( 3 \right)}^2} + 3{{\left( { - 4} \right)}^2}{{\left( 3 \right)}^4}}}{{{{\left( {{{\left( { - 4} \right)}^2} + {{\left( 3 \right)}^2}} \right)}^2}}} \cr & {f_y}\left( { - 4,3} \right) = \frac{{24624}}{{{{\left( {25} \right)}^2}}} \cr & {f_y}\left( { - 4,3} \right) = \frac{{24624}}{{625}} \cr} $$