Chapter 9 - Multiveriable Calculus - 9.2 Partial Derivatives - 9.2 Exercises - Page 478: 14

$$\eqalign{ & {f_x}\left( {x,y} \right) = \frac{{16{x^3} - 4x{y^2}}}{{4{x^4} - 2{x^2}{y^2}}},\,\,\,\,{f_y}\left( {x,y} \right) = - \frac{{4{x^2}y}}{{4{x^4} - 2{x^2}{y^2}}},\,\,\,\,\,{f_x}\left( {2, - 1} \right) = \frac{{15}}{7} \cr & {\text{ }}{f_y}\left( { - 4,3} \right) = \frac{6}{{23}} \cr} $$

$$\eqalign{ & f\left( {x,y} \right) = \ln \left| {4{x^4} - 2{x^2}{y^2}} \right| \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[ {\ln \left| {4{x^4} - 2{x^2}{y^2}} \right|} \right] \cr & {\text{treating y as a constant and }}x{\text{ as a variable use the rule }}\left[ {\ln u} \right]' = \frac{1}{u}\left( {u'} \right) \cr & {f_x}\left( {x,y} \right) = \frac{1}{{4{x^4} - 2{x^2}{y^2}}}\frac{\partial }{{\partial x}}\left[ {4{x^4} - 2{x^2}{y^2}} \right] \cr & {\text{solving derivatives}} \cr & {f_x}\left( {x,y} \right) = \frac{1}{{4{x^4} - 2{x^2}{y^2}}}\left( {16{x^3} - \left( {4x} \right)\left( {{y^2}} \right)} \right) \cr & {\text{simplifying}} \cr & {f_x}\left( {x,y} \right) = \frac{{16{x^3} - 4x{y^2}}}{{4{x^4} - 2{x^2}{y^2}}} \cr & \cr & {\text{evaluate }}{f_x}\left( {2, - 1} \right) \cr & {f_x}\left( {2, - 1} \right) = \frac{{16{{\left( 2 \right)}^3} - 4\left( 2 \right){{\left( { - 1} \right)}^2}}}{{4{{\left( 2 \right)}^4} - 2{{\left( 2 \right)}^2}{{\left( { - 1} \right)}^2}}} \cr & {f_x}\left( {2, - 1} \right) = \frac{{120}}{{56}} \cr & {f_x}\left( {2, - 1} \right) = \frac{{15}}{7} \cr & \cr & {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {\ln \left| {4{x^4} - 2{x^2}{y^2}} \right|} \right] \cr & {\text{treating x as a constant and }}y{\text{ as a variable use the rule }}\left[ {\ln u} \right]' = \frac{1}{u}\left( {u'} \right) \cr & {f_y}\left( {x,y} \right) = \frac{1}{{4{x^4} - 2{x^2}{y^2}}}\frac{\partial }{{\partial y}}\left[ {4{x^4} - 2{x^2}{y^2}} \right] \cr & {\text{solving derivatives}} \cr & {f_y}\left( {x,y} \right) = \frac{1}{{4{x^4} - 2{x^2}{y^2}}}\left( { - 4{x^2}y} \right) \cr & {\text{multiply }} \cr & {f_y}\left( {x,y} \right) = - \frac{{4{x^2}y}}{{4{x^4} - 2{x^2}{y^2}}} \cr & \cr & {\text{evaluate }}{f_y}\left( { - 4,3} \right) \cr & {f_y}\left( { - 4,3} \right) = - \frac{{4{{\left( { - 4} \right)}^2}\left( 3 \right)}}{{4{{\left( { - 4} \right)}^4} - 2{{\left( { - 4} \right)}^2}{{\left( 3 \right)}^2}}} \cr & {f_y}\left( { - 4,3} \right) = - \frac{{192}}{{1024 - 288}} \cr & {f_y}\left( { - 4,3} \right) = \frac{{192}}{{736}} \cr & {f_y}\left( { - 4,3} \right) = \frac{6}{{23}} \cr} $$