## Calculus with Applications (10th Edition)

Published by Pearson

# Chapter 9 - Multiveriable Calculus - 9.2 Partial Derivatives - 9.2 Exercises - Page 478: 19

#### Answer

\eqalign{ & {f_x}\left( {x,y} \right) = \frac{{6xy{e^{xy}} + 12xy - 3{x^2}{y^2}{e^{xy}}}}{{{{\left( {{e^{xy}} + 2} \right)}^2}}},\,\,\,\,{f_y}\left( {x,y} \right) = \frac{{3{x^2}{e^{xy}} + 6{x^2} - 3{x^3}y{e^{xy}}}}{{{{\left( {{e^{xy}} + 2} \right)}^2}}},\,\,\,\,\, \cr & {f_x}\left( {2, - 1} \right) = \frac{{ - 24{e^{ - 2}} - 24}}{{{{\left( {{e^{ - 2}} + 2} \right)}^2}}}{\text{ and }}{f_y}\left( { - 4,3} \right) = \frac{{96 + 624{e^{ - 12}}}}{{{{\left( {{e^{ - 12}} + 2} \right)}^2}}} \cr}

#### Work Step by Step

\eqalign{ & f\left( {x,y} \right) = \frac{{3{x^2}y}}{{{e^{xy}} + 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}}{{{e^{xy}} + 2}}} \right] \cr & {\text{treating y as a constant and }}x{\text{ as a variable and use the quotient rule}} \cr & {f_x}\left( {x,y} \right) = \frac{{\left( {{e^{xy}} + 2} \right)\frac{\partial }{{\partial x}}\left[ {3{x^2}y} \right] - \left( {3{x^2}y} \right)\frac{\partial }{{\partial x}}\left[ {{e^{xy}} + 2} \right]}}{{{{\left( {{e^{xy}} + 2} \right)}^2}}} \cr & {\text{solving derivatives}} \cr & {f_x}\left( {x,y} \right) = \frac{{\left( {{e^{xy}} + 2} \right)\left( {6xy} \right) - \left( {3{x^2}y} \right)\left( {y{e^{xy}}} \right)}}{{{{\left( {{e^{xy}} + 2} \right)}^2}}} \cr & {f_x}\left( {x,y} \right) = \frac{{6xy{e^{xy}} + 12xy - 3{x^2}{y^2}{e^{xy}}}}{{{{\left( {{e^{xy}} + 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){e^{\left( 2 \right)\left( { - 1} \right)}} + 12\left( 2 \right)\left( { - 1} \right) - 3{{\left( 2 \right)}^2}{{\left( { - 1} \right)}^2}{e^{\left( 2 \right)\left( { - 1} \right)}}}}{{{{\left( {{e^{\left( 2 \right)\left( { - 1} \right)}} + 2} \right)}^2}}} \cr & {f_x}\left( {2, - 1} \right) = \frac{{ - 12{e^{ - 2}} - 24 - 12{e^{ - 2}}}}{{{{\left( {{e^{ - 2}} + 2} \right)}^2}}} \cr & {f_x}\left( {2, - 1} \right) = \frac{{ - 24{e^{ - 2}} - 24}}{{{{\left( {{e^{ - 2}} + 2} \right)}^2}}} \cr & \cr & {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {\frac{{3{x^2}y}}{{{e^{xy}} + 2}}} \right] \cr & {\text{treating }}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( {{e^{xy}} + 2} \right)\frac{\partial }{{\partial y}}\left[ {3{x^2}y} \right] - \left( {3{x^2}y} \right)\frac{\partial }{{\partial y}}\left[ {{e^{xy}} + 2} \right]}}{{{{\left( {{e^{xy}} + 2} \right)}^2}}} \cr & {\text{solving derivatives}} \cr & {f_y}\left( {x,y} \right) = \frac{{\left( {{e^{xy}} + 2} \right)\left( {3{x^2}} \right) - \left( {3{x^2}y} \right)\left( {x{e^{xy}}} \right)}}{{{{\left( {{e^{xy}} + 2} \right)}^2}}} \cr & {\text{multiply and simplify}} \cr & {f_y}\left( {x,y} \right) = \frac{{3{x^2}{e^{xy}} + 6{x^2} - 3{x^3}y{e^{xy}}}}{{{{\left( {{e^{xy}} + 2} \right)}^2}}} \cr & {\text{evaluate }}{f_y}\left( { - 4,3} \right) \cr & {f_y}\left( { - 4,3} \right) = \frac{{3{{\left( { - 4} \right)}^2}{e^{\left( { - 4} \right)\left( 3 \right)}} + 6{{\left( { - 4} \right)}^2} - 3{{\left( { - 4} \right)}^3}\left( 3 \right){e^{\left( { - 4} \right)\left( 3 \right)}}}}{{{{\left( {{e^{\left( { - 4} \right)\left( 3 \right)}} + 2} \right)}^2}}} \cr & {f_y}\left( { - 4,3} \right) = \frac{{48{e^{ - 12}} + 96 + 576{e^{ - 12}}}}{{{{\left( {{e^{ - 12}} + 2} \right)}^2}}} \cr & {f_y}\left( { - 4,3} \right) = \frac{{96 + 624{e^{ - 12}}}}{{{{\left( {{e^{ - 12}} + 2} \right)}^2}}} \cr}

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