Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 9 - Multiveriable Calculus - 9.2 Partial Derivatives - 9.2 Exercises - Page 478: 13

Answer

$$\eqalign{ & {f_x}\left( {x,y} \right) = \frac{{15{x^2}{y^2}}}{{1 + 5{x^3}{y^2}}},\,\,\,\,{f_y}\left( {x,y} \right) = \frac{{10{x^3}y}}{{1 + 5{x^3}{y^2}}},\,\,\,\,\,{f_x}\left( {2, - 1} \right) = \frac{{60}}{{41}} \cr & {\text{ }}{f_y}\left( { - 4,3} \right) = \frac{{1920}}{{2879}} \cr} $$

Work Step by Step

$$\eqalign{ & f\left( {x,y} \right) = \ln \left| {1 + 5{x^3}{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| {1 + 5{x^3}{y^2}} \right|} \right] \cr & {\text{treat 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}{{1 + 5{x^3}{y^2}}}\frac{\partial }{{\partial x}}\left[ {1 + 5{x^3}{y^2}} \right] \cr & {\text{solving derivatives}} \cr & {f_x}\left( {x,y} \right) = \frac{1}{{1 + 5{x^3}{y^2}}}\left( {15{x^2}{y^2}} \right) \cr & {\text{simplifying}} \cr & {f_x}\left( {x,y} \right) = \frac{{15{x^2}{y^2}}}{{1 + 5{x^3}{y^2}}} \cr & \cr & {\text{evaluate }}{f_x}\left( {2, - 1} \right) \cr & {f_x}\left( {2, - 1} \right) = \frac{{15{{\left( 2 \right)}^2}{{\left( { - 1} \right)}^2}}}{{1 + 5{{\left( 2 \right)}^3}{{\left( { - 1} \right)}^2}}} \cr & {f_x}\left( {2, - 1} \right) = \frac{{60}}{{41}} \cr & \cr & {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {\ln \left| {1 + 5{x^3}{y^2}} \right|} \right] \cr & {\text{treat 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}{{1 + 5{x^3}{y^2}}}\frac{\partial }{{\partial y}}\left[ {1 + 5{x^3}{y^2}} \right] \cr & {\text{solving derivatives}} \cr & {f_y}\left( {x,y} \right) = \frac{1}{{1 + 5{x^3}{y^2}}}\left( {5{x^3}} \right)\left( {2y} \right) \cr & {\text{multiply }} \cr & {f_y}\left( {x,y} \right) = \frac{{10{x^3}y}}{{1 + 5{x^3}{y^2}}} \cr & \cr & {\text{evaluate }}{f_y}\left( { - 4,3} \right) \cr & {f_y}\left( { - 4,3} \right) = \frac{{10{{\left( { - 4} \right)}^3}\left( 3 \right)}}{{1 + 5{{\left( { - 4} \right)}^3}{{\left( 3 \right)}^2}}} \cr & {f_y}\left( { - 4,3} \right) = \frac{{ - 1920}}{{ - 2879}} \cr & {f_y}\left( { - 4,3} \right) = \frac{{1920}}{{2879}} \cr} $$
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