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: 11

Answer

$$\eqalign{ & {f_x}\left( {x,y} \right) = \frac{{ - {x^4} - 2x{y^2} - 3{x^2}{y^3}}}{{{{\left( {{x^3} - {y^2}} \right)}^2}}},\,\,\,\,{f_y}\left( {x,y} \right) = \frac{{3{x^3}{y^2} + 2{x^2}y - {y^4}}}{{{{\left( {{x^3} - {y^2}} \right)}^2}}},\,\,\,\,\,{f_x}\left( {2, - 1} \right) = \frac{{ - 8}}{{49}}{\text{ }} \cr & {\text{ }}{f_y}\left( { - 4,3} \right) = - \frac{{1713}}{{5329}} \cr} $$

Work Step by Step

$$\eqalign{ & f\left( {x,y} \right) = \frac{{{x^2} + {y^3}}}{{{x^3} - {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{{{x^2} + {y^3}}}{{{x^3} - {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^3} - {y^2}} \right)\frac{\partial }{{\partial x}}\left[ {{x^2} + {y^3}} \right] - \left( {{x^2} + {y^3}} \right)\frac{\partial }{{\partial x}}\left[ {{x^3} - {y^2}} \right]}}{{{{\left( {{x^3} - {y^2}} \right)}^2}}} \cr & {\text{solving derivatives}} \cr & {f_x}\left( {x,y} \right) = \frac{{\left( {{x^3} - {y^2}} \right)\left( {2x} \right) - \left( {{x^2} + {y^3}} \right)\left( {3{x^2}} \right)}}{{{{\left( {{x^3} - {y^2}} \right)}^2}}} \cr & {\text{multiply and simplify}} \cr & {f_x}\left( {x,y} \right) = \frac{{2{x^4} - 2x{y^2} - 3{x^4} - 3{x^2}{y^3}}}{{{{\left( {{x^3} - {y^2}} \right)}^2}}} \cr & {f_x}\left( {x,y} \right) = \frac{{ - {x^4} - 2x{y^2} - 3{x^2}{y^3}}}{{{{\left( {{x^3} - {y^2}} \right)}^2}}} \cr & {\text{evaluate }}{f_x}\left( {2, - 1} \right) \cr & {f_x}\left( {2, - 1} \right) = \frac{{ - {{\left( 2 \right)}^4} - 2\left( 2 \right){{\left( { - 1} \right)}^2} - 3{{\left( 2 \right)}^2}{{\left( { - 1} \right)}^3}}}{{{{\left( {{{\left( 2 \right)}^3} - {{\left( { - 1} \right)}^2}} \right)}^2}}} \cr & {f_x}\left( {2, - 1} \right) = \frac{{ - 8}}{{49}} \cr & \cr & {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {\frac{{{x^2} + {y^3}}}{{{x^3} - {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^3} - {y^2}} \right)\frac{\partial }{{\partial y}}\left[ {{x^2} + {y^3}} \right] - \left( {{x^2} + {y^3}} \right)\frac{\partial }{{\partial y}}\left[ {{x^3} - {y^2}} \right]}}{{{{\left( {{x^3} - {y^2}} \right)}^2}}} \cr & {\text{solving derivatives}} \cr & {f_y}\left( {x,y} \right) = \frac{{\left( {{x^3} - {y^2}} \right)\left( {3{y^2}} \right) - \left( {{x^2} + {y^3}} \right)\left( { - 2y} \right)}}{{{{\left( {{x^3} - {y^2}} \right)}^2}}} \cr & {\text{multiply and simplify}} \cr & {f_y}\left( {x,y} \right) = \frac{{3{x^3}{y^2} - 3{y^4} + 2{x^2}y + 2{y^4}}}{{{{\left( {{x^3} - {y^2}} \right)}^2}}} \cr & {f_y}\left( {x,y} \right) = \frac{{3{x^3}{y^2} + 2{x^2}y - {y^4}}}{{{{\left( {{x^3} - {y^2}} \right)}^2}}} \cr & {\text{evaluate }}{f_x}\left( { - 4,3} \right) \cr & {f_y}\left( { - 4,3} \right) = \frac{{3{{\left( { - 4} \right)}^3}{{\left( 3 \right)}^2} + 2{{\left( { - 4} \right)}^2}\left( 3 \right) - {{\left( 3 \right)}^4}}}{{{{\left( {{{\left( { - 4} \right)}^3} - {{\left( 3 \right)}^2}} \right)}^2}}} \cr & {f_y}\left( { - 4,3} \right) = \frac{{ - 1713}}{{{{\left( { - 73} \right)}^2}}} = - \frac{{1713}}{{5329}} \cr} $$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.