Calculus with Applications (10th Edition)

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

Chapter 6 - Applications of the Derivative - Chapter Review - Review Exercises - Page 351: 22

Answer

$$\frac{{dy}}{{dx}} = \frac{{ - 2x{y^3} - 4y}}{{3{x^2}{y^2} + 4x}}$$

Work Step by Step

$$\eqalign{ & {x^2}{y^3} + 4xy = 2 \cr & {\text{take the derivative on both sides with respect to }}x \cr & \frac{d}{{dx}}\left( {{x^2}{y^3}} \right) + \frac{d}{{dx}}\left( {4xy} \right) = \frac{d}{{dx}}\left( 2 \right) \cr & {\text{use product rule for differentiation as follows}} \cr & {x^2}\frac{d}{{dx}}\left( {{y^3}} \right) + {y^3}\frac{d}{{dx}}\left( {{x^2}} \right) + 4x\frac{d}{{dx}}\left( y \right) + 4y\frac{d}{{dx}}\left( x \right) = \frac{d}{{dx}}\left( 2 \right) \cr & {\text{solve the derivatives using the power rule and the chain rule}} \cr & {x^2}\left( {3{y^2}} \right)\frac{{dy}}{{dx}} + {y^3}\left( {2x} \right) + 4x\frac{{dy}}{{dx}} + 4y\left( 1 \right) = 0 \cr & {\text{simplifying}} \cr & 3{x^2}{y^2}\frac{{dy}}{{dx}} + 2x{y^3} + 4x\frac{{dy}}{{dx}} + 4y = 0 \cr & {\text{combine terms and solve for }}\frac{{dy}}{{dx}} \cr & 3{x^2}{y^2}\frac{{dy}}{{dx}} + 4x\frac{{dy}}{{dx}} = - 2x{y^3} - 4y \cr & \left( {3{x^2}{y^2} + 4x} \right)\frac{{dy}}{{dx}} = - 2x{y^3} - 4y \cr & \frac{{dy}}{{dx}} = \frac{{ - 2x{y^3} - 4y}}{{3{x^2}{y^2} + 4x}} \cr} $$
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