Calculus with Applications (10th Edition)

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

Chapter 6 - Applications of the Derivative - 6.4 Implicit Differentiation - 6.4 Exercises - Page 334: 28



Work Step by Step

$y^{3}+2x^{2}y-8y= x^{3}+19$ Find the y value: $y^{3}+4y-8y= 8+19$ $y=3$ Calculate by implicit differentiation $3y^{2}\frac{dy}{dx}+2x^{2}\frac{dy}{dx}+4xy -8\frac{dy}{dx}=3x^{2}$ $(3y^{2}+2x^{2}-8)\frac{dy}{dx}=3x^{2}-4xy$ $\frac{dy}{dx}=\frac{3x^{2}-4xy}{3y^{2}+2x^{2}-8}$ $m=\frac{-4}{9}$ The equation of the tangent line is then found by using the point-slope form of the equation of a line. $y-y_{1}=m(x-x_{1})$ $y-3=\frac{-4}{9}(x-2)$ $y=\frac{-4}{9}x+\frac{35}{9}$
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.