Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - 2.6 Implicit Differentiation - 2.6 Exercises - Page 167: 32



Work Step by Step

$y^2(y^2-4)=x^2(x^2-5)\\ (y^2)(2yy')+(y^2-4)(2yy')=(x^2)(2x)+(x^2-5)(2x)\\ 2y^3y'+2y^3y'-8yy'=2x^3+2x^3-10x\\ 4y^3y'-8yy'=4x^3-10x\\ 2y'(2y^3-4y)=2(2x^3-5x)\\ y'=\frac{2x^3-5x}{2y^3-4y}$ Plug in $(0,-2)$ into $y′$ to find the gradient $y'=\frac{2(0)^3-5(0)}{2(-2)^3-4(-2)}\\ y'=0$ Equation of the tangent line at $(0,-2)$ $y-(-2)=0(x-0)\\ y=-2$
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.