Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 2 - Differentiation - 2.5 Exercises: 40

Answer

$y=\dfrac{x+2}{3}.$

Work Step by Step

$y^2x^2+y^4=2x^2\rightarrow$ $\dfrac{d}{dx}(y^2x^2)+\dfrac{d}{dx}(y^4)=\dfrac{d}{dx}(2x^2)\rightarrow$ $(2y^2x+\dfrac{dy}{dx}(2x^2y))+\dfrac{dy}{dx}(4y^3)=4x\rightarrow$ $\dfrac{dy}{dx}=\dfrac{4x-2y^2x}{4y^3+2x^2y}=\dfrac{2x-y^2x}{2y^3+x^2y}.$ At $(1, 1)\rightarrow\dfrac{dy}{dx}=\dfrac{2(1)-(1^2)(1)}{2(1^3)+(1^2)(1)}=\dfrac{1}{3}.$ Equation of tangent: $(y-y_0)=m(x-x_0)$ at point $(x_0, y_0)$ and slope $m$. $(y-1)=\frac{1}{3}(x-1)\rightarrow y=\dfrac{x+2}{3}.$
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.