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: 40

Answer

$y'''$ = $42$

Work Step by Step

if $x$ = $1$ then $x^{2}+xy+y^{3}$ = $1$ $(1)^{2}+1(y)+y^{3}$ = $1$ $y^{3}+y$ = $0$ $y(y^{2}+1)$ = $0$ $y$ = $0$ Differentiating implicitly with respect to $x$ $2x+xy'+y+3y^{2}y'$ = $0$ Substituting $1$ for $x$ and $0$ for $y$ $2(1)+(1)y'+0+3(0)^{2}y'$ = $0$ $2+y'$ = $0$ $y'$ = $-2$ Differentiating $2+xy''+y'+y'+3y^{2}y''+6yy'y'$ = $0$ $2+xy''+2y'+3y^{2}y''+6yy'y'$ = $0$ Substituting $1$ for $x$, $0$ for $y$ and $-2$ for $y'$ $2+(1)y''+2(-2)+3(0)^2y''+6(0)(-2)(-2)$ = $0$ $y''-2$ = $0$ $y''$ = $2$ Differentiating $xy'''+y''+2y''+3y^{2}y'''+6yy'y''+6yy'y''+y'(6yy''+6y'y')$ = $0$ $xy'''+y''+2y''+3y^{2}y'''+6yy'y''+6yy'y''+6yy'y''+6y'y'y'$ = $0$ $xy'''+3y''+3y^{2}y'''+18yy'y''+6y'y'y'$ = $0$ Substituting $1$ for $x$, $0$ for $y$, $-2$ for $y'$ and $2$ for $y''$ $1y'''+3(2)+3(0)^2y'''+18(0)(-2)(2)+6(-2)(-2)(-2)$ = $0$ $y'''-42$ = $0$ $y'''$ = $42$
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.