If x2+xy+y3=1 find the value of y

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If x2+xy+y3=1  find the value of y

scomparve5j · Accepted Answer

Derivative with respect to &#039;x&#039;2x+1⋅y+x⋅y′+3y2y′=0⇒y′(x+3y2)=−(2x+y)⇒y′=−(2x+y)x+3y2Now: Derivating with &#039;x&#039; we get2+y′+y′+xy″+6y(y′)2+3y2y″=02+2y′+6y(y′)2+(x+3y2)y″=0derivative it one more time wait &#039;x&#039;2y″+6(y′)3+12yy′y″+y″+xy‴+6yy′y″+3y2y‴=0y&#039; at x=1⇒1+y+y3=1y(1+y2)=0y=0y&#039; at x=1⇒y′=−(2cdo1+0)1+2⋅0y′=−2y&#039;&#039; at x=1⇒2+2(−2)+6t(0)(−2)2+(1+3(0)2)y″=0y″=2y&#039;&#039;&#039; at x=1⇒2(2)+6(−2)3+18(0)(−2)(2)+(+2)+(1+3(0))y‴=0y&#039;&#039;&#039; at x=1⇒2(2)+6(−2)3+18(0)(−2)(2)+(+2)+(1+3(0))y‴=04−48+2+y‴=0y‴=42

Elaine Verrett · Accepted Answer

x2+xy+y3=1x=1; y+y3=0⇒y(1+y2)=0⇒y=0differentiating w.r.t. x2x+y+xy′+3y2y′=0x=1; y=0⇒2+0+y′+0=0⇒y′=−2again differentiating w.r.t. x2+y′+y′+xy″+6y(y′)2+3y2y″=0x=1; y=0; y′=−22−2−2+y″+0+0=0⇒y″=2again differentiating w.r.t. xy″+y″+xy‴+y″+6(y′)3+3y2y‴=0x=1; y=0; y′=−2;y″=22+2+y‴+2−48+0=0⇒y‴=42

user_27qwe · Accepted Answer

Equation is x2+xy+y3=1Differentiating the equation w.r.t x2x+xy′+y+3y2y′=0Differentiating againxy‴+y″+y″+6y‴=0y‴(x+6)+2y″=0y‴=−2y″x+6At x=1y‴=−2y″1+6y‴=−2y″7

If x^2+xy+y^3=1 find the value of y''' at the point where x=1

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