calculus homework help

Question

Find d^2y / dx², where y² - x³ = y

Answer 1

y^2 -x^3 =y

y^2-y-x^3=0

use the quadratic formula

y= .5+.5sqr(1+4x^3)

use the chain rule

y'= +/-(1/2)(1/2)(12x^2)/sqr(1+4x^3)

y' = +3x^2/sqr(1+4x^3)

chain rule again

y" = +(42x^4+6x)/sqr(1+4x^3)

or

y^2 +x^3 = y

2yy'+3x^2=y'

y' = 3x^2/(1-2y)

y"= (6x^2)y'/(1-2y)^2

y"= (6x^2)(3x^2/(1-2y))/(1-2y)

y" =18x^4/(1-2y)^3

or

y^2+x^3=y

2yy'+3x^2=y'

product rule on 1st term

2yy" +2y'^2 +6x =y"

y"= (2y'^2 +6x)/(1-2y)

y'=3x^2/(1-2y)

y"=[2(9x^4/(1-2y)^2 +6x)/(1-2y)

y"= (18x^4)/(1-2y)^3 +6x/(1-2y)

mistake(s) somewhere in the last one or two

Answer 2

can say

y²-y-x³=0

y₁=(1/2)*(1+(1+4x³)^(1/2))

dy₁/dx=(1/4)*12x²(1+4x³)^(-1/2)

d²y₁/dx²=6x(1+4x³)^(-1/2)-18x⁴(1+4x³)^(-3/2)

^y₂=(1/2)*(1-(1+4x³)^(1/2))

find y₂'' similarly