Question

find y'' by implicit differentiation.

2x^3 + 3y^3 = 8


I got the first derivative as you but the problem was asking for second derivative by implicit diff. this is where i got confused.

Thank you!!!

Answer 1

oops. Well, if you got y', y" is just a bit down the road

There are two ways to do it. First, use y' directly:

y' = -2x^2 / 3y^2
by the quotient rule,

y" = ((-4x)(3y^2) - (-2x^2)(6yy')] / 9y^4
= ((-12xy^2 + 12x^2y(-2x^2/3y^2)) / 9y^4
= -12xy(y - x(-2x^2/3y^2))/9y^4
= -12xy(3y^3 + 2x^3)/27y^6
= -4x(3y^3 + 2x^3)/9y^5

Or, since we have

2x^2 + 3y^2 y' = 0
use the product rule to get
4x + 6y (y')^2 + 3y^2 y" = 0
y" = -(4x + 6y(y')^2)/3y^2
= -2(2x+3y(4x^4/9y^4))/3y^2
= -2(18xy^4 + 12x^4y)/27y^6
= -2x(3y^3 + 4x^3)/9y^5

Hmm. I seem to have made an arithmetic error somewhere, but you get the idea, no?