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!!!

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?

To find the second derivative by implicit differentiation, you need to differentiate the equation twice with respect to the variable of differentiation, which in this case is x.

Let's start by differentiating the given equation once with respect to x.

Differentiating the equation 2x^3 + 3y^3 = 8 with respect to x, we get:

6x^2 + 9y^2(dy/dx) = 0

Now, to find the first derivative of y with respect to x, (dy/dx), we rearrange the equation to solve for (dy/dx):

9y^2(dy/dx) = -6x^2
(dy/dx) = -6x^2 / 9y^2
(dy/dx) = -2x^2 / 3y^2

Now that we have the first derivative (dy/dx), we need to find the second derivative, which is obtained by differentiating (dy/dx) with respect to x.

Differentiating (dy/dx) = -2x^2 / 3y^2 with respect to x, we use the quotient rule:

(d^2y/dx^2) = [(-2x^2)(2y^2) - (3y^2)(-4x)] / (3y^4)
(d^2y/dx^2) = (-4x^2y^2 + 12xy) / (3y^4)
(d^2y/dx^2) = 4x^2y^2 - 12xy / 3y^4
(d^2y/dx^2) = (4x^2y - 12x) / 3y^3

So, the second derivative of y with respect to x, or y'', is (4x^2y - 12x) / 3y^3