# Thread: Second derivative of implicit differentiation

1. ## Second derivative of implicit differentiation

If x^2 + x + y^3 + y =1 then find y". Your answer must not contain y'. Do not simplify at all.

So I found the first derivative because I dont understand why the answer cannot contain y' if you have to find y" :S but i found the first derivative and got:

2x + 1/ 3y^2 = dy/dt

then my second deriv was y"= 6y(y')(2x-1) - (2)(3y^2)/ (3y^2)^2

I got 1.5/3 and I dont understand why, any help would be greatly appreciated

2. It "can't contain y' " because that was a condition set by the problem!

$x^2 + x + y^3 + y =1$
so $2x+ 1+ 3y^2y'+ y'= 0$

Your first derivative is wrong because you forgot the y' needed for the derivative of $y^3$ with respect to x.

I have no idea in the world why you think the answer should be a number!

From $2x+ 1+ (3y^2+ 1)y'= 0$ I would differentiate with respect to x again to get
$2+ (6y)y'+ (3y^2+ 1)y''= 0$ so that $y''= -\frac{(2+ 6y)y'}{3y^2+ 1}$

Now, simply because of the requirement that y' not appear in that formula, go back to $2x+1+ (3y^2+ 1)y'= 0$ to get $y'= -\frac{2+ 1}{3y^2+ 1}$ and use that to replace y' in that formula.

3. OHHH
wow, how could I not see that?

ookay thank you very much!