Thread: 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

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


so

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

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

From I would differentiate with respect to x again to get
so that

Now, simply because of the requirement that y' not appear in that formula, go back to to get and use that to replace y' in that formula.

OHHH
wow, how could I not see that?

ookay thank you very much!