# Implicit differentiation - Help

## Homework Statement

find y' given x3+y2+x+y=2

## The Attempt at a Solution

dy/dx=3x2+2y(y')+1+y'
I know thats wrong because my 89 gives it as y'=(-3x2+1)/(2y+1), I don't know how to get there though and what happened to the =2.
Thanks

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## The Attempt at a Solution

dy/dx=3x2+2y(y')+1+y'
I know thats wrong because my 89 gives it as y'=(-3x2+1)/(2y+1), I don't know how to get there though and what happened to the =2.
Thanks
You turned the 2 in the original equation to dy/dx, instead of d(2)/dx, but the rest is correct. You should have 3x2+2y(y')+1+y'=0, then put the non-y terms on the right and factor out y' from 2y(y')+y' and you'll have the book's answer.

It also seems like the correct answer should have (-3x2-1) rather than (-3x2+1).

I've gotten closer to the solution but now I'm stuck at
3x^2+1=1-1-y'-2y(y')
3x^2+1=-y'-2y(y')????????

Thanks

Now factor y' out of -y'-2y(y'). You'll have (-1-2y)y'. Then divide both sides by -1-2y.

Another thing..
in your last reply, you have =1-1-y'-2y(y'). Where did the 1-1 come from? If this is d(2)/dx, then it's sort of incorrect. The derivative of a constant is always 0; 1-1 is of course equal to zero, but writing 1-1 is sort of confusing for the person who's error checking.

yeah, I see what you mean - ill avoid doing that... Thanks for your help.