Implicit differentiation - Help

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Homework Help Overview

The discussion revolves around finding the derivative y' using implicit differentiation for the equation x³ + y² + x + y = 2.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss their attempts at differentiating the equation and express confusion regarding the correct manipulation of terms. Questions arise about the treatment of the constant on the right side of the equation and how it affects the differentiation process.

Discussion Status

Some participants have provided guidance on factoring and rearranging terms to isolate y', while others are questioning specific steps taken in the differentiation process. There is an ongoing exploration of the correct approach without a clear consensus on the final answer.

Contextual Notes

Participants note potential errors in the differentiation process, particularly regarding the treatment of the constant and the resulting expressions for y'.

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Homework Statement



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

Homework Equations





The Attempt at a Solution


dy/dx=3x2+2y(y')+1+y'
I know that's 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 that's 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.
 

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