$\displaystyle (x+y)^3 = x^3 + y^3$ at the point (-1,1). We need to find y' ... what do?

Here's what I did, please tell me where I went wrong.

3(x+y)^2 (1+y') = 3x^2 + 3y^2 y'

How do I proceed?

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- Dec 6th 2009, 09:37 AMArchduke01implicit diff. question
$\displaystyle (x+y)^3 = x^3 + y^3$ at the point (-1,1). We need to find y' ... what do?

Here's what I did, please tell me where I went wrong.

3(x+y)^2 (1+y') = 3x^2 + 3y^2 y'

How do I proceed? - Dec 6th 2009, 09:53 AMArturo_026
$\displaystyle 3(x+y)^2 (1+y') = 3x^2 + 3y^2 y' $

$\displaystyle 3(x^2 + 2xy + y^2) (1+y') = 3x^2 + 3y^2 y'$

Keep solving and then you want to isolate the y' in one side and the x and y in the other. Then you'll solve for y'. - Dec 6th 2009, 10:03 AMArchduke01
- Dec 6th 2009, 10:18 AMArturo_026
Hold on, This is wrong. give me a bit more time to correct it

- Dec 6th 2009, 10:25 AMArchduke01
Look, buddy. How did you go from

$\displaystyle 3(x^2 + 2xy + y^2) (1+y') = 3x^2 + 3y^2 y'

$

to

$\displaystyle 3[x^2 + 2xy + y^2 + y'(x^2 + 2xy + y^2)] = 3x^2 + 3y^2 y'

$

???

EDIT: Nevermind, I'll wait until your edit. - Dec 6th 2009, 10:33 AMArturo_026
$\displaystyle (x^2 + 2xy + y^2)(1 + y') = x^2 + y^2 y'$

$\displaystyle x^2 + 2xy + y^2 + x^2 y' + 2xy y' + y^2 y' = x^2 + y^2 y'$

$\displaystyle y' (x^2 + 2xy) = -2xy - y^2$

$\displaystyle y' = (-2xy - y^2)/(x^2 + 2xy)$

This should be correct.

I apologize for my mistake, if it serves as an excuse, I was solving it in my head because i couldn't find paper. - Dec 6th 2009, 10:40 AMArchduke01
The problem I was having is that I didn't expand ASAP. Your step by step solution helped me see my mistakes. Thanks for the help, bromosapien. And no worries about the error.