# implicit diff. question

• Dec 6th 2009, 09:37 AM
Archduke01
implicit 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 AM
Arturo_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 AM
Archduke01
Quote:

Originally Posted by Arturo_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'.

I got as far as \$\displaystyle 6xy+3y^2 = 3y^2 y' / 1+ y'\$. How do I proceed from there?
• Dec 6th 2009, 10:18 AM
Arturo_026
Hold on, This is wrong. give me a bit more time to correct it
• Dec 6th 2009, 10:25 AM
Archduke01
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 AM
Arturo_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 AM
Archduke01
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.