Question

Can someone check my work for finding y' for (x+y)^2 +y = 2 at (0,1). I know i messed up at a step

Answer 1

\[(x+y)^{2}+y=2\]
\[2(x+y)(1+y')+y'=0\]
\[2+2y'(x+y)+y'=0\]
\[2y'(x+y)+y'=-2\]
\[y'(2(x+y)+1)=-2\]
\[y'=\frac{ -2 }{ 2x+2y+1 }\]

Answer 2

I know i have messed up somewhere, nut I don't know where

Answer 3

\[\large\rm \color{red}{2+2y'(x+y)+y'=0}\]Hmmm here maybe... thinking

Answer 4

\[\large\rm \color{royalblue}{2(x+y)}(1+y')\quad=\quad \color{royalblue}{2(x+y)}(1)+\color{royalblue}{2(x+y)}y'\]

Answer 5

okay i can see that, moving around the numbers confused me somewhat. thank you :)

Answer 6

cool c: