implicit differentiation: find y" for y = (y+x)ln(y+x)

[Guest]

y=(y+x)ln(y+x)
Find y'' !!!

y' = (y'+1)(ln(y+x))+(y'+1)
y' = -1-(1/ln(y+x))

y''= -0+(y'+1)/(y+x)(ln(y+x))^2
y''= -1/(y+x)(ln(y+x))^3

right or wrong ? :?

 

[skeeter]

y''= -1/(y+x)(ln(y+x))^3

if all in red is the denominator, then yes, it's correct.

y' = -1 - [ln(y+x)]^-1

y" = [ln(y+x)]^-2*(y' + 1)/(y + x)

y" = 1/[ln(y+x)]²*{-1/[ln(y+x)]}*1/(y+x)

y" = -1/{(y+x)[ln(y+x)]³}