Question
If xy log(x+y) = 1, prove that dy/dx = -y(x^2y +x+ y)/x(xy^2 + x + y)
Answers
bobpursley
(ydx+xdy)log(x+y)+ xy/(x+y) * (dx+dy)=0
(y+xy')log(x+y)=-xy(1+y')/(x+y)
but log(x+y)=1/xy
(y+xy')/xy=-xy((1+y')/(x+y)
(y+xy')(x+y)=-(x^2y^2)(1+y')
xy'(x+y)+y'(x^2y^2)=-y(x+y)-(xy)^2
y'= -y(x+y)-(xy)^2 ]/(x^2+xy+x^2y^2)
which reduces to what you want.
(y+xy')log(x+y)=-xy(1+y')/(x+y)
but log(x+y)=1/xy
(y+xy')/xy=-xy((1+y')/(x+y)
(y+xy')(x+y)=-(x^2y^2)(1+y')
xy'(x+y)+y'(x^2y^2)=-y(x+y)-(xy)^2
y'= -y(x+y)-(xy)^2 ]/(x^2+xy+x^2y^2)
which reduces to what you want.