Calculus : Derivative

If xy log(x+y) = 1, prove that dy/dx = -y(x^2y +x+ y)/x(xy^2 + x + y)

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  1. (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.

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