Question

Differentiation xy log(x+y)=1

Answer 1

Differentiating w.r.t x
xy/(x+y)×(1+dy/dx)+xdy/dx×log(x+y)+ylog(x+y)=0
(xy+xy(dy/dx))/(x+y) + x(dy/dx)log(x+y) + ylog(x+y)=0
dy/dx(xy/(x+y) + xlog(x+y)=-(xy/(x+y) + ylog(x+y))
dy/dx=-(xy/(x+y) + ylog(x+y))/ (xy/(x+y) + xlog(x+y))