Question

find dy/dx of xy log (x+y) = 1​

Answer 1

Step-by-step explanation:

xy log(x+y)=1

y=log(X+y)+X(dy/dx)log(x+y)+(xy/X+y)(1+dy/dx)=0

since log(X+y)=1/xy

y/xy+dy/dx(x/xy)+(xy/x+y)(1+dy/dx)=0

1/X+1/y(dy/dx)+(xy/x+y)+(xy/X+y)(dy/dx)=0

dy/dx{(1/y)+(xy/x+y)} = -{(xy/x+y)+(1/X)}

dy/dx{x+y+xy^2/y(x+y)} = -1 {(x^2y+x+y)/(x+y)x}

dy/dx = -y(x^2y+x+y)/(x+y+xy^2)