Question

Let $xy=x+y$ then prove that $\frac{dy}{dx}+\frac{1}{(x-1{)}^2}=0$.

Answer 1

Given,

$xy=x+y$

or, $(x-1)y=x$

or, $y=\frac{x}{x-1}$

Now differentiating both sides with respect to $x$ we get,

$\frac{dy}{dx}=\frac{(x-1).1-x.1}{(x-1{)}^2}=-\frac{1}{(x-1{)}^2}$

or, $\frac{dy}{dx}+\frac{1}{(x-1{)}^2}=0$.