Question

If $x=y\left(\log xy\right)$ then find $\frac{dy}{dx}$.

Answer 1

Given,

$x=y\left(\log xy\right)$

or, $x=y\{\log x+\log y\}$
$x=y\log x+y\log y$

Now differentiate both sides with respect to $x$ we get,
$1=\frac{y}{x}+\frac{dy}{dx}\log x+\log y\frac{dy}{dx}+y\left(\frac{1}{y}\right)\frac{dy}{dx}$
$1=\frac{y}{x}+\frac{dy}{dx}(\log x+\log y+1)$
$\frac{dy}{dx}(\log x+\log y+1)=1-\frac{y}{x}$
$\therefore \frac{dy}{dx}=\frac{x-y}{x(\log x+\log y+1)}$