Originally Posted by

**kalagota** just note that $\displaystyle y' = \frac{dy}{dx}$

it came out because of your "chain rule"..

if your $\displaystyle y=e^y \ln{x} + xy^2$ and you are looking for $\displaystyle y' = \frac{dy}{dx}$.. here it is..

taking the derivative of both sides yields to

$\displaystyle \frac{dy}{dx} = \underbrace{\frac{e^y}{x} + e^y\, \ln{x} \, \frac{dy}{dx}}_{product \, rule \, and \, chain \, rule \, on \, e^y} + \underbrace{y^2 + 2xy \, \frac{dy}{dx}}_{product \, rule \, and \, chain \, rule \, on \, y^2}$