Im pretty sure i did the derivative part right but i just want a second take on my algerbra

$\displaystyle \frac {e^{x^2}}{1+ \ln x}$

$\displaystyle \frac {dy}{dx} = 1 + \ln x * e^{x^2} *2x - e^{x^2} * \frac {1}{x} $

$\displaystyle = 2x(1 + \ln x ) * e^{x^2} - \frac {e^{x^2}}{x}$