I am asked to differentiate $y = 2x^x$

taking logs to base e

$\ln y = x \ln 2x $

implicit differentiation

$\dfrac{1}{y} \dfrac{dy}{dx} = x \dfrac{2}{2x} + 1. \ln 2x $

$\dfrac{1}{y} \dfrac{dy}{dx} = 1 + \ln 2x $

$\dfrac{dy}{dx} = 2x^x (1 + \ln 2x) $

I think this is the right answer. The answer given is $2x^x ( 1 + \ln x) $ am I doing something wrong is it a mistake in the answer?