Differentiate (w.r.t x ) $\displaystyle e^{2logx + 3x}$

I tried to solve it as follows:-

$\displaystyle e^{2logx} + e^{3x}$

$\displaystyle \frac{\mbox{d}}{\mbox{dx}}(e^{2logx}) + \frac{\mbox{d}}{\mbox{dx}}(e^{3x}) $

Using chain rule my answer is:-

$\displaystyle e^{2logx}. \frac{\ 2}{x} + e^{3x}.3 $

Is it correct.? The actual answer is different.