#### Answer

$$2{x^{2x}}\left( {\ln x + 1} \right)$$

#### Work Step by Step

$$\eqalign{
& \frac{d}{{dx}}\left( {{x^{2x}}} \right) \cr
& {\text{we use the inverse relationship }}{e^{\ln x}} = x \cr
& = {e^{\ln {x^{2x}}}} \cr
& = {e^{2x\ln x}} \cr
& {\text{evaluate the derivative}} \cr
& = \frac{d}{{dx}}\left( {{e^{2x\ln x}}} \right) \cr
& {\text{by }}\frac{d}{{dx}}\left( {{e^{u\left( x \right)}}} \right) = {e^{u\left( x \right)}}u'\left( x \right) \cr
& = {e^{2x\ln x}}\frac{d}{{dx}}\left( {2x\ln x} \right) \cr
& {\text{product rule}} \cr
& = {e^{2x\ln x}}\left( {2\ln x + \frac{{2x}}{x}} \right) \cr
& {\text{simplify}} \cr
& = {x^{2x}}\left( {2\ln x + 2} \right) \cr
& = 2{x^{2x}}\left( {\ln x + 1} \right) \cr} $$