How do you differentiate e^(2x^2-4x)  using the chain rule?

$= 4 \left({e}^{2 {x}^{2} - 4 x}\right) \left(x - 1\right)$
$\frac{d}{\mathrm{dx}} \left({e}^{2 {x}^{2} - 4 x}\right) = \left({e}^{2 {x}^{2} - 4 x}\right) \frac{d}{\mathrm{dx}} \left(2 {x}^{2} - 4 x\right)$
$= \left({e}^{2 {x}^{2} - 4 x}\right) \left(\frac{d}{\mathrm{dx}} \left(2 {x}^{2}\right) - \frac{d}{\mathrm{dx}} \left(4 x\right)\right)$
=(e^(2x^2-4x))(2*2x^(2-1)-4))
$= \left({e}^{2 {x}^{2} - 4 x}\right) \left(4 x - 4\right)$
$= 4 \left({e}^{2 {x}^{2} - 4 x}\right) \left(x - 1\right)$