Answer
$s^{\prime}(x)=2xe^{2x-1}(1+x)$
Work Step by Step
$s(x) $ is a product$ u(x)v(x)$ where
$u(x)=x^{2},\qquad u^{\prime}(x)=2$
$v(x)=e^{2x-1},$
$v^{\prime}(x)= \displaystyle \frac{d}{dx}[e^{w}]=e^{u}\cdot\frac{dw}{dx}$,
with $w=2x-1\displaystyle \quad\frac{dw}{dx}=2x$
$v^{\prime}(x)=\displaystyle \frac{d}{dx}[e^{2x-1}]=e^{2x-1}\cdot(2)=2e^{2x-1}$
So
$s(x)=u^{\prime}(x)v(x)+u(x)v^{\prime}(x)$
$=2xe^{2x-1}+x^{2}(2e^{2x-1})$
$s^{\prime}(x)=2xe^{2x-1}(1+x)$
