Answer
$r^{\prime}(x)=4e^{4x-2}$
Work Step by Step
With $u=e^{2x-1}$,
we use the generaalized power rule,
$\displaystyle \frac{d}{dx}[u^{n}]=nu^{n-1}\frac{du}{dx}$
$r^{\prime}(x)=2u\displaystyle \cdot\frac{du}{dx}=2e^{2x-1}\cdot\frac{du}{dx}$
$\displaystyle \frac{du}{dx}=\frac{d}{dx}[e^{2x-1}]=e^{2x-1}\cdot\frac{d}{dx}[2x-1]=2e^{2x-1}$
So,
$r^{\prime}(x)=2e^{2x-1}\cdot 2e^{2x-1}$
$r^{\prime}(x)=4e^{4x-2}$
