Answer
$r^{\prime}(x)=12xe^{6x^{2}}$
Work Step by Step
With $u=e^{2x^{2}}$,
we use the generaalized power rule,
$\displaystyle \frac{d}{dx}[u^{n}]=nu^{n-1}\frac{du}{dx}$
$r^{\prime}(x)=3u^{2}\displaystyle \cdot\frac{du}{dx}=3(e^{2x^{2}})^{2}\cdot\frac{du}{dx}=3e^{4x^{2}}\cdot\frac{du}{dx}$
$\displaystyle \frac{du}{dx}=\frac{d}{dx}[e^{2x^{2}}]=e^{2x^{2}}\cdot\frac{d}{dx}[2x^{2}]$
$=e^{2x^{2}}(2\cdot 2x)$
$=4xe^{2x^{2}}$
So,
$r^{\prime}(x)=3e^{4x^{2}}\cdot 4xe^{2x^{2}}$
$r^{\prime}(x)=12xe^{6x^{2}}$
