Answer
$v^{\prime}(x)=2 \ln (4) \cdot 4^{2x} e^{2x}+2 \cdot 4^{2x}e^{2x}$
Work Step by Step
We have: $v(x)=e^{2x} 4^{2x}$
We differentiate both sides with respect to $x$.
$v^{\prime}(x)=\dfrac{d}{dx}[e^{2x}]+4^{2x}\dfrac{d}{dx}[e^{2x}]$
Use rule: $\displaystyle \frac{d}{dx}[a^{u}]=a^u \ln (a) \frac{du}{dx}$
Now, $v^{\prime}(x)=e^{2x} \cdot 4^{2x} \ln (4) \dfrac{d}{dx}(2x)+(4^{2x}) e^{2x} \dfrac{d}{dx} (2x)$
So,
$v^{\prime}(x)=2 \ln (4) \cdot 4^{2x} e^{2x}+2 \cdot 4^{2x}e^{2x}$
