Answer
$v^{\prime}(x)=2 \cdot 3^{2x+1} \ln (3)+3 \cdot e^{3x+1}$
Work Step by Step
We have: $v(x)=3^{2x+1}+e^{3x+1}$
We differentiate both sides with respect to $x$.
$v^{\prime}(x)=\dfrac{d}{dx}[3^{2x+1}]+\dfrac{d}{dx}[e^{3x+1}]$
Use rule: $\displaystyle \frac{d}{dx}[a^{u}]=a^u \ln (a) \frac{du}{dx}$
Now, $v^{\prime}(x)=2 \cdot 3^{2x+1} \ln (3)+(3) e^{3x+1}$
So,
$v^{\prime}(x)=2 \cdot 3^{2x+1} \ln (3)+3 \cdot e^{3x+1}$
