# Question #28efc

By the power rule, the derivative of ${f}^{n} \left(x\right) = n {f}^{n - 1} \left(x\right) f ' \left(x\right)$.
In your case, $f \left(x\right) = 3 {x}^{2} + 5 x + 1$, and $n = \frac{3}{2}$
This means that $n - 1 = \frac{1}{2}$, and the derivative $f ' \left(x\right)$ is also easy to calculate, since it is $6 x + 5$.
Putting the pieces together, we can translate the general formula $n {f}^{n - 1} \left(x\right) f ' \left(x\right)$ into
$\frac{3}{2} {\left(3 {x}^{2} + 5 x + 1\right)}^{\frac{1}{2}} \left(6 x + 5\right)$