# How do you find the derivative of (3x+1)^(3/2) (2x+4)?

Sep 17, 2015

$2 {\left(3 x + 1\right)}^{\frac{3}{2}} + \left(9 \left(x + 2\right) {\left(3 x + 1\right)}^{\frac{1}{2}}\right)$

#### Explanation:

it can be found by uv rule
let us assume $u = {\left(3 x + 1\right)}^{\frac{3}{2}}$ and $v = 2 x + 4$
from basic formula of diferentiation
$d \frac{u v}{\mathrm{dx}}$=$u \frac{d}{\mathrm{dx}} v + v \frac{d}{\mathrm{dx}} u$=$u v ' + v u '$
$u = {\left(3 x + 1\right)}^{\frac{3}{2}} v = 2 x + 4$
$u ' = \left(\frac{3}{2}\right) {\left(3 x + 1\right)}^{\frac{3}{2} - 1} 3$
$\implies u ' = \frac{9}{2} {\left(3 x + 1\right)}^{\frac{1}{2}}$
$v ' = 2$
substituiting all the values
$\frac{d}{\mathrm{dx}} {\left(3 x + 1\right)}^{\frac{3}{2}} \left(2 x + 4\right)$=$2 {\left(3 x + 1\right)}^{\frac{3}{2}} + \left(\left(2 x + 4\right) \frac{9}{2} {\left(3 x + 1\right)}^{\frac{1}{2}}\right)$
=$2 {\left(3 x + 1\right)}^{\frac{3}{2}} + \left(9 \left(x + 2\right) {\left(3 x + 1\right)}^{\frac{1}{2}}\right)$