# How do you differentiate g(x) =(2x+1)^(5/2) (4x-1)^(3/4)  using the product rule?

Mar 29, 2017

g'(x) =(2(2x+1)^(3/2) (13x - 1))/(root(4)((4x-1))

#### Explanation:

Use a combination of the Power Rule $\left({u}^{n}\right) ' = n {u}^{n - 1} u '$
and the Product Rule $\left(u v\right) ' = u v ' + v u '$:

Let $u = {\left(2 x + 1\right)}^{\frac{5}{2}} , u ' = \frac{5}{2} {\left(2 x + 1\right)}^{\frac{3}{2}} \left(2\right) = 5 {\left(2 x + 1\right)}^{\frac{3}{2}}$
Let $v = {\left(4 x - 1\right)}^{\frac{3}{4}} , v ' = \frac{3}{4} {\left(4 x - 1\right)}^{- \frac{1}{4}} \left(4\right) = 3 {\left(4 x - 1\right)}^{- \frac{1}{4}}$

$g ' \left(x\right) = {\left(2 x + 1\right)}^{\frac{5}{2}} \cdot 3 {\left(4 x - 1\right)}^{- \frac{1}{4}} + {\left(4 x - 1\right)}^{\frac{3}{4}} \cdot 5 {\left(2 x + 1\right)}^{\frac{3}{2}}$

Factor remembering ${x}^{\frac{5}{2}} = {x}^{\frac{3}{2}} {x}^{\frac{2}{2}} = {x}^{\frac{3}{2}} x$

$g ' \left(x\right) = {\left(2 x + 1\right)}^{\frac{3}{2}} \left[\frac{3 \left(2 x + 1\right)}{4 x - 1} ^ \left(\frac{1}{4}\right) + 5 {\left(4 x - 1\right)}^{\frac{3}{4}}\right]$

Find a common denominator for the sum (${x}^{\frac{3}{4}} {x}^{\frac{1}{4}} = x$):

$g ' \left(x\right) = {\left(2 x + 1\right)}^{\frac{3}{2}} \left[\frac{6 x + 3 + 5 \left(4 x - 1\right)}{4 x - 1} ^ \left(\frac{1}{4}\right)\right]$

Simplify:
$g ' \left(x\right) = {\left(2 x + 1\right)}^{\frac{3}{2}} \left[\frac{6 x + 3 + 20 x - 5}{4 x - 1} ^ \left(\frac{1}{4}\right)\right]$

$g ' \left(x\right) = \frac{{\left(2 x + 1\right)}^{\frac{3}{2}} \left(26 x - 2\right)}{4 x - 1} ^ \left(\frac{1}{4}\right)$

g'(x) =(2(2x+1)^(3/2) (13x - 1))/(root(4)((4x-1))