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

Oct 30, 2015

See pointer for calculation method in Explanation

#### Explanation:

Let $u = {\left(2 x + 1\right)}^{\frac{5}{2}} \text{ }$ then we have $\frac{\mathrm{du}}{\mathrm{dx}} = \frac{5}{2} {\left(2 x + 1\right)}^{\frac{3}{2}} \left(2\right)$

Let $v = {\left(4 x - 1\right)}^{\frac{3}{4}} \text{ }$ then $\frac{\mathrm{dv}}{\mathrm{dx}} = \frac{3}{4} {\left(4 x - 1\right)}^{- \frac{1}{4}} \left(4\right)$

So
$\frac{\mathrm{du}}{\mathrm{dx}} = \frac{5}{\cancel{2}} {\left(2 x + 1\right)}^{\frac{3}{2}} \cancel{\left(2\right)}$

$\frac{\mathrm{dv}}{\mathrm{dx}} = \frac{3}{\cancel{4}} {\left(4 x - 1\right)}^{- \frac{1}{4}} \left(\cancel{4}\right) = \frac{3}{{\left(4 x - 1\right)}^{\frac{1}{4}}}$

Substitute into:

${f}^{'} \left(x\right) = v \frac{\mathrm{du}}{\mathrm{dx}} + u \frac{\mathrm{dv}}{\mathrm{dx}}$

Once that is done it is a matter of simplification!