# How do you find the derivative of x^3(2x-5)^4?

Jan 29, 2016

$> {x}^{2} {\left(2 x - 5\right)}^{3} \left(14 x - 15\right)$

#### Explanation:

Differentiate using the$\textcolor{b l u e}{\text{ product and chain rule}}$

Product rule:$\frac{d}{\mathrm{dx}} \left(f \left(x\right) . g \left(x\right)\right) = f \left(x\right) . g ' \left(x\right) + g \left(x\right) . f ' \left(x\right)$

$\frac{d}{\mathrm{dx}} \left({x}^{3} {\left(2 x - 5\right)}^{4}\right)$

$= {x}^{3} \frac{d}{\mathrm{dx}} {\left(2 x - 5\right)}^{4} + {\left(2 x - 5\right)}^{4} \frac{d}{\mathrm{dx}} \left({x}^{3}\right)$

$= {x}^{3} \left[4 {\left(2 x - 5\right)}^{3} \frac{d}{\mathrm{dx}} \left(2 x - 5\right)\right] + {\left(2 x - 5\right)}^{4} .3 {x}^{2}$

$= {x}^{3} \left[4 {\left(2 x - 5\right)}^{3} .2\right] + 3 {x}^{2} {\left(2 x - 5\right)}^{4}$

$= 8 {x}^{3} {\left(2 x - 5\right)}^{3} + 3 {x}^{2} {\left(2 x - 5\right)}^{4}$

$= {x}^{2} {\left(2 x - 5\right)}^{3} \left[8 x + 3 \left(2 x - 5\right)\right]$

$= {x}^{2} {\left(2 x - 5\right)}^{3} \left(14 x - 15\right)$