# How do you differentiate f(x)=x(x^3-3)  using the product rule?

Feb 7, 2016

$f ' \left(x\right) = 4 {x}^{3} - 3$

#### Explanation:

product rule : if f(x) = g(x).h(x)

then f'(x) = g(x).h'(x) + h(x).g'(x)

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

$f ' \left(x\right) = x \frac{d}{\mathrm{dx}} \left({x}^{3} - 3\right) + \left({x}^{3} - 3\right) \frac{d}{\mathrm{dx}} \left(x\right)$

$= x \left(3 {x}^{2}\right) + \left({x}^{3} - 3\right) .1$

$\Rightarrow f ' \left(x\right) = 3 {x}^{3} + {x}^{3} - 3 = 4 {x}^{3} - 3$