What is the second derivative of f(x)=3x^3-18x^2+6x-2 ?

Jan 7, 2016

$f ' ' \left(x\right) = 18 x - 36 = 18 \left(x - 2\right)$

Explanation:

First, you have to obtain the first derivate:

$f ' \left(x\right) = 3 \cdot 3 {x}^{3 - 1} - 18 \cdot 2 \cdot {x}^{2 - 1} + 6 {x}^{1 - 1} + 0 =$
$= 9 {x}^{2} - 36 x + 6 {x}^{0} = 9 {x}^{2} - 36 x + 6$

Now you derivate again in $x$

$f ' ' \left(x\right) = 9 \cdot 2 {x}^{2 - 1} - 36 \cdot 1 {x}^{1 - 1} + 0 = 18 x - 36 {x}^{0} =$

$= 18 x - 36 \cdot 1 = 18 x - 36 = 18 \left(x - 2\right)$