# What is the second derivative of? : 2(x^2-1)^3

Oct 5, 2017

$12 {\left({x}^{2} - 1\right)}^{2} + 48 {x}^{2} \left({x}^{2} - 1\right)$

#### Explanation:

To find the second derivative, you need to find the first derivative, so let's do that immediately:

$\frac{d}{\mathrm{dx}} 2 {\left({x}^{2} - 1\right)}^{3} = 6 {\left({x}^{2} - 1\right)}^{2} \cdot \left(2 x\right) = 12 x {\left({x}^{2} - 1\right)}^{2}$

I used the chain rule and product rule to find the first derivative. Now let's find the second derivative:

d/dx 12x(x^2-1)^2 = 12(x^2-1)^2+12x*[2(x^2-1)*2x] = 12(x^2-1)^2 + 48x^2(x^2-1)

For the second derivative, I used the product rule and chain rule.

Oct 5, 2017

$\frac{{d}^{2}}{{\mathrm{dx}}^{2}} 2 {\left({x}^{2} - 1\right)}^{3} = 12 \left({x}^{2} - 1\right) \left(5 {x}^{2} - 1\right)$

#### Explanation:

First we generate the first derivative using the chain rule:

$\frac{d}{\mathrm{dx}} 2 {\left({x}^{2} - 1\right)}^{3} = 2 \left(3\right) 2 {\left({x}^{2} - 1\right)}^{2} \frac{d}{\mathrm{dx}} \left({x}^{2} - 1\right)$

$\text{ } = 6 {\left({x}^{2} - 1\right)}^{2} \left(2 x\right)$

$\text{ } = 12 x {\left({x}^{2} - 1\right)}^{2}$

Then, we generate the second derivative using the product rule and the chain rule:

$\frac{{d}^{2}}{{\mathrm{dx}}^{2}} 2 {\left({x}^{2} - 1\right)}^{3} = \frac{d}{\mathrm{dx}} 12 x {\left({x}^{2} - 1\right)}^{2}$
$\text{ } = 12 x \left(\frac{d}{\mathrm{dx}} {\left({x}^{2} - 1\right)}^{2}\right) + \left(\frac{d}{\mathrm{dx}} 12 x\right) {\left({x}^{2} - 1\right)}^{2}$
$\text{ } = 12 x \left(2 \left({x}^{2} - 1\right) \left(\frac{d}{\mathrm{dx}} \left({x}^{2} - 1\right)\right)\right) + 12 {\left({x}^{2} - 1\right)}^{2}$
$\text{ } = 24 x \left({x}^{2} - 1\right) \left(2 x\right) + 12 {\left({x}^{2} - 1\right)}^{2}$

$\text{ } = 12 \left({x}^{2} - 1\right) \left\{4 {x}^{2} + \left({x}^{2} - 1\right)\right\}$
$\text{ } = 12 \left({x}^{2} - 1\right) \left(5 {x}^{2} - 1\right)$