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

Jan 7, 2017

$\frac{{d}^{2}}{{\mathrm{dx}}^{2}} {\left({x}^{2} - 1\right)}^{3} = 6 \left({x}^{2} - 1\right) \left(5 {x}^{2} - 1\right)$

Explanation:

Let's calculate the first derivative using the chain rule:

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

Now using the product rule:

$\frac{{d}^{2}}{{\mathrm{dx}}^{2}} {\left({x}^{2} - 1\right)}^{3} = \frac{d}{\mathrm{dx}} \left[6 x {\left({x}^{2} - 1\right)}^{2}\right] = 6 {\left({x}^{2} - 1\right)}^{2} + 24 {x}^{2} \left({x}^{2} - 1\right) = 6 \left({x}^{2} - 1\right) \left({x}^{2} - 1 + 4 {x}^{2}\right) = 6 \left({x}^{2} - 1\right) \left(5 {x}^{2} - 1\right)$

Jan 7, 2017

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = 6 \left(5 {x}^{4} - 6 {x}^{2} + 1\right) .$

Explanation:

Let $y = {\left({x}^{2} - 1\right)}^{3}$

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

$\therefore \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = \frac{d}{\mathrm{dx}} \left(\frac{\mathrm{dy}}{\mathrm{dx}}\right) = \frac{d}{\mathrm{dx}} \left\{6 x {\left({x}^{2} - 1\right)}^{2}\right\}$

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

=6[x{d/dx(x^2-1)^2}+(x^2-1)^2{d/dxx}]...["Product Rule]"

$= 6 \left[x \left\{2 \left({x}^{2} - 1\right) \frac{d}{\mathrm{dx}} \left({x}^{2} - 1\right)\right\} + {\left({x}^{2} - 1\right)}^{2} \left(1\right)\right]$

$= 6 \left[x \left\{4 x \left({x}^{2} - 1\right)\right\} + {\left({x}^{2} - 1\right)}^{2}\right]$

$= 6 \left({x}^{2} - 1\right) \left\{4 {x}^{2} + \left({x}^{2} - 1\right)\right\} = 6 \left({x}^{2} - 1\right) \left(5 {x}^{2} - 1\right)$

$\therefore \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = 6 \left(5 {x}^{4} - 6 {x}^{2} + 1\right) .$

Alternatively, we can expand

${\left({x}^{2} - 1\right)}^{3} \text{ as } {\left({x}^{2}\right)}^{3} - {1}^{3} - 3 {x}^{2} \left({x}^{2} - 1\right)$

and get, $y = {x}^{6} - 1 - 3 {x}^{4} + 3 {x}^{2}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = 6 {x}^{5} - 12 {x}^{3} + 6 x$

$\therefore \frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = 30 {x}^{4} - 36 {x}^{2} + 6 = 6 \left(5 {x}^{4} - 6 {x}^{2} + 1\right) ,$ as before!

Enjoy Maths.!