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

Apr 20, 2017

$f ' ' \left(x\right) = \frac{4 \left(3 {x}^{8} + 7 {x}^{6} + 3 {x}^{4} - 15 {x}^{2} + 2\right)}{{x}^{2} + 1} ^ 3$

#### Explanation:

Use the quotient rule to find the first derivative $\left(\frac{u}{v}\right) ' = \frac{v u ' - u v '}{v} ^ 2$

Let $u = {\left({x}^{2} - 1\right)}^{3} , u ' = 3 {\left({x}^{2} - 1\right)}^{2} \left(2 x\right) = 6 x {\left({x}^{2} - 1\right)}^{2}$

Let $v = {x}^{2} + 1 , v ' = 2 x$

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

Factor: $f \left(x '\right) = \frac{2 x {\left({x}^{2} - 1\right)}^{2} \left[3 \left({x}^{2} + 1\right) - \left({x}^{2} - 1\right)\right]}{{x}^{2} + 1} ^ 2$

Simplify: $f ' \left(x\right) = \frac{2 x {\left({x}^{2} - 1\right)}^{2} \left[3 {x}^{2} + 3 - {x}^{2} + 1\right]}{{x}^{2} + 1} ^ 2$

Simplify: $f ' \left(x\right) = \frac{2 x {\left({x}^{2} - 1\right)}^{2} \left(2 {x}^{2} + 4\right)}{{x}^{2} + 1} ^ 2$

Simplify: $f ' \left(x\right) = \frac{4 x {\left({x}^{2} - 1\right)}^{2} \left({x}^{2} + 2\right)}{{x}^{2} + 1} ^ 2$

To find the second derivative , one method is to multiply out the numerator of the first derivative and then use the quotient rule:

$f ' \left(x\right) = \frac{4 x \left({x}^{4} - 2 {x}^{2} + 1\right) \left({x}^{2} + 2\right)}{{x}^{2} + 1} ^ 2$

$f ' \left(x\right) = \frac{\left(4 {x}^{5} - 8 {x}^{3} + 4 x\right) \left({x}^{2} + 2\right)}{{x}^{2} + 1} ^ 2$

$f ' \left(x\right) = \frac{4 {x}^{7} + \left(8 {x}^{5}\right) - 8 {x}^{5} - 16 {x}^{3} + 4 {x}^{3} + 8 x}{{x}^{2} + 1} ^ 2$

$f ' \left(x\right) = \frac{4 {x}^{7} - 12 {x}^{3} + 8 x}{{x}^{2} + 1} ^ 2$

Let $u = 4 {x}^{7} - 12 {x}^{3} + 8 x = 4 x \left({x}^{6} - 3 {x}^{2} + 2\right)$

$u ' = 28 {x}^{6} - 36 {x}^{2} + 8 = 4 \left(7 {x}^{6} - 9 {x}^{2} + 2\right)$

Let $v = {\left({x}^{2} + 1\right)}^{2} , v ' = 2 \left({x}^{2} + 1\right) \left(2 x\right) = 4 x \left({x}^{2} + 1\right)$

$f ' ' \left(x\right) =$
$\frac{{\left({x}^{2} + 1\right)}^{2} \cdot 4 \left(7 {x}^{6} - 9 {x}^{2} + 2\right) - 4 x \left({x}^{6} - 3 {x}^{2} + 2\right) 4 x \left({x}^{2} + 1\right)}{{x}^{2} + 1} ^ 4$

Factor:$f ' ' \left(x\right) =$
$\frac{4 \left({x}^{2} + 1\right) \left[\left({x}^{2} + 1\right) \left(7 {x}^{6} - 9 {x}^{2} + 2\right) - 4 {x}^{2} \left({x}^{6} - 3 {x}^{2} + 2\right)\right]}{{x}^{2} + 1} ^ 4$

Cancel one if the $\left({x}^{2} + 1\right)$ from both numerator & denominator:

$f ' ' \left(x\right) = \frac{4 \left[\left({x}^{2} + 1\right) \left(7 {x}^{6} - 9 {x}^{2} + 2\right) - 4 {x}^{2} \left({x}^{6} - 3 {x}^{2} + 2\right)\right]}{{x}^{2} + 1} ^ 3$

Simplify the numerator through distribution:

$f ' ' \left(x\right) =$
$\frac{4 \left[7 {x}^{8} - 9 {x}^{4} + 2 {x}^{2} + 7 {x}^{6} - 9 {x}^{2} + 2 - 4 {x}^{8} + 12 {x}^{4} - 8 {x}^{2}\right]}{{x}^{2} + 1} ^ 3$

Simplify:

$f ' ' \left(x\right) = \frac{4 \left(3 {x}^{8} + 7 {x}^{6} + 3 {x}^{4} - 15 {x}^{2} + 2\right)}{{x}^{2} + 1} ^ 3$