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

Jan 17, 2016

$\frac{18 \left(1 - {x}^{2}\right)}{{x}^{2} + 3} ^ 3$

#### Explanation:

Using the quotient rule which states that:

If $f \left(x\right) = \frac{u \left(x\right)}{v \left(x\right)}$ then $f ' \left(x\right) = \frac{v u ' - u v '}{v} ^ 2$

$u = {x}^{2} , u ' = 2 x$
$v = {x}^{2} + 3 , v ' = 2 x$

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

$= \frac{6 x}{{x}^{2} + 3} ^ 2$

Reset u & v.

$u = 6 x , u ' = 6$
$v = {\left({x}^{2} + 3\right)}^{2} , v ' = 4 x \left({x}^{2} + 3\right)$

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

$= \frac{\left({x}^{2} + 3\right) \left[6 {x}^{2} + 18 - 24 {x}^{2}\right]}{{x}^{2} + 3} ^ 4 = \frac{18 \left(1 - {x}^{2}\right)}{{x}^{2} + 3} ^ 3$