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

Dec 26, 2015

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

#### Explanation:

Use the quotient rule: for a function $f \left(x\right) = \frac{g \left(x\right)}{h \left(x\right)}$,

$f ' \left(x\right) = \frac{g ' \left(x\right) h \left(x\right) - h ' \left(x\right) g \left(x\right)}{h \left(x\right)} ^ 2$

$g \left(x\right) = x$
$h \left(x\right) = {\left({x}^{2} + 1\right)}^{2}$

$g ' \left(x\right) = 1$
$h ' \left(x\right) = 4 x \left({x}^{2} + 1\right)$

Thus,

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

Simplify.

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

$f ' \left(x\right) = \frac{1 - 3 {x}^{2}}{{x}^{2} + 1} ^ 3 \textcolor{w h i t e}{s s s}$ First Derivative

To find the second derivative, use the quotient rule again.

$g \left(x\right) = 1 - 3 {x}^{2}$
$h \left(x\right) = {\left({x}^{2} + 1\right)}^{3}$

$g ' \left(x\right) = - 6 x$
$h ' \left(x\right) = 6 x {\left({x}^{2} + 1\right)}^{2}$

Thus,

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

Simplify.

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

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

$f ' ' \left(x\right) = \frac{12 x \left({x}^{2} - 1\right)}{{x}^{2} + 1} ^ 4 \textcolor{w h i t e}{s s s}$ Second Derivative

The main two pitfalls here are

1. algebra
2. remembering to use the chain rule when differentiating things like ${\left({x}^{2} + 1\right)}^{2}$