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

Sep 27, 2017

$f ' ' \left(x\right) = = \frac{2 {x}^{5} - 4 {x}^{3} - 6 x}{{x}^{2} + 1} ^ 4$

graph{(2x^5-4x^3-6x)/(x^2+1)^4 [-10, 10, -5, 5]}

#### Explanation:

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

graph{x/(x^2+1) [-10, 10, -5, 5]}

color(red)(f(x)=g(x)/(h(x)color(red)(=>f'(x)= (g'(x).h(x)-h'(x).g(x))/(f(x))^2
color(red)(g(x)=x
color(red)(h(x)=x^2+1

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

$= \frac{{x}^{2} + 1 - 2 {x}^{2}}{{x}^{2} + 1} ^ 2 = \frac{{x}^{2} - 2 {x}^{2} + 1}{{x}^{2} + 1} ^ 2$$= \frac{- {x}^{2} + 1}{{x}^{2} + 1} ^ 2$ $\implies$

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

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

color(red)(A(x)=(x^n+r)^m=>A'(x)=m(nx^(n-1))(x^n+r)^(m-1)

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

color(red)((a^2+1)(-a^2+1)=1-a^4

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

$= \frac{\textcolor{g r e e n}{- 2 {x}^{5}} - 4 {x}^{3} - \textcolor{b l u e}{2 x} - \textcolor{b l u e}{4 x} + \textcolor{g r e e n}{4 {x}^{5}}}{{x}^{2} + 1} ^ 4$
$= \frac{\textcolor{red}{2 {x}^{5}} - 4 {x}^{3} - \textcolor{b l u e}{6 x}}{{x}^{2} + 1} ^ 4$

$f ' ' \left(x\right) = = \frac{\textcolor{g r e e n}{2 {x}^{5}} - 4 {x}^{3} - \textcolor{b l u e}{6 x}}{{x}^{2} + 1} ^ 4$