How to prove that the second derivative of (1/(s^2+1)) = 2(-1+3s^2)/(s^2+1)^3?

I cannot get power 3 in the denominator

Simple.

d/dx [ 1/u] = -1/u^2 * u'.

Let u = s^2 +1

u' = 2s.

derivative = -1/(s^2+1)^2 (2s)

= -2s/(s^2+1)^2

_________________________-

Now, use quotient rule if you like.

(s^2+1)^2(-2) - (-2s)(2(s^2+1)(2s))
____________
(s^2+1)^4

Divide top and bottom by (s^2+1)

-2(s^2+1) + 8s^2
______________
(s^2+1)^3

++++++++++++++++++++++++++

6s^2 - 2
_________
(s^2+1)^3

++++++++++++++++++++++++

2(3s^2 -1)
_________
(s^2+1)^3

[1/(s^2+1)]'=-2s/(s^2+1)^2
[1/(s^2+1)]''=([-2s]'(s^2+1)^2-(-2s)[(s...
=(-2(s^2+1)^2+4s(s^2+1)2s)/(s^2+1)^4=
=(-2(s^2+1)+8s^2)(s^2+1)/(s^2+1)^4=
=(-2s^2-2+8s^2)/(s^2+1)^3=
=(-2+6s^2)/(s^2+1)^3
=2(-1+3s^2)/(s^2+1)^3

Try to simplify every derivative after the first one by calculating the gcd between both terms of the fraction (usually a power of the denominator)