I'm having a hard time find 2nd derivative of the following function:

$\displaystyle \frac{p^3 }{ p^3+(1-p)^3}$

I have the first derivative:

$\displaystyle \frac{3p^2[p^3+(1-p)^3][-3p^2-3(1-p)^2](p^3)}{[(p^3+(1-p)^3]^2}$

2

$\displaystyle \frac{3p^5+3p^2(1-p)^3-3p^5+3p^3(1-p)^2}{[(p^3+(1-p)^3]^2}$

3

$\displaystyle \frac{3p^2(1-p)^3+3p^3(1-p)^2}{[p^3+(1-p)^3]^2}$

4

$\displaystyle \frac{3p^2(1-p)^3(1-p)+p}{[p^3+(1-p)^3]^2}$

5

$\displaystyle \frac{3p^2(1-p)^2}{p^3+(1-p)^3]^2}$

F

$\displaystyle 3p^2(1-p)^2$

$\displaystyle 6p(1-p)^2+2(1-p)(-1)3p^2$

$\displaystyle 6p(1-p)^2-6(1-p)p^2$

$\displaystyle 6p(1-p)[(1-p)-p]$

$\displaystyle 6p(1-p)(1-2p)$

g

$\displaystyle 2(p^3+(1-p)^3)[3p^2+3(1-p)^2(-1)]$

$\displaystyle 2[p^3+(1-p)^3][3p^2-3(1-p)^2]$

$\displaystyle 2[p^3+(1-p)^3[3p^2-3+6p-3p^2]$

$\displaystyle 2[p^3+(1-p)^3](-3+6p)$

$\displaystyle 6(p^3+(1-p)^3](2p-1)$

h2

$\displaystyle \frac{6p(1-p)(1-2p)[p^3+(1-p)^3]^2-[6(p^3+(1-p)^3](3p^2)(1-p)^2}{[p^3+(1-p)^3]^4}$

...

$\displaystyle \frac{6(1-p)[p^3+1-p)^3[p^3+(1-p)^3](1-2p)-3p^2(1-p)}{p^3+[(1-p)^3]^4}$

...

$\displaystyle \frac{6(1-p)[p(p^3+(1-p)^3](1-p)^3(1-2p)-3p^2(1-p)}{[p^3+(1-p)3]^3}$

$\displaystyle (p^3+[1-p)^3]^2$

$\displaystyle \frac{.61^3}{.61+(1-.61)^3}$