1. ## Derivative help!

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

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

...

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

...

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

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

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

2. Using MathCad, I get $\displaystyle \frac{{12p^3 - 18p^2 + 6p}}{{27p^6 - 81p^5 + 108p^4 - 81p^3 + 36p^2 - 9p + 1}}$

3. [quote=Plato;212682]Using MathCad, I get $\displaystyle \frac{{12p^3 - 18p^2 + 6p}}{{27p^6 - 81p^5 + 108p^4 - 81p^3 + 36p^2 - 9p + 1}}$[/

quote]

Thanks for your help plato. Your results is for 2nd derivative right? If indeed it's 2nd derivative can you please double check my answer for first derivative.

$\displaystyle \frac{61^3}{39^3}$

$\displaystyle 3.826 + 1$

$\displaystyle \frac{100}{4.826}$

$\displaystyle (20.72)(3.826)$

4. Originally Posted by minus
Thanks for your help plato. Your results is for 2nd derivative right? If indeed it's 2nd derivative can you please double check my answer for first derivative.
That is not my result; I used a computer algebra system to do it.
I think doing this sort of a problem without a CAS is a pointless exercise!
The first derivative is
$\displaystyle \frac{{3p^4 - 6p^3 + 3p^2 }}{{9p^4 - 18p^3 + 15p^2 - 6p + 1}}$