Derivative help!

**minus** · October 31st 2008, 02:04 PM

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

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

I have the first derivative:

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

2

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

3

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

4

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

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

F

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

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

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

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

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

g

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

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

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

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

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

h2

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

...

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

...

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

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

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

**Plato** · October 31st 2008, 02:19 PM

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

**minus** · October 31st 2008, 02:27 PM

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

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.

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

$3.826 + 1$

$\frac{100}{4.826}$

$(20.72)(3.826)$

**Plato** · October 31st 2008, 03:20 PM

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
$\frac{{3p^4 - 6p^3 + 3p^2 }}{{9p^4 - 18p^3 + 15p^2 - 6p + 1}}$

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