Math Help - Finding Second Derivatives

1. Finding Second Derivatives

This is a continuation of a problem which I have the derivative of. It is -2x^3+6x^2-6x+2/(x^3-3x^2+2x)^2

The question is to find the second derivative but I keep getting caught up in the simplifying areas around it after taking the derivative of it. Can anyone help me out by even posting a couple of main steps and the final answer part?

I'm not sure if I'm doing it right either.

Thanks again.

2. Simplify first

This is a continuation of a problem which I have the derivative of. It is -2x^3+6x^2-6x+2/(x^3-3x^2+2x)^2

The question is to find the second derivative but I keep getting caught up in the simplifying areas around it after taking the derivative of it. Can anyone help me out by even posting a couple of main steps and the final answer part?

I'm not sure if I'm doing it right either.

Thanks again.
Have you tried factorising the expression first?

$\frac{-2x^3+6x^2-6x+2}{(x^3-3x^2+2x)^2}$

$= \frac{-2(x^3-3x^2+3x-1)}{x^2(x^2-3x+2)^2}$

$=\frac{-2(x-1)^3}{x^2(x-1)^2(x-2)^2}$

$=\frac{-2(x-1)}{x^2(x-2)^2}$

Have another go, starting from here.

Hello ShadoownedHave you tried factorising the expression first?

$\frac{-2x^3+6x^2-6x+2}{(x^3-3x^2+2x)^2}$

$= \frac{-2(x^3-3x^2+3x-1)}{x^2(x^2-3x+2)^2}$

$=\frac{-2(x-1)^3}{x^2(x-1)^2(x-2)^2}$

$=\frac{-2(x-1)}{x^2(x-2)^2}$

Have another go, starting from here.