Calculate the second derivative of this rational function

Question

I'm studying for my exam Math and I came across a problem with one exercise.

Calculate the second derivative of $$f'(x)=\frac{2x^3-3x^2}{(x^2-1)^2}$$

I just can't seem to calculate the second derivative of this rational function. If someone could help me . I just can't seem to get the right answer

See my calculations and what the right answer should be according to my teacher

Thanks!

Answer 1

In your very first step, where you have $6x^2 - 6x \cdot (x^2 - 1)^2 \ldots$ in magenta, the first part -- the $6x^2 - 6x$ -- should be in parentheses.

Answer 2

I split the fraction into two parts, then used the product rule on both parts.

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

After product rule on both terms we have

$\dfrac{12x^3-8x^4}{(x^2-1)^3} + \dfrac{6x^2-6x}{(x^2-1)^2}$

$\dfrac{12x^3-8x^4}{(x^2-1)^3} + \dfrac{(6x^2-6x)*(x^2-1)}{(x^2-1)^{2}*(x^2-1)}$

After simplifying the numerators and factoring out a $-2x$, we are left with

$f''(x)=\dfrac{-2x(x^3-3x^2+3x-3)}{(x^2-1)^{3}}$