How to calculate the second derivative of this example? I am trying to calculate the second derivative of the following equation:
$$f(x)=\frac{6}{(x^2+3)}$$
$$f'(x)=\frac{-12x}{(x^2+3)^2}$$
The correct answer given was $$f''(x) = \frac{36(x+1)(x-1)}{(x^2+3)^3}$$ but the second derivative that I have gotten was different from the answer.
I have shown the calculations below and would like to understand where is the mistake. Would anyone be able to point out my mistake?
$$f'(x) =\frac{-12x}{(x^2+3)^2} = (-12x)(x^2+3)^{-2}$$
$$= (-12x)\frac{d}{dx} [(x^2+3)^{-2}]+(x^2+3)^{-2}\frac{d}{dx}[-12x]$$
$$=         48x^2(x^2+3)^{-3} -12(x^2+3)^{-2}$$

Keep going...
$$48x^2(x^2+3)^{-3}-12(x^2+3)^{-2}$$
$$=\frac{48x^2}{(x^2+3)^3}-\frac{12}{(x^2+3)^2}$$
$$=\frac{48x^2-12(x^2+3)}{(x^2+3)^3}$$
$$=\frac{36x^2-36}{(x^2+3)^3}$$
Factorise the numerator to match the given answer.
