<s>[INST] Simplifying derivative result I am doing the derivative of
$$f(x) = \frac{x^2 -4x +3}{x^2-1}$$
So my result is the following
$$f'(x) = \frac{4x^2 -8x +4}{(x^2-1)^2}$$
I am sure the answer is correct, but in my solutions book and In Wolfram Alpha they simplify until
$$f'(x) = \frac{4}{(x+1)^2}$$
And I don't know why, which steps are they doing?
 [/INST] You are correct. Now note that
$$\frac{4x^2-8x+4}{(x^2-1)^2}=\frac{4(x^2-2x+1)}{((x-1)(x+1))^2}=\frac{4(x-1)^2}{(x-1)^2(x+1)^2}=\frac{4}{\color{red}{(x+1)^2}}$$
This is not $$\frac{4}{(x^2-1)^2}$$
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