<s>[INST] What goes wrong in this derivative? $$ f(x) = \frac{2}{3} x (x^2-1)^{-2/3} $$
and f'(x) is searched.
So, by applying the product rule $ (uv)' = u'v + uv' $ with $ u=(x^2-1)^{-2/3} $ and $ v=\frac{2}{3} x $, so $ u'=-\frac{4}{3} x (x^2-1)^{-5/3} $ and $ v' = \frac{2}{3} $, I obtain
$$ f'(x) = - \frac{2}{9} (x^2-1)^{-5/3} x^2 + \frac{2}{3} (x^2-1)^{-2/3} $$ 
whereas according to Wolfram Alpha (see alternate form), the correct result is:
$$ f'(x) = - \frac{2}{9} (x^2-1)^{-5/3} x^2 - \frac{2}{3} (x^2-1)^{-5/3} $$
So apparently, my calculation for $u'v$ is correct, but $uv'$ is wrong. What am I missing here?
 [/INST] Doing what you said, you should find 
$$f'(x)=-\frac{8}{9}(x^2-1)^{-5/3}x^2+\frac{2}{3}(x^2-1)^{-2/3}.$$
Now
$$
(x^2-1)^{-2/3}=(x^2-1)(x^2-1)^{-5/3}.
$$
So
$$
f'(x)=-\frac{8}{9}(x^2-1)^{-5/3}x^2+\frac{2}{3}(x^2-1)(x^2-1)^{-5/3}
$$
$$
= \left(\frac{2}{3}-\frac{8}{9}\right)(x^2-1)^{-5/3}x^2-\frac{2}{3}(x^2-1)^{-5/3}
$$
$$
=\mbox{Wolfram Alpha's answer}.
$$
</s>