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?
 A: You're so close, but you have simply multiplied incorrectly.
Note that if $u' = -\frac{4}{3}x(x^2 - 1)^{-5/3}$ and $v = \frac{2}{3}x$, then $$u'v = -\frac{8}{9}x^2(x^2-1)^{-5/3}$$
This will make your answer correct.  To match W|A, you just need to combine fractions carefully.
A: 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
$$
f'(x)=-\frac{8}{9}(x^2-1)^{-5/3}x^2+\frac{2}{3}(x^2-1)(x^2-1)^{-5/3}
$$
$$
= \left(\frac{2}{3}-\frac{8}{9}\right)(x^2-1)^{-5/3}x^2-\frac{2}{3}(x^2-1)^{-5/3}
$$
$$
=\mbox{Wolfram Alpha's answer}.
$$
A: Your $u'v$ is wrong. Check your calculation.
A: So
$$\begin{align}
(uv)' = u'v + uv' &= \frac{-4}{3}x(x^2 - 1)^{-5/3}\frac{2}{3}x  + (x^2 - 1)^{-2/3}\frac{2}{3} \\ 
&= \frac{-8}{9}x^2(x^2 - 1)^{-5/3} + \frac{2}{3}(x^2 - 1)^{-2/3}\\
&= \frac{-\color{green}8}{\color{green}9}\color{green}x^\color{green}2(x^2 - 1)^{-5/3} + \frac{\color{red}2}{\color{red}3}(\color{red}x^\color{red}2-1)(x^2-1)^{-5/3} \\
&= \left[\frac{-\color{green}8}{\color{green}9}\color{green}x^\color{green}2 + \frac{\color{red}2}{\color{red}3}\color{red}x^\color{red}2\right](x^2 - 1)^{-5/3} - \frac{\color{red}2}{\color{red}3}(x^2 - 1)^{-5/3}\\
&= -\frac{2}{9}x^2(x^2 - 1)^{-5/3} - \frac{2}{3}(x^2 - 1)^{-5/3}.
\end{align}
$$
