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$$ f(x) = \frac{2}{3} x (x^2-1)^{-2/3} $$

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?

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4 Answers 4

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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.

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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 $$ f'(x)=-\frac{8}{9}(x^2-1)^{-5/3}x^2+\frac{2}{3}(x^2-1)(x^2-1)^{-5/3} $$ $$ = \left(\frac{2}{3}-\frac{8}{9}\right)(x^2-1)^{-5/3}x^2-\frac{2}{3}(x^2-1)^{-5/3} $$ $$ =\mbox{Wolfram Alpha's answer}. $$

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Your $u'v$ is wrong. Check your calculation.

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So $$\begin{align} (uv)' = u'v + uv' &= \frac{-4}{3}x(x^2 - 1)^{-5/3}\frac{2}{3}x + (x^2 - 1)^{-2/3}\frac{2}{3} \\ &= \frac{-8}{9}x^2(x^2 - 1)^{-5/3} + \frac{2}{3}(x^2 - 1)^{-2/3}\\ &= \frac{-\color{green}8}{\color{green}9}\color{green}x^\color{green}2(x^2 - 1)^{-5/3} + \frac{\color{red}2}{\color{red}3}(\color{red}x^\color{red}2-1)(x^2-1)^{-5/3} \\ &= \left[\frac{-\color{green}8}{\color{green}9}\color{green}x^\color{green}2 + \frac{\color{red}2}{\color{red}3}\color{red}x^\color{red}2\right](x^2 - 1)^{-5/3} - \frac{\color{red}2}{\color{red}3}(x^2 - 1)^{-5/3}\\ &= -\frac{2}{9}x^2(x^2 - 1)^{-5/3} - \frac{2}{3}(x^2 - 1)^{-5/3}. \end{align} $$

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