What is $\frac{d}{dx}\left(\frac{1}{(3x^{2} + 1)^{2}}\right)$?

Explanation

1. 01

   Write it as $u^{-2}$ with $u = 3x^{2} + 1$: the derivative is $-2u^{-3} \cdot u'$, $u' = 6x$.

2. 02

   $-2(3x^{2} + 1)^{-3} \cdot 6x = -\frac{12x}{(3x^{2} + 1)^{3}}$.