Answer
$u' = \frac{-3}{x^{4}}$
$u'' = \frac{12}{x^{5}}$
Work Step by Step
$u = \frac{x(x+1)(x^{2}-x+1)}{x^{4}}$
$u = \frac{(x+1)(x^{2}-x+1)}{x^{3}}$
$u = \frac{x^{3}+1}{x^{3}}$
$u' = \frac{x^{3}(3x^{2}) - 3x^{2}(x^{3}+1)}{x^{6}}$
$u' = \frac{-3x^{2}}{x^{6}}$
$u' = \frac{-3}{x^{4}}$
$u'' = \frac{12}{x^{5}}$