Answer
$$\frac{{du}}{{dx}} = - 3{x^{ - 4}}{\text{ and }}\frac{{{d^2}u}}{{d{x^2}}} = 12{x^{ - 5}}$$
Work Step by Step
$$\eqalign{
& u = \frac{{\left( {{x^2} + x} \right)\left( {{x^2} - x + 1} \right)}}{{{x^4}}} \cr
& {\text{Use the distributive property}} \cr
& u = \frac{{{x^2}\left( {{x^2} - x + 1} \right) + x\left( {{x^2} - x + 1} \right)}}{{{x^4}}} \cr
& u = \frac{{{x^4} - {x^3} + {x^2} + {x^3} - {x^2} + x}}{{{x^4}}} \cr
& u = \frac{{{x^4} + x}}{{{x^4}}} \cr
& u = \frac{{{x^4}}}{{{x^4}}} + \frac{x}{{{x^4}}} \cr
& u = 1 + \frac{1}{{{x^3}}} \cr
& u = 1 + {x^{ - 3}} \cr
& \cr
& {\text{Find the first derivative}} \cr
& \frac{{du}}{{dx}} = \frac{d}{{dx}}\left[ {1 + {x^{ - 3}}} \right] \cr
& {\text{use }}\frac{d}{{dx}}\left[ {{x^n}} \right] = n{x^{n - 1}} \cr
& \frac{{du}}{{dx}} = 0 - 3{x^{ - 4}} \cr
& \frac{{du}}{{dx}} = - 3{x^{ - 4}} \cr
& \cr
& {\text{Find the second derivative}} \cr
& \frac{{{d^2}u}}{{d{x^2}}} = \frac{d}{{dx}}\left[ { - 3{x^{ - 4}}} \right] \cr
& {\text{use }}\frac{d}{{dx}}\left[ {{x^n}} \right] = n{x^{n - 1}} \cr
& \frac{{{d^2}u}}{{d{x^2}}} = - 3\left( { - 4{x^{ - 5}}} \right) \cr
& \frac{{{d^2}u}}{{d{x^2}}} = 12{x^{ - 5}} \cr} $$
