Answer
$$\frac{{dy}}{{dx}} = 1 - 4\left( {{x^2} - x} \right)\left( {1 - \frac{1}{{{x^2}}}} \right) - 4\left( {x + \frac{1}{x}} \right)\left( {2x - 1} \right)$$
Work Step by Step
$$\eqalign{
& y = \left( {x + 2} \right) - 4\left( {{x^2} - x} \right)\left( {x + \frac{1}{x}} \right) \cr
& {\text{Differentiate}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\left( {x + 2} \right) - 4\left( {{x^2} - x} \right)\left( {x + \frac{1}{x}} \right)} \right] \cr
& {\text{Use sum rule}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {x + 2} \right] - 4\frac{d}{{dx}}\left[ {\left( {{x^2} - x} \right)\left( {x + \frac{1}{x}} \right)} \right] \cr
& {\text{Use product rule}} \cr
& \frac{{dy}}{{dx}} = 1 - 4\left( {{x^2} - x} \right)\frac{d}{{dx}}\left[ {x + \frac{1}{x}} \right] - 4\left( {x + \frac{1}{x}} \right)\frac{d}{{dx}}\left[ {{x^2} - x} \right] \cr
& \frac{{dy}}{{dx}} = 1 - 4\left( {{x^2} - x} \right)\left( {1 - \frac{1}{{{x^2}}}} \right) - 4\left( {x + \frac{1}{x}} \right)\left( {2x - 1} \right) \cr} $$
