Answer
$$u'\left( x \right) = - \left( {2x - 1} \right)\sin \left( {{x^2} - x} \right)$$
Work Step by Step
$$\eqalign{
& u\left( x \right) = \cos \left( {{x^2} - x} \right) \cr
& {\text{Differentiate}} \cr
& u'\left( x \right) = \frac{d}{{dx}}\left[ {\cos \left( {{x^2} - x} \right)} \right] \cr
& {\text{Use the chain rule}} \cr
& u'\left( x \right) = - \sin \left( {{x^2} - x} \right)\frac{d}{{dx}}\left[ {{x^2} - x} \right] \cr
& u'\left( x \right) = - \sin \left( {{x^2} - x} \right)\left( {2x - 1} \right) \cr
& u'\left( x \right) = - \left( {2x - 1} \right)\sin \left( {{x^2} - x} \right) \cr} $$
