Answer
$$dy = \left( {1 - \cos x + x\sin x} \right)dx$$
Work Step by Step
$$\eqalign{
& y = x\left( {1 - \cos x} \right) \cr
& {\text{Differentiate both sides with respect to }}x \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {x\left( {1 - \cos x} \right)} \right] \cr
& \frac{{dy}}{{dx}} = \left( {1 - \cos x} \right)\frac{d}{{dx}}\left[ x \right] + x\frac{d}{{dx}}\left[ {1 - \cos x} \right] \cr
& \frac{{dy}}{{dx}} = \left( {1 - \cos x} \right) + x\left( {\sin x} \right) \cr
& \frac{{dy}}{{dx}} = 1 - \cos x + x\sin x \cr
& {\text{Write in differential form }}dy = f'\left( x \right)dx \cr
& dy = \left( {1 - \cos x + x\sin x} \right)dx \cr} $$
