Answer
$$r'\left( x \right) = \cos x - x\sin x + 2x$$
Work Step by Step
$$\eqalign{
& r\left( x \right) = x\cos x + {x^2} + 1 \cr
& {\text{Differentiate}} \cr
& r'\left( x \right) = \frac{d}{{dx}}\left[ {x\cos x + {x^2} + 1} \right] \cr
& r'\left( x \right) = \frac{d}{{dx}}\left[ {x\cos x} \right] + \frac{d}{{dx}}\left[ {{x^2} + 1} \right] \cr
& {\text{By the product rule}} \cr
& r'\left( x \right) = \cos x\frac{d}{{dx}}\left[ x \right] + x\frac{d}{{dx}}\left[ {\cos x} \right] + \frac{d}{{dx}}\left[ {{x^2} + 1} \right] \cr
& {\text{Computing derivatives}} \cr
& r'\left( x \right) = \cos x\left( 1 \right) + x\left( { - \sin x} \right) + 2x \cr
& r'\left( x \right) = \cos x - x\sin x + 2x \cr} $$
