Answer
$y' = 2x - {\csc ^2}x$
Work Step by Step
$$\eqalign{
& y = {x^2} + \cot x \cr
& {\text{Differentiating}} \cr
& y' = \frac{d}{{dx}}\left[ {{x^2} + \cot x} \right] \cr
& {\text{Use the sum rule for derivatives }} \cr
& y' = \frac{d}{{dx}}\left[ {{x^2}} \right] + \frac{d}{{dx}}\left[ {\cot x} \right] \cr
& \cr
& {\text{Use the Derivatives of Trigonometric Functions }} \cr
& {\text{and }}\frac{d}{{dx}}\left[ {{x^n}} \right] = n{x^{n - 1}},{\text{ then}} \cr
& y' = 2x + \left( { - {{\csc }^2}x} \right) \cr
& {\text{Simplify}} \cr
& y' = 2x - {\csc ^2}x \cr} $$
