Answer
$$\frac{{dy}}{{dx}} = 1$$
Work Step by Step
$$\eqalign{
& y = {\cos ^{ - 1}}\left( {\cos x} \right) \cr
& {\text{differentiate using the formula }} \cr
& \frac{d}{{dx}}\left[ {{{\cos }^{ - 1}}u} \right] = - \frac{1}{{\sqrt {1 - {u^2}} }}\frac{{du}}{{dx}}{\text{ }}\left( {{\text{see page 467}}} \right) \cr
& \cr
& \frac{{dy}}{{dx}} = - \frac{1}{{\sqrt {1 - {{\left( {\cos x} \right)}^2}} }}\frac{d}{{dx}}\left( {\cos x} \right) \cr
& \frac{{dy}}{{dx}} = \frac{1}{{\sqrt {1 - {{\cos }^2}x} }}\left( {\sin x} \right) \cr
& {\text{simplifying}}{\text{, use identity si}}{{\text{n}}^2}x + {\cos ^2}x = 1 \cr
& \frac{{dy}}{{dx}} = \frac{1}{{\sqrt {{{\sin }^2}x} }}\left( {\sin x} \right) \cr
& \frac{{dy}}{{dx}} = \frac{{\sin x}}{{\sin x}} \cr
& \frac{{dy}}{{dx}} = 1 \cr
& \cr
& or \cr
& y = {\cos ^{ - 1}}\left( {\cos x} \right) \cr
& y = x \cr
& \frac{{dy}}{{dx}} = 1 \cr} $$