Answer
$$\frac{{dy}}{{dx}} = - \frac{1}{{\sqrt {4 - {{\left( {x + 1} \right)}^2}} }}$$
Work Step by Step
$$\eqalign{
& y = {\cos ^{ - 1}}\left( {\frac{{x + 1}}{2}} \right) \cr
& {\text{differentiate using the formula }} \cr
& \frac{d}{{dx}}\left[ {{{\cos }^{ - 1}}u} \right] = - \frac{1}{{\sqrt {1 - {u^2}} }}\frac{{du}}{{dx}}{\text{ }}\left( {{\text{see page 467}}} \right) \cr
& \cr
& \frac{{dy}}{{dx}} = - \frac{1}{{\sqrt {1 - {{\left( {\frac{{x + 1}}{2}} \right)}^2}} }}\frac{d}{{dx}}\left( {\frac{{x + 1}}{2}} \right) \cr
& \frac{{dy}}{{dx}} = - \frac{1}{{\sqrt {1 - {{\left( {\frac{{x + 1}}{2}} \right)}^2}} }}\frac{d}{{dx}}\left( {\frac{x}{2} + \frac{1}{2}} \right) \cr
& \frac{{dy}}{{dx}} = - \frac{1}{{\sqrt {1 - {{\left( {\frac{{x + 1}}{2}} \right)}^2}} }}\left( {\frac{1}{2}} \right) \cr
& {\text{simplifying}} \cr
& \frac{{dy}}{{dx}} = - \frac{1}{{\sqrt {\frac{{4 - {{\left( {x + 1} \right)}^2}}}{4}} }}\left( {\frac{1}{2}} \right) \cr
& \frac{{dy}}{{dx}} = - \frac{1}{{\left( {1/2} \right)\sqrt {4 - {{\left( {x + 1} \right)}^2}} }}\left( {\frac{1}{2}} \right) \cr
& \frac{{dy}}{{dx}} = - \frac{1}{{\sqrt {4 - {{\left( {x + 1} \right)}^2}} }} \cr} $$
