Answer
$$\frac{{dy}}{{dx}} = - \frac{1}{{{{\cos }^{ - 1}}x\sqrt {1 - {x^2}} }}$$
Work Step by Step
$$\eqalign{
& y = \ln \left( {{{\cos }^{ - 1}}x} \right) \cr
& {\text{differentiate using }}\left( {\ln u} \right)' = \frac{1}{u}u' \cr
& \frac{{dy}}{{dx}} = \frac{1}{{{{\cos }^{ - 1}}x}}\left( {{{\cos }^{ - 1}}x} \right)' \cr
& {\text{compute the derivative}} \cr
& \frac{d}{{dx}}\left[ {{{\cos }^{ - 1}}u} \right] = - \frac{1}{{\sqrt {1 - {u^2}} }}\frac{{du}}{{dx}}{\text{ }}\left( {{\text{see page 467}}} \right) \cr
& \frac{{dy}}{{dx}} = \frac{1}{{{{\cos }^{ - 1}}x}}\left( { - \frac{1}{{\sqrt {1 - {x^2}} }}} \right) \cr
& {\text{simplify}} \cr
& \frac{{dy}}{{dx}} = - \frac{1}{{{{\cos }^{ - 1}}x\sqrt {1 - {x^2}} }} \cr} $$
