Answer
$$\frac{{dy}}{{dx}} = - \frac{x}{{\sqrt {\left( {1 - {x^4}} \right)\left( {{{\cos }^{ - 1}}{x^2}} \right)} }}$$
Work Step by Step
$$\eqalign{
& y = \sqrt {{{\cos }^{ - 1}}{x^2}} \cr
& {\text{write the function as}} \cr
& y = {\left( {{{\cos }^{ - 1}}{x^2}} \right)^{1/2}} \cr
& {\text{differentiate both sides}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {{{\left( {{{\cos }^{ - 1}}{x^2}} \right)}^{1/2}}} \right] \cr
& {\text{by the chain rule}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{2}{\left( {{{\cos }^{ - 1}}{x^2}} \right)^{ - 1/2}}\frac{d}{{dx}}\left[ {{{\cos }^{ - 1}}{x^2}} \right] \cr
& {\text{Using the formula on the page 467}}\,\,\frac{d}{{dx}}\left[ {{{\cos }^{ - 1}}u} \right] = - \frac{1}{{\sqrt {1 - {u^2}} }}\frac{{du}}{{dx}} \cr
& {\text{Setting }}u = {x^2}{\text{ we have}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{2}{\left( {{{\cos }^{ - 1}}{x^2}} \right)^{ - 1/2}}\left( { - \frac{1}{{\sqrt {1 - {{\left( {{x^2}} \right)}^2}} }}} \right)\frac{d}{{dx}}\left[ {{x^2}} \right] \cr
& {\text{Then}}{\text{,}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{2}{\left( {{{\cos }^{ - 1}}{x^2}} \right)^{ - 1/2}}\left( { - \frac{1}{{\sqrt {1 - {{\left( {{x^2}} \right)}^2}} }}} \right)\left( {2x} \right) \cr
& {\text{Simplifying}} \cr
& \frac{{dy}}{{dx}} = {\left( {{{\cos }^{ - 1}}{x^2}} \right)^{ - 1/2}}\left( { - \frac{x}{{\sqrt {1 - {x^4}} }}} \right) \cr
& \frac{{dy}}{{dx}} = - \frac{x}{{\sqrt {{{\cos }^{ - 1}}{x^2}} \sqrt {1 - {x^4}} }} \cr
& \frac{{dy}}{{dx}} = - \frac{x}{{\sqrt {\left( {1 - {x^4}} \right)\left( {{{\cos }^{ - 1}}{x^2}} \right)} }} \cr} $$
