Answer
$$ - \frac{1}{{\sqrt {1 - {x^2}} }}$$
Work Step by Step
$$\eqalign{
& {\text{Identity}}\left( 5 \right) \cr
& {\sin ^{ - 1}}x + {\cos ^{ - 1}}x = \frac{\pi }{2} \cr
& {\text{Solve for }}{\cos ^{ - 1}}x \cr
& {\cos ^{ - 1}}x = \frac{\pi }{2} - {\sin ^{ - 1}}x \cr
& {\text{Differentiate}} \cr
& \frac{d}{{dx}}\left[ {{{\cos }^{ - 1}}x} \right] = \frac{d}{{dx}}\left[ {\frac{\pi }{2} - {{\sin }^{ - 1}}x} \right] \cr
& \frac{d}{{dx}}\left[ {{{\cos }^{ - 1}}x} \right] = \frac{d}{{dx}}\left[ {\frac{\pi }{2}} \right] - \frac{d}{{dx}}\left[ {{{\sin }^{ - 1}}x} \right] \cr
& {\text{By the formula }}\left( {{\text{14}}} \right){\text{ }}\frac{d}{{dx}}\left[ {{{\sin }^{ - 1}}u} \right] = \frac{1}{{\sqrt {1 - {u^2}} }}\frac{{du}}{{dx}},{\text{ then}} \cr
& \frac{d}{{dx}}\left[ {{{\cos }^{ - 1}}x} \right] = 0 - \frac{1}{{\sqrt {1 - {x^2}} }} \cr
& \frac{d}{{dx}}\left[ {{{\cos }^{ - 1}}x} \right] = - \frac{1}{{\sqrt {1 - {x^2}} }} \cr} $$
