Answer
$$\frac{{dy}}{{dx}} = 0$$
Work Step by Step
$$\eqalign{
& y = {\sin ^{ - 1}}x + {\cos ^{ - 1}}x \cr
& {\text{differentiate}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {{{\sin }^{ - 1}}x + {{\cos }^{ - 1}}x} \right] \cr
& {\text{sum rule}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {{{\sin }^{ - 1}}x} \right] + \frac{d}{{dx}}\left[ {{{\cos }^{ - 1}}x} \right] \cr
& {\text{compute the derivatives}}{\text{, use formulas of the page 467}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{{\sqrt {1 - {x^2}} }} + \left( { - \frac{1}{{\sqrt {1 - {x^2}} }}} \right) \cr
& \frac{{dy}}{{dx}} = \frac{1}{{\sqrt {1 - {x^2}} }} - \frac{1}{{\sqrt {1 - {x^2}} }} \cr
& {\text{simplify}} \cr
& \frac{{dy}}{{dx}} = 0 \cr
& \cr
& or \cr
& y = {\sin ^{ - 1}}x + {\cos ^{ - 1}}x \cr
& y = \frac{\pi }{2},{\text{ }} - 1 \leqslant x \leqslant 1 \cr
& \frac{{dy}}{{dx}} = 0 \cr} $$
