Answer
$$\frac{{dy}}{{dx}} = 1$$
Work Step by Step
$$\eqalign{
& y = \int_0^{\sin x} {\frac{{dt}}{{\sqrt {1 - {t^2}} }}},\,\,\,\,\,\,\,\,\left| x \right| < \frac{\pi }{2} \cr
& {\text{differentiate both sides with respect to }}x \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left( {\int_0^{\sin x} {\frac{{dt}}{{\sqrt {1 - {t^2}} }}} } \right) \cr
& {\text{use the fundamental theorem of calculus part 1 with the chain rule }} \cr
& \frac{{dy}}{{dx}} = \frac{1}{{\sqrt {1 - {{\left( {\sin x} \right)}^2}} }}\frac{d}{{dx}}\left( {\sin x} \right) \cr
& {\text{compute the derivative and simplify}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{{\sqrt {1 - {{\sin }^2}x} }}\left( {\cos x} \right) \cr
& \frac{{dy}}{{dx}} = \frac{1}{{\sqrt {{{\cos }^2}x} }}\left( {\cos x} \right) \cr
& \frac{{dy}}{{dx}} = 1 \cr} $$
