Calculus of a Single Variable, Chapter 5, 5.6, Section 5.6, Problem 47

This function is a composite one, it may be expressed as h(t) = u(v(t)), where v(t) = arccos(t) and u(y) = sin(y). The chain rule is applicable here, h'(t) = u'(v(t))*v'(t).
This gives us h'(t) = -cos(arccos(t))*1/sqrt(1-t^2).
Note that cos(arccos(t)) = t for all t in [-1, 1], and the final answer is
h'(t) = -t/sqrt(1-t^2).
We may obtain the same result if note that sin(arccos(t)) = sqrt(1-t^2) for all t in [-1, 1] (arccos(t) is non-negative and therefore square root should be taken with plus sign only).