\( \left. \begin{array} { l } { h ( x ) = \sqrt { 1 - x ^ { 2 } } \sin ^ { - 1 } ( x ) } \\ { h ^ { \prime } ( x ) = \square } \end{array} \right. \)

To differentiate the function \( h(x) = \sqrt{1 - x^2} \sin^{-1}(x) \), we can use the product rule: if \( u(x) = \sqrt{1 - x^2} \) and \( v(x) = \sin^{-1}(x) \), then \[ h'(x) = u'(x)v(x) + u(x)v'(x). \] Calculating \( u'(x) \) gives \( u'(x) = \frac{-x}{\sqrt{1 - x^2}} \), and \( v'(x) = \frac{1}{\sqrt{1 - x^2}} \). Plugging these into the product rule formula yields: \[ h'(x) = \left(\frac{-x}{\sqrt{1 - x^2}}\right) \sin^{-1}(x) + \sqrt{1 - x^2} \left(\frac{1}{\sqrt{1 - x^2}}\right) = \frac{-x \sin^{-1}(x)}{\sqrt{1 - x^2}} + 1. \] So, the derivative \( h'(x) \) is: \[ h'(x) = 1 - \frac{x \sin^{-1}(x)}{\sqrt{1 - x^2}}. \]