**Solution.**Let y = \sin^{-1}(x). Then x = \sin(y) and -\frac{\pi}{2} \leq y \leq \frac{\pi}{2}. We will differentiate with respect to x:

\begin{align*} \frac{d}{dx} x = \frac{d}{dx} \sin(y) &\iff 1 = \frac{d(sin(y))}{dy} \frac{dy}{dx} \\ &\iff 1 = \cos(y) \frac{dy}{dx} \\ &\iff \frac{dy}{dx} = \frac{1}{\cos(y)}, \end{align*}

\begin{align*} \cos^2(y) = 1 - \sin^2(y) \iff \cos(y) = \sqrt{1 - \sin^2(y)}. \end{align*}

\begin{align*} \frac{d}{dx} \arcsin(x) = \frac{d}{dx} \sin^{-1}(x) = \frac{dy}{dx} = \frac{1}{\cos(y)} = \frac{1}{\sqrt{1-x^2}}, \quad x \in (-1,1). \end{align*}