**Solution.**Let y = \cos^{-1}(x). Then x = \cos(y) and 0 \leq y \leq \pi. We want to differentiate with respect to x:

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

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

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