Prove the inverse-cosine derivative using its principal range

Let $y=\cos^{-1}x$, so $\cos y=x$ and $0\le y\le\pi$. On the interior $-1<x<1$, implicit differentiation gives $-\sin y\,y^{\prime}=1$, hence $y^{\prime}=-1/\sin y$.

The source uses $\sin^2y+\cos^2y=1$ and $\sin y\ge0$ on $[0,\pi]$ to obtain $\sin y=\sqrt{1-\cos^2y}=\sqrt{1-x^2}$. On the strict interior this is positive, so $y^{\prime}=-1/\sqrt{1-x^2}$. The real function includes endpoints, but this finite derivative formula excludes them.