Derivative of Arcsine Function

Theorem
Let $$x \in \mathbb{R}$$ be a real number such that $$-1 < x < 1$$.

Let $$\arcsin x$$ be the arcsine of $$x$$.

Then $$\frac{d\left({\arcsin x}\right)}{dx} = \frac 1 {\sqrt {1 - x^2}}$$.

Proof
Let $$y = \arcsin x$$ where $$-1 < x < 1$$.

Then $$x = \sin y$$.

Then $$\frac {dx} {dy} = \cos y$$ from Derivative of Sine Function.

Hence from Derivative of an Inverse Function, $$\frac {dy} {dx} = \frac 1 {\cos y}$$.

From Sum of Squares of Sine and Cosine, we have $$\cos^2 y + \sin^2 y = 1 \Longrightarrow \cos y = \pm \sqrt {1 - \sin^2 y}$$.

Now $$\cos y \ge 0$$ on the range of $$\arcsin x$$, i.e. $$\left[{-\frac \pi 2 \,. \, . \, \frac \pi 2}\right]$$.

Thus it follows that we need to take the positive root of $$\sqrt {1 - \sin^2 y}$$.

So $$\frac {dy} {dx} = \frac 1 {\sqrt {1 - \sin^2 y}}$$ and hence the result.