Derivative of Arcsine Function

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

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

Then:
 * $\dfrac {\mathrm d \left({\arcsin x}\right)}{\mathrm dx} = \dfrac 1 {\sqrt {1 - x^2}}$

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

Then $x = \sin y$.

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

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

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

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

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

So: