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:
 * $\dfrac {\map \d {\arcsin x} } {\d x} = \dfrac 1 {\sqrt {1 - x^2} }$

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

Then:
 * $x = \sin y$

Then from Derivative of Sine Function:
 * $\dfrac {\d x} {\d y} = \cos y$

Hence from Derivative of Inverse Function:
 * $\dfrac {\d y} {\d x} = \dfrac 1 {\cos y}$

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 $\arcsin x$, that is:
 * $y \in \closedint {-\dfrac \pi 2} {\dfrac \pi 2}$

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

So:

Also see

 * Derivative of Arccosine Function
 * Derivative of Arctangent Function
 * Derivative of Arccotangent Function
 * Derivative of Arcsecant Function
 * Derivative of Arccosecant Function