Derivative of Arccosine Function

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

Let $\arccos x$ be the arccosine of $x$.

Then:
 * $\displaystyle D_x \left({\arccos x}\right) = \frac {-1} {\sqrt {1 - x^2}}$

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

Then $x = \cos y$.

Then $\displaystyle \frac {\mathrm dx} {\mathrm dy} = -\sin y$ from Derivative of Cosine Function.

Hence from Derivative of Inverse Function:
 * $\displaystyle \frac {\mathrm dy} {\mathrm dx} = \frac {-1} {\sin y}$

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

Now $\sin y \ge 0$ on the range of $\arccos x$, i.e. for $y \in \left[{0 \,.\,.\, \pi}\right]$.

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

So:
 * $\displaystyle \frac {\mathrm dy} {\mathrm dx} = \frac {-1} {\sqrt {1 - \cos^2 y}}$

and hence the result.