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

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

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. $$\left[{1 \,. \, . \, \pi}\right]$$.

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

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