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:
 * $\map {D_x} {\arccos x} = \dfrac {-1} {\sqrt {1 - x^2}}$

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

Then:
 * $x = \cos y$

Then from Derivative of Cosine Function:
 * $\dfrac {\d x} {\d y} = -\sin y$

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

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 $\arccos x$, that is, for $y \in \closedint 0 \pi$.

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

So:
 * $\dfrac {\d y} {\d x} = \dfrac {-1} {\sqrt {1 - \cos^2 y} }$

and hence the result.

Also see

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