# A calculus of the absurd

##### 10.10.4 Derivative of $$\arccos (x)$$

The method for finding the derivative of $$\arccos (x)$$ is pretty much the same as the method for finding the derivative of $$\arctan (x)$$ (as in Section $$\ref {derivative of arctan}$$).

\begin{equation} \arccos (\cos (x)) = x, \end{equation}

which we then differentiate. The result of this is that

\begin{align} \left ( \frac {d}{d(\cos (x))} \arccos (\cos (x)) \right ) \frac {d}{dx} (\cos (x)) = \frac {d}{dx} (x), \end{align}

which we can simplify by computing the parts we know how to compute which allows us to establish the equation

\begin{equation} \left ( \frac {d}{d(\cos (x))} \arccos (\cos (x)) \right ) (-\sin (x)) = 1. \end{equation}

Therefore, we know that

\begin{equation} \left ( \frac {d}{d(\cos (x))} \arccos (\cos (x)) \right ) = -\frac {1}{\sin (x)}. \end{equation}

Now, as when differentiating $$\arctan (x)$$, we can substitute. In this case we will use $$u = \cos (x)$$ rather than $$\tan (x)$$, and then write that

\begin{equation} \frac {d}{du}\arccos (u) = -\frac {1}{\sqrt {1-u^2}}. \end{equation}

This gives us the answer, and if you prefer the variable name $$x$$ to $$u$$, then we can just rename it, which means that

\begin{equation} \frac {d}{dx}\arccos (x) = -\frac {1}{\sqrt {1-x^2}}. \end{equation}