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 \(\ref {derivative of arctan}\)).

We start with the identity

\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}