A calculus of the absurd

11.10.3 Derivative of \(\arctan (x)\)

From the definition of \(\arctan (x)\) we know that

\begin{equation} \arctan (\tan (x)) = x \end{equation}

We can differentiate both sides (using the chain rule). Performing this leaves us with

\begin{align*} & \frac {d}{d\rbrackets {\tan (x)}}\left ( \arctan (\tan (x)) \right ) \frac {d}{dx} \left ( \tan (x) \right ) = 1 \\ & \frac {d}{d\rbrackets {\tan (x)}}\left ( \arctan (\tan (x)) \right ) \left ( \sec ^2(x) \right ) = 1 \end{align*}


\begin{equation} \frac {d}{d\rbrackets {\tan (x)}}\left ( \arctan (\tan (x)) \right ) = \frac {1}{\sec ^2(x)} \end{equation}

If we substitute \(u=\tan (x)\), we get that

\begin{align*} \frac {d}{du} (\arctan (u)) &= \frac {1}{1 + \tan ^2(x)} \text { by the Pythagorean identity} \\ &= \frac {1}{1+u^2} \end{align*}

Or, equivalently (as renaming the variable \(u\) has no effect on the inequality),

\begin{equation} \frac {d}{dx} (\arctan (x)) = \frac {1}{1+x^2} \end{equation}