A calculus of the absurd

10.10.3 Derivative of \(\arctan (x)\)

This requires a “smart” idea here, which is to relate the derivative of \(\arctan (x)\), which we don’t know, to the derivative of \(\tan (x)\), which we do know. We can start with this formula, which exploits the property that \(\arctan \) is the inverse function of \(\tan \) (and vice versa),

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

Then, we can differentiate both sides (using the chain rule), from which we can deduce that

\begin{align} & \frac {d}{d\rbrackets {\tan (x)}}\left ( \arctan (\tan (x)) \right ) \frac {d}{dx} \left ( \tan (x) \right ) = 1 \\ & \implies \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}

Then, we can use a substitution (another clever idea), in this case by setting \(u=\tan (x)\) we then can deduce 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 equation),

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