# 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),

$$\arctan (\tan (x)) = x.$$

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}

Therefore,

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

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),

$$\frac {d}{dx} (\arctan (x)) = \frac {1}{1+x^2}$$