# A calculus of the absurd

##### 11.9.3 Derivative of $$\arctan (x)$$

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

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

Using the chain rule, we then obtain that

\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*}

Therefore,

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

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),

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