Arc tangent function

This article is about a particular function from a subset of the real numbers to the real numbers. Information about the function, including its domain, range, and key data relating to graphing, differentiation, and integration, is presented in the article.
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For functions involving angles (trigonometric functions, inverse trigonometric functions, etc.) we follow the convention that all angles are measured in radians. Thus, for instance, the angle of $90\,^{\circ }$ is measured as $\pi /2$.

Definition

The arc tangent function, denoted $\arctan$ or $\tan ^{-1}$, is a function defined as follows: for $x\in \mathbb {R}$, $\arctan x$ is the unique number $y$ in the open interval $(-\pi /2,\pi /2)$ such that $\tan y=x$.

Equivalently, the arc tangent function is the inverse function to the restriction of the tangent function to the interval $(-\pi /2,\pi /2)$.

Key data

Item	Value
default domain	all real numbers, i.e., all of $\mathbb {R}$
range	the open interval $(-\pi /2,\pi /2)$, i.e., the set $\{y\mid -\pi /2\leq y\leq \pi /2\}$ no absolute minimum or absolute maximum, because the extremes are attained asymptotically at infinity.
local minimum values and points of attainment	no local minimum values
local maximum values and points of attainment	no local maximum values
points of inflection (both coordinates)	$(0,0)$ (the origin) only
horizontal asymptotes	The line $y=\pi /2$ as $x\to \infty$ The line $y=-\pi /2$ as $x\to -\infty$
derivative	$x\mapsto {\frac {1}{1+x^{2}}}$
second derivative	$x\mapsto {\frac {-2x}{(1+x^{2})^{2}}}$
higher derivatives	The $n^{th}$ derivative is a rational function of $x$ of degree $-n-1$.
antiderivative	$x\arctan x-{\frac {1}{2}}\ln(1+x^{2})+C$. We use integration by parts (see #Integration).
higher antiderivatives	The function can be antidifferentiated any number of times in terms of elementary functions.

Differentiation

First derivative

We use the inverse function theorem, and the fact that the derivative of $\tan$ is $\sec ^{2}$.

By the inverse function theorem, we have:

$\arctan 'x={\frac {1}{\tan '(\arctan x)}}={\frac {1}{\sec ^{2}(\arctan x)}}$

If $\arctan x=\theta$, then $x=\tan \theta$ and we get:

$\!\sec ^{2}\theta =1+\tan ^{2}\theta =1+x^{2}$

Plugging this into the above, we get:

$\arctan 'x={\frac {1}{1+x^{2}}}$

Second derivative

The second derivative is given as:

$\arctan ''x={\frac {d}{dx}}\arctan 'x={\frac {d}{dx}}\left({\frac {1}{1+x^{2}}}\right)={\frac {d}{dx}}[(1+x^{2})^{-1}]=(2x)(-1)(1+x^{2})^{-2}={\frac {-2x}{(1+x^{2})^{2}}}$

Higher derivatives

For higher derivatives, we use the quotient rule for differentiation, combined with the chain rule for differentiation to deal with powers of $1+x^{2}$.