# Derivative of arctan x: Formula, Proof and Examples

> The derivative of arctan x is 1/(1 + x²). See the proof from tan y = x, chain rule examples like arctan(2x) and arctan(x²), and a graph.

Source: https://practicecalculus.com/derivatives/arctan-x
Updated: 2026-11-06

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

## At a glance

- **Derivative:** $\dfrac{1}{1 + x^2}$ for every real $x$
- **Chain rule form:** $\dfrac{d}{dx}\arctan u = \dfrac{u'}{1 + u^2}$
- **Domain of arctan x:** all real numbers
- **Range of arctan x:** $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$
- **Integral version:** $\int \frac{1}{1 + x^2}\,dx = \arctan x + C$

## Key takeaways

- The derivative of $\arctan x$ is $\frac{1}{1 + x^2}$, and it works for every real $x$.
- The proof uses $\tan y = x$ and the identity $\sec^2 y = 1 + \tan^2 y$.
- With the chain rule, $\frac{d}{dx}\arctan u = \frac{u'}{1 + u^2}$.
- The slope is largest at $x = 0$, where it equals $1$, and fades to $0$ as $x \to \pm\infty$.

The **derivative of $\arctan x$ is $\dfrac{1}{1 + x^2}$**. Unlike arcsin and arccos, there is no square root and no domain restriction: the formula works for every real number.

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

## Graph of arctan x and its derivative

```graph
title: 'Graph of y = arctan x and its derivative y = 1/(1 + x²)'
caption: 'The inverse tangent climbs fastest at the origin, with slope $1$, then levels off toward its horizontal asymptotes $y = \pm\frac{\pi}{2}$.'
x: [-6, 6]
y: [-2, 2]
functions:
  - expr: arctan(x)
    label: y = arctan x
  - expr: 1/(1+x^2)
    label: y = 1/(1 + x²) (derivative)
    dashed: true
points:
  - [0, 0, '(0, 0)']
  - [0, 1, 'slope 1']
```

The derivative is a bell-shaped curve that is always positive, so $\arctan x$ always increases. Because the derivative goes to $0$ at both ends, the graph flattens out toward $\pm\frac{\pi}{2}$.

## Proof of the derivative of arctan x

Let $y = \arctan x$, so $\tan y = x$ with $-\frac{\pi}{2} < y < \frac{\pi}{2}$. Differentiate both sides with [implicit differentiation](/rules/implicit-differentiation), using the [derivative of tan x](/derivatives/tan-x):

$$\sec^2 y \cdot \frac{dy}{dx} = 1 \quad\Longrightarrow\quad \frac{dy}{dx} = \frac{1}{\sec^2 y}$$

The Pythagorean identity $\sec^2 y = 1 + \tan^2 y$ turns this into $x$:

$$\frac{dy}{dx} = \frac{1}{1 + \tan^2 y} = \frac{1}{1 + x^2}$$

No sign choice is needed this time, since $1 + x^2$ is always positive.

## Chain rule version: derivative of arctan(u)

With the [chain rule](/rules/chain-rule):

$$\frac{d}{dx}\arctan u = \frac{u'}{1 + u^2}$$

| Function | Inside $u$ | Derivative |
| --- | --- | --- |
| $\arctan(2x)$ | $2x$ | $\frac{2}{1 + 4x^2}$ |
| $\arctan(x^2)$ | $x^2$ | $\frac{2x}{1 + x^4}$ |
| $\arctan(e^x)$ | $e^x$ | $\frac{e^x}{1 + e^{2x}}$ |
| $\arctan(\sqrt{x})$ | $\sqrt{x}$ | $\frac{1}{2\sqrt{x}(1 + x)}$ |
| $\arctan\!\left(\frac{x}{a}\right)$ | $\frac{x}{a}$ | $\frac{a}{a^2 + x^2}$ |

The last row gives the standard integral $\int \frac{dx}{a^2 + x^2} = \frac{1}{a}\arctan\frac{x}{a} + C$.

## Worked examples

**Example 1: arctan of 1/x.** For $x \ne 0$,

$$\frac{d}{dx}\arctan\frac{1}{x} = \frac{-1/x^2}{1 + 1/x^2} = \frac{-1}{x^2 + 1}$$

This is the negative of the arctan derivative, so $\arctan x + \arctan\frac{1}{x}$ is constant on each side of $0$. It equals $\frac{\pi}{2}$ for $x > 0$ and $-\frac{\pi}{2}$ for $x < 0$.

**Example 2: product rule.** For $f(x) = x\arctan x$:

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

**Example 3: tangent line.** At $x = 1$, $\arctan 1 = \frac{\pi}{4}$ and the slope is $\frac{1}{2}$. The tangent line is $y = \frac{\pi}{4} + \frac{1}{2}(x - 1)$.

## Common mistakes

1. **Writing $\sec^2 x$.** That is the derivative of $\tan x$, not of $\arctan x$.
2. **Writing $\frac{1}{1 + x}$.** The $x$ is squared. Check: at $x = 1$ the slope must be $\frac{1}{2}$.
3. **Forgetting to square the whole inside.** $\arctan(3x)$ gives $1 + 9x^2$, not $1 + 3x^2$.
4. **Mixing it up with $\frac{1}{\tan x} = \cot x$.** $\tan^{-1} x$ is the inverse function.

Read backwards, this derivative gives the [integral of 1/(1 + x²)](/integrals/1-over-1-plus-x-squared). The [integral of arctan x](/integrals/arctan-x) itself uses integration by parts. See also the [arctangent function](/functions/arctangent) and [inverse trig derivative problems](/practice-problems/inverse-trig-derivatives).

## Frequently asked questions

### What is the derivative of arctan x?

The derivative of $\arctan x$ (also written $\tan^{-1} x$) is $\frac{1}{1 + x^2}$.

### What is the derivative of arctan(2x)?

By the chain rule it is $\frac{2}{1 + (2x)^2} = \frac{2}{1 + 4x^2}$.

### What is the derivative of arctan(x²)?

The inside derivative is $2x$, so the answer is $\frac{2x}{1 + x^4}$.

### What is the integral of 1/(1 + x²)?

It is $\arctan x + C$. This is the derivative of $\arctan x$ read backwards.