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Inverse trig derivatives

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.

Inverse trig derivatives Updated 2 min read

At a glance

\frac{d}{dx}\,\arctan x = \frac{1}{1 + x^2}
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
On this page
  1. Graph of arctan x and its derivative
  2. Proof of the derivative of arctan x
  3. Chain rule version: derivative of arctan(u)
  4. Worked examples
  5. Common mistakes

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 of y = arctan x and its derivative y = 1/(1 + x²)
The inverse tangent climbs fastest at the origin, with slope 1, then levels off toward its horizontal asymptotes y = \pm\frac{\pi}{2}.

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, using the derivative of 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:

\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). The integral of arctan x itself uses integration by parts. See also the arctangent function and inverse trig derivative problems.

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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.

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