Derivative of arctan(x) = 1/(1+x²): Proof and Examples

Quick answer:

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

This holds for every real number \( x \), which makes arctangent one of the friendliest functions to differentiate. The answer is a plain rational function with no trig left in it, and that surprising fact is what makes arctangent so important for integration. Below you will find the intuition, the proof, seven worked examples, the common mistakes and practice problems.

Why it works: the intuition

Arctangent undoes tangent. If \( \tan y = x \), then \( y = \arctan x \), with \( y \) restricted to \( -\frac{\pi}{2} < y < \frac{\pi}{2} \) so that each \( x \) gives only one angle.

Graphically, the curve \( y = \arctan x \) is the reflection of one branch of \( y = \tan x \) across the line \( y = x \). Reflection swaps rise and run, so the slope of the inverse is the reciprocal of the original slope at the matching point. Tangent’s slope is \( \sec^2 y = 1 + \tan^2 y \), and on the reflected graph \( \tan y \) is just \( x \). So the slope of arctangent is \( \frac{1}{1 + x^2} \).

That formula also describes the shape you see. Near the origin the slope is about 1, so the graph starts out looking like the line \( y = x \). As \( |x| \) grows, \( 1 + x^2 \) grows quickly and the slope shrinks, so the curve flattens and creeps toward its horizontal asymptotes at \( \pm\frac{\pi}{2} \) without ever reaching them.

Proof using implicit differentiation

Let \( y = \arctan x \), so \( \tan y = x \) with \( -\frac{\pi}{2} < y < \frac{\pi}{2} \). Differentiate both sides with respect to \( x \). The left side needs the chain rule because \( y \) depends on \( x \):

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

Use the identity \( \sec^2 y = 1 + \tan^2 y = 1 + x^2 \):

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

(This is the same technique explained in implicit differentiation.) The key move is the last one: the answer first comes out in terms of \( y \), and a trig identity turns it back into \( x \). Every inverse trig derivative is found this way.

Another way to see it: the inverse function rule

There is a general shortcut for any inverse function. If \( f \) and \( f^{-1} \) are inverses, then the slope of the inverse at \( x \) is one over the slope of \( f \) at the matching point:

$$\big(f^{-1}\big)'(x) = \frac{1}{f'\big(f^{-1}(x)\big)}$$

Apply it with \( f = \tan \). The derivative of tangent is \( 1 + \tan^2 \), and evaluating it at \( \arctan x \) gives \( 1 + x^2 \) because tangent and arctangent cancel. Taking the reciprocal gives the same answer as before. This version is worth knowing because it works for every inverse function you will meet, including \( \ln x \), arcsine and arcsecant.

Chain rule form

When the input is a function \( g(x) \), the chain rule multiplies by the derivative of the inside:

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

In words: derivative of the inside on top, one plus the inside squared on the bottom.

Worked examples

1. \( \arctan(2x) \): the inside derivative is 2 and the inside squared is \( 4x^2 \), so the answer is \( \frac{2}{1 + 4x^2} \).

2. \( \arctan(x^2) \): the inside derivative is \( 2x \) and the inside squared is \( x^4 \), giving \( \frac{2x}{1 + x^4} \).

3. \( \arctan(1/x) \): the inside has derivative \( -\frac{1}{x^2} \), so

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

Interesting: \( \arctan x \) and \( \arctan(1/x) \) have opposite derivatives, which means their sum is constant on each side of zero (it equals \( \frac{\pi}{2} \) for \( x > 0 \)).

4. \( x\arctan x \): by the product rule, \( \arctan x + \frac{x}{1+x^2} \).

5. Tangent line at \( x = 1 \). The point is \( \left(1, \frac{\pi}{4}\right) \) since \( \tan\frac{\pi}{4} = 1 \), and the slope is \( \frac{1}{1 + 1} = \frac12 \). The tangent line is

$$y - \frac{\pi}{4} = \frac12(x - 1)$$

6. A surprising derivative. Differentiate \( \arctan\frac{x - 1}{x + 1} \). The inside has derivative \( \frac{2}{(x+1)^2} \) by the quotient rule, and one plus the inside squared simplifies to \( \frac{2x^2 + 2}{(x+1)^2} \). Dividing,

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

It has the same derivative as \( \arctan x \), so on \( x > -1 \) the two functions differ by a constant, which turns out to be \( -\frac{\pi}{4} \). Exam writers love this kind of question because the algebra hides a simple answer.

7. \( \arctan(\ln x) \): the inside derivative is \( \frac1x \), so the result is \( \frac{1}{x\left(1 + (\ln x)^2\right)} \) for \( x > 0 \).

Why this derivative matters for integrals

Reading the formula backwards gives one of the most used integrals in calculus:

$$\int\frac{dx}{1+x^2} = \arctan x + C, \qquad \int_0^1\frac{dx}{1+x^2} = \frac{\pi}{4}$$

The more general version, \( \int\frac{dx}{a^2 + x^2} = \frac{1}{a}\arctan\frac{x}{a} + C \), appears constantly in partial fraction problems. For example, with \( a = 2 \),

$$\int_0^2\frac{dx}{4 + x^2} = \frac12\arctan 1 = \frac{\pi}{8}$$

You can derive the general formula yourself with u-substitution, setting \( u = \frac{x}{a} \). It is remarkable that a rational function with no trig in it integrates to an angle; that is the reason \( \pi \) shows up in so many answers.

A series for arctan and for pi

The derivative also gives a quick route to the power series. For \( |x| < 1 \), the geometric series says

$$\frac{1}{1 + x^2} = 1 - x^2 + x^4 - x^6 + \cdots$$

Integrating term by term from 0 to \( x \):

$$\arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots$$

Setting \( x = 1 \) (where the series still converges) gives the Leibniz formula \( \frac{\pi}{4} = 1 - \frac13 + \frac15 - \cdots \). See Taylor series for more on building series this way.

Shape of the graph

The derivative \( \frac{1}{1+x^2} \) is always positive and largest at \( x = 0 \) (where it equals 1). So arctan is always increasing, steepest at the origin, and flattens out toward its horizontal asymptotes \( y = \pm\frac{\pi}{2} \). The second derivative, \( -\frac{2x}{(1+x^2)^2} \), is positive for \( x < 0 \) and negative for \( x > 0 \), so the origin is an inflection point.

Where it’s used

Arctangent turns a ratio into an angle, so its derivative appears whenever you ask how fast an angle is changing. Suppose you stand 500 m from a straight road and watch a car drive past. If the car is \( x \) meters along the road from the closest point, your viewing angle is \( \theta = \arctan\frac{x}{500} \). By the chain rule,

$$\frac{d\theta}{dt} = \frac{500}{500^2 + x^2}\,\frac{dx}{dt}$$

When the car is 500 m along and moving at 30 m/s, your head turns at \( \frac{500\cdot 30}{500^2 + 500^2} = 0.03 \) radians per second. The turning rate is fastest when the car is right in front of you (\( x = 0 \)) and fades as it drives away, exactly the shape of \( \frac{1}{1 + x^2} \).

The same function describes the direction of a vector in physics and navigation, the phase angle of a signal in electronics, and the angle of a ramp from its rise and run. In every case, the derivative tells you how sensitive the angle is to a change in the ratio, and that sensitivity is always greatest when the ratio is near zero.

Common mistakes

• Using the arcsine formula. \( \frac{1}{\sqrt{1 - x^2}} \) belongs to arcsine. Arctangent has a plus sign and no square root: \( \frac{1}{1 + x^2} \).
• Squaring the wrong thing. \( \frac{1}{(1 + x)^2} \) is not the same as \( \frac{1}{1 + x^2} \). Only \( x \) is squared.
• Forgetting to square the inside. For \( \arctan(3x) \), the bottom is \( 1 + 9x^2 \), not \( 1 + 3x^2 \) or \( 1 + 3x \).
• Reading \( \tan^{-1}x \) as a reciprocal. It is the inverse function. The reciprocal \( \frac{1}{\tan x} \) is \( \cot x \), whose derivative is \( -\csc^2 x \).
• Mixing in degrees. The formula gives the rate of change in radians per unit of \( x \). If a problem wants degrees, multiply by \( \frac{180}{\pi} \) at the very end rather than converting halfway through.
• Stopping before simplifying. Answers like \( \frac{-1/x^2}{1 + 1/x^2} \) are correct but hide the structure. Multiply top and bottom by \( x^2 \) to reach the cleaner form, which often reveals a constant difference or a cancellation.

Try it yourself

Type any arctangent expression below to see each step, or use the full derivative calculator.

Practice problems

Try these before checking the answers.

1. \( \frac{d}{dx}\arctan(3x) \)
2. \( \frac{d}{dx}\arctan(e^x) \)
3. \( \frac{d}{dx}\arctan\sqrt{x} \)
4. \( \frac{d}{dx}\arctan\frac{x}{2} \)
5. \( \frac{d}{dx}(\arctan x)^2 \)

Answers: (1) \( \frac{3}{1 + 9x^2} \); (2) \( \frac{e^x}{1 + e^{2x}} \); (3) \( \frac{1}{2\sqrt{x}(1 + x)} \); (4) \( \frac{2}{4 + x^2} \); (5) \( \frac{2\arctan x}{1 + x^2} \).

FAQ

Is arctan the same as tan⁻¹?

Yes. \( \tan^{-1}x \) means the inverse function, not \( \frac{1}{\tan x} \). The reciprocal is \( \cot x \).

What is the second derivative of arctan x?

\( \frac{d}{dx}(1+x^2)^{-1} = -\frac{2x}{(1+x^2)^2} \).

What is the integral of arctan x?

Integration by parts gives \( x\arctan x - \frac12\ln(1 + x^2) + C \). Differentiating it returns \( \arctan x \), using Example 4.

Why is the derivative of arctan x defined for every x?

The denominator \( 1 + x^2 \) is never zero, so the formula works for all real numbers. Compare arcsine, whose derivative only exists for \( |x| < 1 \).

What is the derivative of arctan x at 0?

It is \( \frac{1}{1 + 0} = 1 \), the steepest slope anywhere on the graph.

Related: derivative of arcsin x, derivative of tan x.

Further reading

• Paul’s Online Notes: Derivatives of Inverse Trig Functions — derivations of all six inverse trig derivatives and more examples.
• Paul’s Online Notes: Implicit Differentiation — the technique used in the proof, practiced on many other curves.