Derivative of arctan(x)

Q: What is the derivative of tan^-1(x)?

\tan^{-1}x is the same as \arctan x, so the derivative is \frac{1}{1+x^2}.

Q: What is the second derivative of arctan(x)?

Differentiate (1+x^2)^{-1} with the chain rule: -\frac{2x}{(1+x^2)^2}.

Q: What is the derivative of arctan(2x)?

\frac{2}{1+4x^2} – the inner derivative 2 goes to the numerator, the whole argument 2x is squared.

Q: Why is the integral of 1/(1+x²) equal to arctan(x)?

Because differentiation and integration are inverse operations: since (\arctan x)' = \frac{1}{1+x^2}, an antiderivative of \frac{1}{1+x^2} is \arctan x + C.

The derivative of the inverse tangent is $\dfrac{d}{dx}\arctan x = \dfrac{1}{1+x^2}$. It is defined for every real number $x$ – unlike arcsin and arccos, arctan has no domain restriction. The same function is also written $\tan^{-1}x$ or atan(x).

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

See the derivative of arctan(x) step by step in the calculator →

📈 Graph of arctan(x) and its derivative

The derivative is always positive, so arctan is increasing everywhere. Its largest value is $1$ at $x=0$, where the graph of arctan is steepest; far out on both sides the slope tends to $0$ as the graph approaches the horizontal asymptotes $y = \pm\frac{\pi}{2}$.

🔍 Why is the derivative of arctan(x) equal to $\frac{1}{1+x^2}$?

Let $y = \arctan x$. Then $\tan y = x$ with $-\frac{\pi}{2} < y < \frac{\pi}{2}$. Differentiate both sides with respect to $x$, using $(\tan y)' = (1+\tan^2 y)\,y'$:

$$ (1+\tan^2 y)\,y' = 1 \quad\Longrightarrow\quad y' = \frac{1}{1+\tan^2 y} = \frac{1}{1+x^2}. $$

The key step is replacing $\tan y$ by $x$ – that is why the result contains no trigonometric function at all.

🔗 Derivative of arctan(x) with an inner function

If the argument is a function $u(x)$, the chain rule adds the inner derivative:

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

✏️ Worked examples

Function	Derivative	Note
$\arctan(2x)$	$\frac{2}{1+4x^2}$	Inner derivative $2$; the whole argument is squared: $(2x)^2 = 4x^2$.	steps →
$\arctan(x^2)$	$\frac{2x}{1+x^4}$	$u = x^2$, $u' = 2x$, $u^2 = x^4$.	steps →
$\arctan\left(\frac{x}{3}\right)$	$\frac{3}{9+x^2}$	$\frac{1/3}{1+x^2/9}$ – multiply numerator and denominator by $9$.	steps →
$\arctan\left(\frac{1}{x}\right)$	$-\frac{1}{1+x^2}$	Exactly the negative of the derivative of $\arctan x$ (for $x\neq 0$), because $\arctan x + \arctan\frac1x$ is constant on each half-line.	steps →

⚠️ Common mistakes

Writing $\frac{1}{1+x}$ – the $x$ in the denominator is squared.

Reading $\tan^{-1}x$ as $\frac{1}{\tan x} = \cot x$; that function has the derivative $-\csc^2 x$.

❓ Frequently asked questions

What is the derivative of tan^-1(x)?

$\tan^{-1}x$ is the same as $\arctan x$, so the derivative is $\frac{1}{1+x^2}$.

What is the second derivative of arctan(x)?

Differentiate $(1+x^2)^{-1}$ with the chain rule: $-\frac{2x}{(1+x^2)^2}$.

What is the derivative of arctan(2x)?

$\frac{2}{1+4x^2}$ – the inner derivative $2$ goes to the numerator, the whole argument $2x$ is squared.

Why is the integral of 1/(1+x²) equal to arctan(x)?

Because differentiation and integration are inverse operations: since $(\arctan x)' = \frac{1}{1+x^2}$, an antiderivative of $\frac{1}{1+x^2}$ is $\arctan x + C$.