# Derivative of arccot x: Formula, Proof and Examples

> The derivative of arccot x is −1/(1 + x²). See the proof from cot y = x, the link to arctan, chain rule examples and a graph.

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

$$ \frac{d}{dx}\,\operatorname{arccot} x = -\frac{1}{1 + x^2} $$

## At a glance

- **Derivative:** $-\dfrac{1}{1 + x^2}$ for every real $x$
- **Chain rule form:** $\dfrac{d}{dx}\operatorname{arccot} u = -\dfrac{u'}{1 + u^2}$
- **Domain of arccot x:** all real numbers
- **Range of arccot x:** $(0, \pi)$
- **Key identity:** $\operatorname{arccot} x = \frac{\pi}{2} - \arctan x$

## Key takeaways

- The derivative of $\operatorname{arccot} x$ is $-\frac{1}{1 + x^2}$.
- It is the negative of the arctan derivative, because $\operatorname{arccot} x = \frac{\pi}{2} - \arctan x$.
- With the chain rule, $\frac{d}{dx}\operatorname{arccot} u = -\frac{u'}{1 + u^2}$.
- Arccot is always decreasing, with its steepest slope of $-1$ at $x = 0$.

The **derivative of $\operatorname{arccot} x$ is $-\dfrac{1}{1 + x^2}$**. It is the [derivative of arctan x](/derivatives/arctan-x) with a minus sign, and it holds for every real number.

$$\frac{d}{dx}\,\operatorname{arccot} x = -\frac{1}{1 + x^2}$$

## Graph of arccot x and its derivative

```graph
title: 'Graph of y = arccot x and its derivative'
caption: 'The inverse cotangent falls from $\pi$ toward $0$. Its slope is $-1$ at $x = 0$ and fades to $0$ far from the origin.'
x: [-6, 6]
y: [-1.5, 3.5]
functions:
  - expr: pi/2 - arctan(x)
    label: y = arccot x
  - expr: -1/(1+x^2)
    label: y = −1/(1 + x²) (derivative)
    dashed: true
points:
  - [0, 1.5708, '(0, π/2)']
  - [0, -1, 'slope −1']
```

## Proof from cot y = x

Let $y = \operatorname{arccot} x$, so $\cot y = x$ with $0 < y < \pi$. Differentiate with [implicit differentiation](/rules/implicit-differentiation), using $\frac{d}{dy}\cot y = -\csc^2 y$ (see the [derivative of cot x](/derivatives/cot-x)):

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

The identity $\csc^2 y = 1 + \cot^2 y = 1 + x^2$ finishes it:

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

## Proof from arctan

With the range $(0, \pi)$, arccot and arctan are related by

$$\operatorname{arccot} x = \frac{\pi}{2} - \arctan x$$

The derivative of the constant $\frac{\pi}{2}$ is $0$, so the derivative of arccot is just minus the derivative of arctan. This identity is also how the graph above is drawn.

## A note on conventions

Some books and software define $\operatorname{arccot} x = \arctan\frac{1}{x}$, which has range $\left(-\frac{\pi}{2}, 0\right) \cup \left(0, \frac{\pi}{2}\right]$ and jumps at $x = 0$. The derivative still comes out the same wherever the function is defined:

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

So the formula is safe to use either way. Only the values of the function differ.

## Chain rule version and examples

$$\frac{d}{dx}\operatorname{arccot} u = -\frac{u'}{1 + u^2}$$

| Function | Derivative |
| --- | --- |
| $\operatorname{arccot}(3x)$ | $-\frac{3}{1 + 9x^2}$ |
| $\operatorname{arccot}(x^2)$ | $-\frac{2x}{1 + x^4}$ |
| $\operatorname{arccot}(\ln x)$ | $-\frac{1}{x(1 + \ln^2 x)}$ |
| $\operatorname{arccot}(e^x)$ | $-\frac{e^x}{1 + e^{2x}}$ |

**Worked example.** Differentiate $f(x) = x\operatorname{arccot} x$ with the [product rule](/rules/product-rule):

$$f'(x) = \operatorname{arccot} x - \frac{x}{1 + x^2}$$

At $x = 1$: $\operatorname{arccot} 1 = \frac{\pi}{4}$, so $f'(1) = \frac{\pi}{4} - \frac{1}{2} \approx 0.285$.

## Common mistakes

1. **Writing $-\csc^2 x$.** That is the derivative of $\cot x$, not of the inverse.
2. **Losing the minus sign.** Arccot decreases, so its slope must be negative.
3. **Forgetting $u'$.** $\operatorname{arccot}(5x)$ gives $-\frac{5}{1 + 25x^2}$.
4. **Assuming the integral needs arccot.** Since the derivatives differ only by sign, $\int \frac{dx}{1 + x^2}$ is usually written as $\arctan x + C$. The answer $-\operatorname{arccot} x + C$ is also correct, because the two differ by a constant.

All six inverse trig derivatives come in pairs: arcsin and [arccos](/derivatives/arccos-x), arctan and arccot, [arcsec](/derivatives/arcsec-x) and arccsc. The second of each pair has a minus sign. Practise them with [inverse trig derivative problems](/practice-problems/inverse-trig-derivatives).

## Frequently asked questions

### What is the derivative of arccot x?

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

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

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

### Is arccot x equal to arctan(1/x)?

Only for $x > 0$. For $x < 0$ they differ by $\pi$. Either way, where both are defined they have the same derivative, $-\frac{1}{1 + x^2}$.

### What is the derivative of arctan x + arccot x?

It is $0$, since the sum is the constant $\frac{\pi}{2}$.