# Derivative of arccos x: Formula, Proof and Examples

> The derivative of arccos x is −1/√(1 − x²). See the proof, why it is the negative of the arcsin derivative, chain rule examples and a graph.

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

$$ \frac{d}{dx}\,\arccos x = -\frac{1}{\sqrt{1 - x^2}} $$

## At a glance

- **Derivative:** $-\dfrac{1}{\sqrt{1 - x^2}}$ for $-1 < x < 1$
- **Chain rule form:** $\dfrac{d}{dx}\arccos u = -\dfrac{u'}{\sqrt{1 - u^2}}$
- **Domain of arccos x:** $[-1, 1]$
- **Range of arccos x:** $[0, \pi]$
- **Key identity:** $\arcsin x + \arccos x = \frac{\pi}{2}$

## Key takeaways

- The derivative of $\arccos x$ is $-\frac{1}{\sqrt{1 - x^2}}$ for $-1 < x < 1$.
- It is exactly the negative of the derivative of $\arcsin x$, because $\arcsin x + \arccos x = \frac{\pi}{2}$.
- With the chain rule, $\frac{d}{dx}\arccos u = -\frac{u'}{\sqrt{1 - u^2}}$.
- The derivative is always negative, so $\arccos x$ is always decreasing.

The **derivative of $\arccos x$ is $-\dfrac{1}{\sqrt{1 - x^2}}$**. It looks just like the [derivative of arcsin x](/derivatives/arcsin-x) with a minus sign in front, and there is a neat reason for that.

$$\frac{d}{dx}\,\arccos x = -\frac{1}{\sqrt{1 - x^2}}, \qquad -1 < x < 1$$

## Graph of arccos x and its derivative

```graph
title: 'Graph of y = arccos x and its derivative'
caption: 'The inverse cosine falls from $\pi$ to $0$. Its slope is $-1$ at $x = 0$ and becomes very steep near $x = \pm 1$.'
x: [-1.5, 1.5]
y: [-4, 3.5]
functions:
  - expr: arccos(x)
    label: y = arccos x
    domain: [-1, 1]
  - expr: -1/sqrt(1-x^2)
    label: y = −1/√(1 − x²) (derivative)
    dashed: true
    domain: [-0.97, 0.97]
points:
  - [0, 1.5708, '(0, π/2)']
  - [0, -1, 'slope −1']
```

## Proof of the derivative of arccos x

Let $y = \arccos x$, so $\cos y = x$ with $0 \le y \le \pi$. Use [implicit differentiation](/rules/implicit-differentiation):

$$-\sin y \cdot \frac{dy}{dx} = 1 \quad\Longrightarrow\quad \frac{dy}{dx} = -\frac{1}{\sin y}$$

On $[0, \pi]$ sine is never negative, so $\sin y = \sqrt{1 - \cos^2 y} = \sqrt{1 - x^2}$. That gives

$$\frac{d}{dx}\arccos x = -\frac{1}{\sqrt{1 - x^2}}$$

## A faster proof using arcsin

For every $x$ in $[-1, 1]$,

$$\arcsin x + \arccos x = \frac{\pi}{2}$$

Picture a right triangle: if one acute angle has sine $x$, the other has cosine $x$, and the two angles add to $\frac{\pi}{2}$. Differentiate both sides. The right side is a constant, so its derivative is $0$:

$$\frac{1}{\sqrt{1 - x^2}} + \frac{d}{dx}\arccos x = 0$$

So the arccos derivative is the arcsin derivative with the sign flipped. This is a good way to remember it.

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

$$\frac{d}{dx}\arccos u = -\frac{u'}{\sqrt{1 - u^2}}$$

| Function | Inside $u$ | Derivative |
| --- | --- | --- |
| $\arccos(3x)$ | $3x$ | $-\frac{3}{\sqrt{1 - 9x^2}}$ |
| $\arccos(x^2)$ | $x^2$ | $-\frac{2x}{\sqrt{1 - x^4}}$ |
| $\arccos(\sqrt{x})$ | $\sqrt{x}$ | $-\frac{1}{2\sqrt{x}\sqrt{1 - x}}$ |
| $\arccos\!\left(\frac{x}{2}\right)$ | $\frac{x}{2}$ | $-\frac{1}{\sqrt{4 - x^2}}$ |

## Worked examples

**Example 1: product rule.** For $f(x) = x^2\arccos x$, the [product rule](/rules/product-rule) gives

$$f'(x) = 2x\arccos x - \frac{x^2}{\sqrt{1 - x^2}}$$

**Example 2: slope at a point.** At $x = \frac{\sqrt{3}}{2}$, the slope of $\arccos x$ is

$$-\frac{1}{\sqrt{1 - \frac{3}{4}}} = -\frac{1}{1/2} = -2$$

**Example 3: tangent line.** At $x = 0$ the curve passes through $\left(0, \frac{\pi}{2}\right)$ with slope $-1$, so the tangent line is $y = \frac{\pi}{2} - x$.

## Common mistakes

1. **Dropping the minus sign.** It is the only difference from arcsin, and the easiest thing to lose.
2. **Thinking $\cos^{-1} x = \sec x$.** The reciprocal of cosine is secant. $\cos^{-1}$ is the inverse function.
3. **Leaving out $u'$.** $\frac{d}{dx}\arccos(5x) = -\frac{5}{\sqrt{1 - 25x^2}}$, not $-\frac{1}{\sqrt{1 - 25x^2}}$.
4. **Using it at $x = \pm 1$.** The tangent is vertical there, so there is no derivative.

You will rarely see $-\frac{1}{\sqrt{1 - x^2}}$ integrated as arccos, since $-\arcsin x + C$ is the same family of answers. See the [arccosine function](/functions/arccosine) page for the graph and range, and practise with [inverse trig derivative problems](/practice-problems/inverse-trig-derivatives).

## Frequently asked questions

### What is the derivative of arccos x?

The derivative of $\arccos x$ (also written $\cos^{-1} x$) is $-\frac{1}{\sqrt{1 - x^2}}$ for $-1 < x < 1$.

### Why is the derivative of arccos x negative?

Because $\arccos x$ falls from $\pi$ to $0$ as $x$ goes from $-1$ to $1$. A decreasing function has a negative derivative.

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

By the chain rule it is $-\frac{2}{\sqrt{1 - 4x^2}}$, valid for $-\frac{1}{2} < x < \frac{1}{2}$.

### What is the derivative of arcsin x + arccos x?

It is $0$. The two derivatives cancel, which matches the identity $\arcsin x + \arccos x = \frac{\pi}{2}$, a constant.