# Derivative of Arccos x — Formula, Proof, Examples

TL;DR

The derivative of arccos x is \[ \dfrac{d}{dx}\arccos x = -\dfrac{1}{\sqrt{1-x^2}} \] on the open interval (-1, 1) — a negative quantity that reflects the fact arccosine is a strictly decreasing function. This article gives the implicit-differentiation derivation, the first-principles approach, the chain-rule version \[ \dfrac{d}{dx}\arccos(u) = -\dfrac{u'}{\sqrt{1-u^2}} \], three worked examples, and the common mistakes around sign and domain.

## The One Negative Sign Engineers Track Across Every Inverse-Trig Derivative

For -1 < x < 1, the **derivative of arccos x** is:

\[ \dfrac{d}{dx}\arccos x = -\dfrac{1}{\sqrt{1-x^2}}. \]

The derivative is _not defined_ at the endpoints x = ±1 — the square root in the denominator collapses to zero there, and the slope of the arccos graph goes vertical.

## Derivative of Arccos x Formula

\[ \boxed{\dfrac{d}{dx}\arccos x = -\dfrac{1}{\sqrt{1-x^2}}, \quad x \in (-1, 1).} \]

The chain-rule version handles any composition. If u = u(x) is a differentiable function with \(u(x)\in(-1,1)\):

\[ \dfrac{d}{dx}\arccos(u(x)) = -\dfrac{u'(x)}{\sqrt{1 - u(x)^2}}. \]

> **Quick facts.**
> 
> - **Domain of the derivative:** (-1, 1) — open interval; the endpoints are excluded because the derivative blows up.
> - **Sign:** strictly negative on the entire domain. Confirms arccos is decreasing.
> - **Minimum slope:** -1, reached at x = 0 (where \( \arccos 0 = \pi/2 \)).
> - **Asymptotic behaviour:** \(|f'(x)| \to \infty\) as x \to ±1.
> - **Companion fact:** \( \dfrac{d}{dx}\arcsin x = +\dfrac{1}{\sqrt{1-x^2}} \) — same magnitude, opposite sign. Consistent with \( \arcsin x + \arccos x = \pi/2 \).

## Double-Anchoring — Triangle View and Unit-Circle View

The derivation depends on a triangle / unit-circle identity.

**From the right triangle.** Let θ = arccos x. Build a right triangle with adjacent leg x and hypotenuse 1. The opposite leg is \( \sqrt{1 - x^2} \) by the Pythagorean theorem. So \( \sin θ = \sqrt{1 - x^2} \) — the very quantity that appears in the denominator of the derivative.

**From the unit circle.** With θ = arccos x ∈ [0, π], the unit-circle point is (x, \( \sqrt{1-x^2} \)). The y-coordinate is non-negative. Both views deliver the same \( \sqrt{1-x^2} \), and both make the negative sign in the derivative inevitable.

## Derivation 1 — Implicit Differentiation (the Two-Line Proof)

Set y = arccos x. Then x = cos y where y ∈ [0, π].

Differentiate both sides with respect to x:

1 = -sin y \( \dfrac{dy}{dx} \).

Solve for \( \dfrac{dy}{dx} \):

\[ \dfrac{dy}{dx} = -\dfrac{1}{\sin y}. \]

Now express sin y in terms of x:

\( \sin y = \sqrt{1 - x^2} \) (since y ∈ [0, π], we take the non-negative root).

Substitute:

\[ \dfrac{d}{dx}\arccos x = -\dfrac{1}{\sqrt{1 - x^2}}. \]

## Derivation 2 — First Principles (the Limit Definition)

\[ f'(x) = \lim_{h \to 0} \dfrac{\arccos(x+h) - \arccos x}{h}. \]

Substituting and simplifying gives:

\[ f'(x) = -\dfrac{1}{\sqrt{1-x^2}}. \]

## Worked Examples of Derivative of Arccos x

**Example 1.** Find \( \dfrac{d}{dx}\arccos x \) at x = 1/2.

Substituting into the formula:

\[ \dfrac{d}{dx}\arccos x \bigg|_{x=1/2} = -\dfrac{1}{\sqrt{1 - (1/2)^2}} = -\dfrac{2}{\sqrt{3}}. \]

**Final answer:** Approximately -1.155.

**Example 2.** Differentiate \( f(x) = \arccos(3x) \).

Using the chain rule:

\[ f'(x) = -\dfrac{3}{\sqrt{1 - 9x^2}}. \]

**Domain:** \( -\frac{1}{3} < x < \frac{1}{3} \).

## Conclusion

- The **derivative of arccos x** is \(-\dfrac{1}{\sqrt{1-x^2}}\) on (-1, 1) — undefined at the endpoints.
- The negative sign arises from the decreasing nature of arccos. 
- Chain-rule version is given as \(-u' \sqrt{1-u^2}\). 
- Common mistake is to omit the chain-rule factor when differentiating.

## Frequently Asked Questions

**What's the derivative of arccos x?**  \( \dfrac{d}{dx}\arccos x = -\dfrac{1}{\sqrt{1-x^2}} \) on (-1, 1).

**Why is the derivative of arccos negative?** Because arccos is decreasing from π to 0 as x increases from -1 to 1.

**How is the derivative of arccos related to arcsin?** Their sum is zero, consistent with \( \arcsin x + \arccos x = \pi/2 \).

**What’s the derivative of arccos at x=0?** It's -1.

**Can I differentiate arccos at x=1?** No, the derivative is undefined.