# Derivative of arcsec x: Formula, Proof and Examples

> The derivative of arcsec x is 1/(|x|√(x² − 1)). See where the absolute value comes from, a proof via arccos(1/x), and chain rule examples.

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

$$ \frac{d}{dx}\,\operatorname{arcsec} x = \frac{1}{\lvert x\rvert\sqrt{x^2 - 1}} $$

## At a glance

- **Derivative:** $\dfrac{1}{\lvert x\rvert\sqrt{x^2 - 1}}$ for $\lvert x\rvert > 1$
- **Chain rule form:** $\dfrac{d}{dx}\operatorname{arcsec} u = \dfrac{u'}{\lvert u\rvert\sqrt{u^2 - 1}}$
- **Domain of arcsec x:** $x \le -1$ or $x \ge 1$
- **Range of arcsec x:** $\left[0, \frac{\pi}{2}\right) \cup \left(\frac{\pi}{2}, \pi\right]$
- **Arccsc derivative:** $-\dfrac{1}{\lvert x\rvert\sqrt{x^2 - 1}}$

## Key takeaways

- The derivative of $\operatorname{arcsec} x$ is $\frac{1}{\lvert x\rvert\sqrt{x^2 - 1}}$ for $\lvert x\rvert > 1$.
- The absolute value comes from the range $[0, \pi]$, which makes arcsec increasing on both pieces of its domain.
- Since $\operatorname{arcsec} x = \arccos\frac{1}{x}$, the derivative follows from the arccos rule and the chain rule.
- The derivative of $\operatorname{arccsc} x$ is the same expression with a minus sign.

The **derivative of $\operatorname{arcsec} x$ is $\dfrac{1}{\lvert x\rvert\sqrt{x^2 - 1}}$**. It is defined for $\lvert x\rvert > 1$, the same values where secant can output a number.

$$\frac{d}{dx}\,\operatorname{arcsec} x = \frac{1}{\lvert x\rvert\sqrt{x^2 - 1}}, \qquad \lvert x\rvert > 1$$

For $x > 1$, where most problems live, this is just $\frac{1}{x\sqrt{x^2 - 1}}$.

## Graph of arcsec x

```graph
title: 'Graph of y = arcsec x, drawn as arccos(1/x)'
caption: 'Both pieces of the inverse secant rise. The right piece starts at $(1, 0)$ and the left piece ends at $(-1, \pi)$. Both approach the line $y = \frac{\pi}{2}$.'
x: [-6, 6]
y: [-0.5, 3.5]
functions:
  - expr: arccos(1/x)
    label: y = arcsec x
    domain: [1, 6]
  - expr: arccos(1/x)
    label: y = arcsec x
    domain: [-6, -1]
  - expr: 1.5708
    label: y = π/2
    dashed: true
points:
  - [1, 0, '(1, 0)']
  - [-1, 3.1416, '(−1, π)']
```

Both branches go upward from left to right, so the derivative is positive everywhere. That is exactly why the formula has $\lvert x\rvert$ in it: without the absolute value, the formula would give negative slopes for $x < -1$, which does not match the picture.

## Proof using arccos

The usual range of arcsec makes this identity true:

$$\operatorname{arcsec} x = \arccos\frac{1}{x}$$

Differentiate with the [chain rule](/rules/chain-rule) and the [derivative of arccos x](/derivatives/arccos-x):

$$\frac{d}{dx}\arccos\frac{1}{x} = -\frac{1}{\sqrt{1 - \frac{1}{x^2}}}\cdot\left(-\frac{1}{x^2}\right) = \frac{1}{x^2\sqrt{1 - \frac{1}{x^2}}}$$

Now simplify the root: $\sqrt{1 - \frac{1}{x^2}} = \frac{\sqrt{x^2 - 1}}{\sqrt{x^2}} = \frac{\sqrt{x^2 - 1}}{\lvert x\rvert}$. So

$$\frac{d}{dx}\operatorname{arcsec} x = \frac{\lvert x\rvert}{x^2\sqrt{x^2 - 1}} = \frac{1}{\lvert x\rvert\sqrt{x^2 - 1}}$$

using $x^2 = \lvert x\rvert^2$. The absolute value appears naturally from $\sqrt{x^2} = \lvert x\rvert$.

## Chain rule version and examples

$$\frac{d}{dx}\operatorname{arcsec} u = \frac{u'}{\lvert u\rvert\sqrt{u^2 - 1}}$$

**Example 1.** $\frac{d}{dx}\operatorname{arcsec}(x^2) = \frac{2x}{x^2\sqrt{x^4 - 1}} = \frac{2}{x\sqrt{x^4 - 1}}$ for $\lvert x\rvert > 1$. Here $\lvert x^2\rvert = x^2$, so no absolute value is left.

**Example 2.** For $x > 0$, $\frac{d}{dx}\operatorname{arcsec}(e^x) = \frac{e^x}{e^x\sqrt{e^{2x} - 1}} = \frac{1}{\sqrt{e^{2x} - 1}}$.

**Example 3: slope at a point.** At $x = 2$, the slope of $\operatorname{arcsec} x$ is $\frac{1}{2\sqrt{3}} \approx 0.289$. Since $\operatorname{arcsec} 2 = \frac{\pi}{3}$, the tangent line is $y = \frac{\pi}{3} + \frac{1}{2\sqrt{3}}(x - 2)$.

## The arccsc derivative

Since $\operatorname{arcsec} x + \operatorname{arccsc} x = \frac{\pi}{2}$, the derivatives are opposites:

$$\frac{d}{dx}\operatorname{arccsc} x = -\frac{1}{\lvert x\rvert\sqrt{x^2 - 1}}$$

## Common mistakes

1. **Dropping the absolute value** and then getting negative slopes for $x < -1$.
2. **Mixing textbook conventions.** Check which range your course uses for arcsec before copying a formula.
3. **Confusing $\sec^{-1} x$ with $\cos x$.** $\sec^{-1}$ is the inverse of secant, while $\frac{1}{\sec x} = \cos x$.
4. **Using it for $-1 < x < 1$.** No angle has a secant between $-1$ and $1$, so arcsec and its derivative do not exist there.

The formula also gives an integral: $\int \frac{dx}{x\sqrt{x^2 - 1}} = \operatorname{arcsec}\lvert x\rvert + C$. For more, see [inverse trig derivative problems](/practice-problems/inverse-trig-derivatives) and the [reciprocal trig functions](/functions/reciprocal-trig-functions).

## Frequently asked questions

### What is the derivative of arcsec x?

The derivative of $\operatorname{arcsec} x$ is $\frac{1}{\lvert x\rvert\sqrt{x^2 - 1}}$ for $\lvert x\rvert > 1$. For $x > 1$ this is simply $\frac{1}{x\sqrt{x^2 - 1}}$.

### Why do some books write x√(x² − 1) without the absolute value?

Those books pick a different range for arcsec, $\left[0, \frac{\pi}{2}\right) \cup \left[\pi, \frac{3\pi}{2}\right)$. With that choice the formula is $\frac{1}{x\sqrt{x^2 - 1}}$. With the more common range $[0, \pi]$ you need $\lvert x\rvert$.

### What is the derivative of arccsc x?

It is $-\frac{1}{\lvert x\rvert\sqrt{x^2 - 1}}$, the negative of the arcsec derivative, because $\operatorname{arcsec} x + \operatorname{arccsc} x = \frac{\pi}{2}$.

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

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