# Derivative of arcsin x: Formula, Proof and Examples

> The derivative of arcsin x is 1/√(1 − x²). See the proof from sin y = x, the chain rule form for arcsin(u), a graph and worked examples.

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

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

## At a glance

- **Derivative:** $\dfrac{1}{\sqrt{1 - x^2}}$ for $-1 < x < 1$
- **Chain rule form:** $\dfrac{d}{dx}\arcsin u = \dfrac{u'}{\sqrt{1 - u^2}}$
- **Domain of arcsin x:** $[-1, 1]$
- **Range of arcsin x:** $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
- **Integral version:** $\int \frac{1}{\sqrt{1 - x^2}}\,dx = \arcsin x + C$

## Key takeaways

- The derivative of $\arcsin x$ is $\frac{1}{\sqrt{1 - x^2}}$, valid for $-1 < x < 1$.
- The proof uses implicit differentiation on $\sin y = x$ and the identity $\cos y = \sqrt{1 - x^2}$.
- With the chain rule, $\frac{d}{dx}\arcsin u = \frac{u'}{\sqrt{1 - u^2}}$.
- The derivative is always positive, so $\arcsin x$ is always increasing, and it blows up at $x = \pm 1$.

The **derivative of $\arcsin x$ is $\dfrac{1}{\sqrt{1 - x^2}}$**. It holds on the open interval $-1 < x < 1$, where the inverse sine is smooth.

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

## Graph of arcsin x and its derivative

```graph
title: 'Graph of y = arcsin x and its derivative'
caption: 'The slope of $\arcsin x$ is smallest at $x = 0$, where it equals $1$, and grows without bound near $x = \pm 1$.'
x: [-1.5, 1.5]
y: [-2, 4]
functions:
  - expr: arcsin(x)
    label: y = arcsin 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, 0, '(0, 0)']
  - [0, 1, 'slope 1']
```

The derivative is never negative, so $\arcsin x$ always rises. Its smallest slope is $1$ at the origin, and the curve turns vertical at the endpoints $(\pm 1, \pm\frac{\pi}{2})$.

## Proof of the derivative of arcsin x

Let $y = \arcsin x$. Then $\sin y = x$ with $-\frac{\pi}{2} \le y \le \frac{\pi}{2}$. Differentiate both sides with [implicit differentiation](/rules/implicit-differentiation):

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

Now write $\cos y$ in terms of $x$. From $\sin^2 y + \cos^2 y = 1$ we get $\cos y = \pm\sqrt{1 - x^2}$. On $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ cosine is never negative, so we take the positive root:

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

The same idea works for every inverse function, see the [inverse function rule](/rules/inverse-function-rule).

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

With the [chain rule](/rules/chain-rule), put the inside derivative on top:

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

| Function | Inside $u$ | Derivative |
| --- | --- | --- |
| $\arcsin(2x)$ | $2x$ | $\frac{2}{\sqrt{1 - 4x^2}}$ |
| $\arcsin(x^2)$ | $x^2$ | $\frac{2x}{\sqrt{1 - x^4}}$ |
| $\arcsin\!\left(\frac{x}{a}\right)$, $a > 0$ | $\frac{x}{a}$ | $\frac{1}{\sqrt{a^2 - x^2}}$ |
| $\arcsin(e^x)$ | $e^x$ | $\frac{e^x}{\sqrt{1 - e^{2x}}}$ |

The third row is worth remembering. It is why $\int \frac{dx}{\sqrt{a^2 - x^2}} = \arcsin\frac{x}{a} + C$.

## Worked examples

**Example 1: a product.** Differentiate $f(x) = x\arcsin x$ with the [product rule](/rules/product-rule):

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

**Example 2: a slope at a point.** The slope of $y = \arcsin x$ at $x = \frac{1}{2}$ is

$$\frac{1}{\sqrt{1 - \frac{1}{4}}} = \frac{1}{\sqrt{3}/2} = \frac{2}{\sqrt{3}} \approx 1.155$$

**Example 3: a power of arcsin.** For $g(x) = (\arcsin x)^2$ the outer function is a square:

$$g'(x) = 2\arcsin x \cdot \frac{1}{\sqrt{1 - x^2}} = \frac{2\arcsin x}{\sqrt{1 - x^2}}$$

## Common mistakes

1. **Reading $\sin^{-1} x$ as $\frac{1}{\sin x}$.** The $-1$ means inverse function, not reciprocal.
2. **Forgetting the inside derivative.** $\frac{d}{dx}\arcsin(3x)$ is $\frac{3}{\sqrt{1 - 9x^2}}$, not $\frac{1}{\sqrt{1 - 9x^2}}$.
3. **Squaring only $x$ in the chain rule.** Inside the root you need $1 - u^2$, so $\arcsin(x^2)$ gives $\sqrt{1 - x^4}$, not $\sqrt{1 - x^2}$.
4. **Mixing up the sign with arccos.** The [derivative of arccos x](/derivatives/arccos-x) is the same expression with a minus sign.

Going the other way, the [integral of 1/√(1 − x²)](/integrals/1-over-sqrt-1-minus-x-squared) is $\arcsin x + C$, and the [integral of arcsin x](/integrals/arcsin-x) uses integration by parts. Practise more with [inverse trig derivative problems](/practice-problems/inverse-trig-derivatives) or read about the [arcsine function](/functions/arcsine).

## Frequently asked questions

### What is the derivative of arcsin x?

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

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

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

### Is the derivative of arcsin x defined at x = 1?

No. At $x = \pm 1$ the denominator $\sqrt{1 - x^2}$ is zero and the graph of $\arcsin x$ has a vertical tangent, so the derivative does not exist there.

### Is sin⁻¹ x the same as 1/sin x?

No. $\sin^{-1} x$ means the inverse function $\arcsin x$. The reciprocal $\frac{1}{\sin x}$ is $\csc x$, whose derivative is $-\csc x\cot x$.