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Inverse trig derivatives

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.

Inverse trig derivatives Updated 2 min read

At a glance

\frac{d}{dx}\,\arcsin x = \frac{1}{\sqrt{1 - x^2}}
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
On this page
  1. Graph of arcsin x and its derivative
  2. Proof of the derivative of arcsin x
  3. Chain rule version: derivative of arcsin(u)
  4. Worked examples
  5. Common mistakes

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 of y = arcsin x and its derivative
The slope of \arcsin x is smallest at x = 0, where it equals 1, and grows without bound near x = \pm 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:

\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.

Chain rule version: derivative of arcsin(u)

With the 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:

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 is the same expression with a minus sign.

Going the other way, the integral of 1/√(1 − x²) is \arcsin x + C, and the integral of arcsin x uses integration by parts. Practise more with inverse trig derivative problems or read about the arcsine function.

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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.

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