AP Calculus AB and BC

Derivative of arcsin x: Answer, Proof, Mistakes

Q: Why is the derivative of arcsin x equal to 1/sqrt(1-x^2)?

Because \arcsin x is the inverse of sine on \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]. Rewrite y = \arcsin x as \sin y = x, differentiate to get \cos y \cdot \frac{dy}{dx} = 1, then use \cos y = \sqrt{1-x^2} (positive on that range) to get \frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}.

Q: What is the derivative of arccos x?

It is -\frac{1}{\sqrt{1-x^2}}, the negative of the arcsin derivative. Since \arcsin x + \arccos x = \frac{\pi}{2} is a constant, differentiating both sides shows the two derivatives must be exact opposites.

Q: What is the derivative of arcsin(u)?

Use the chain rule: \frac{d}{dx}[\arcsin(u)] = \frac{u'}{\sqrt{1-u^2}}, the derivative of the inside over the square root of one minus the inside squared. For example, \frac{d}{dx}[\arcsin(3x)] = \frac{3}{\sqrt{1-9x^2}}.

Q: Where is the derivative of arcsin x undefined?

At x = \pm 1. The denominator \sqrt{1-x^2} is 0 there, which matches the vertical tangents on the graph of \arcsin, so the derivative exists only on the open interval -1 < x < 1 even though \arcsin x itself is defined on [-1, 1].

The derivative of arcsin x is 1/sqrt(1 - x^2), valid for -1 < x < 1. In prime notation, if f(x) = arcsin x then f'(x) = 1/sqrt(1 - x^2). The proof differentiates sin(f(x)) = x implicitly and simplifies cos back to sqrt(1 - x^2) on arcsin's range. The derivative of arccos x is the same but negative.

\frac{d}{dx}\left[\arcsin x\right] = \frac{1}{\sqrt{1-x^2}}

How to prove the derivative of arcsin x

Start from $y = \arcsin x$ . By the definition of the inverse sine, this is the same statement as $\sin y = x$ with $y$ restricted to $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ , the range that keeps $\arcsin$ a genuine function.

\sin y = x

Differentiate both sides with respect to $x$ . The right side is just 1; the left side needs the chain rule, since $y$ depends on $x$ .

\cos y \cdot \frac{dy}{dx} = 1

Solve for $\frac{dy}{dx}$ , then rewrite $\cos y$ in terms of $x$ . From $\sin y = x$ and the identity $\sin^2 y + \cos^2 y = 1$ , you get $\cos y = \pm\sqrt{1 - x^2}$ .

\frac{dy}{dx} = \frac{1}{\cos y}, \qquad \cos y = \pm\sqrt{1 - x^2}

The range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ settles the sign: cosine is never negative on that interval, so take the positive root. Substituting gives the derivative.

\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}}

Watch the domain and the endpoints

$\arcsin x$ is defined on the closed interval $[-1, 1]$ , but the derivative needs $1 - x^2 > 0$ , so $\frac{1}{\sqrt{1-x^2}}$ exists only on the open interval $-1 < x < 1$ . At $x = \pm 1$ the denominator is 0: the graph of $\arcsin$ has vertical tangents there, and the derivative is undefined.

arccos is the mirror image

The same proof on $\cos y = x$ gives $\frac{d}{dx}[\arccos x] = -\frac{1}{\sqrt{1-x^2}}$ , the exact negative. It has to be: $\arcsin x + \arccos x = \frac{\pi}{2}$ is constant, so differentiating both sides forces their derivatives to cancel.

Where arcsin x's derivative shows up on the AP exam

The rule $\frac{d}{dx}[\arcsin x] = \frac{1}{\sqrt{1-x^2}}$ is introduced in Unit 3, Topic 3.4 (Differentiating Inverse Trigonometric Functions). Unit 3, Differentiation: Composite, Implicit, and Inverse Functions, is weighted 5 to 10 percent on both the AB and BC exams.

The proof above is Topic 3.2 (Implicit Differentiation) applied to $\sin y = x$ , and it is a special case of Topic 3.3 (Differentiating Inverse Functions). On the exam the rule is rarely asked bare; it usually appears wrapped in the chain rule (Topic 3.1) on a composite $\arcsin(u)$ .

\frac{d}{dx}\left[\arcsin(u)\right] = \frac{u'}{\sqrt{1 - u^2}}

It also runs in reverse. Recognizing that $\frac{1}{\sqrt{1-x^2}}$ antidifferentiates back to $\arcsin x$ is part of choosing an antidifferentiation technique in Unit 6 (Topic 6.14), so this one derivative pays off in both the differentiation and integration units.

CED topic	How the derivative of arcsin x appears
3.4 Differentiating Inverse Trigonometric Functions	The rule itself, $\frac{d}{dx}[\arcsin x] = \frac{1}{\sqrt{1-x^2}}$
3.2 Implicit Differentiation	The $\sin y = x$ derivation
3.1 The Chain Rule	Composites such as $\arcsin(2x)$ and $\arcsin(x^2)$
6.14 Selecting Techniques for Antidifferentiation	Reverse: $\int \frac{1}{\sqrt{1-x^2}}\,dx = \arcsin x + C$

Common mistakes with the derivative of arcsin x

Putting the minus sign on arcsin. The derivative of $\arcsin x$ is $+\frac{1}{\sqrt{1-x^2}}$ ; the negative belongs to $\frac{d}{dx}[\arccos x] = -\frac{1}{\sqrt{1-x^2}}$ . Same denominator, opposite sign.
Reading $\arcsin x$ as $\frac{1}{\sin x}$ . The notation $\sin^{-1} x$ means the inverse sine, not a reciprocal. The reciprocal $\frac{1}{\sin x} = \csc x$ is a different function with derivative $-\csc x \cot x$ .
Forgetting to square the inside on a composite. For $\arcsin(u)$ the denominator is $\sqrt{1 - u^2}$ , so $\frac{d}{dx}[\arcsin(x^3)] = \frac{3x^2}{\sqrt{1 - x^6}}$ , with $x^6 = (x^3)^2$ under the root, not $x^3$ .
Using the derivative at $x = \pm 1$ . Those endpoints are in the domain of $\arcsin$ , but the derivative blows up there (vertical tangents), so $\frac{1}{\sqrt{1-x^2}}$ is valid only on $-1 < x < 1$ .
Dropping the chain rule factor. $\frac{d}{dx}[\arcsin(2x)] = \frac{2}{\sqrt{1-4x^2}}$ , not $\frac{1}{\sqrt{1-4x^2}}$ ; the inner derivative 2 multiplies, and the inside becomes $(2x)^2 = 4x^2$ .

Worked chain-rule examples

Every composite $\arcsin(u)$ runs on the same template: $\frac{1}{\sqrt{1-u^2}}$ times $u'$ . Track the inside $u$ carefully, because it plays two roles at once: it is squared under the root and differentiated on top.

Example 1. Differentiate $\arcsin(2x)$ . Here $u = 2x$ , so $u' = 2$ and $u^2 = 4x^2$ .

\frac{d}{dx}\left[\arcsin(2x)\right] = \frac{2}{\sqrt{1 - 4x^2}}

Example 2. Differentiate $\arcsin(x^2)$ . Now $u = x^2$ , so $u' = 2x$ and $u^2 = x^4$ .

\frac{d}{dx}\left[\arcsin(x^2)\right] = \frac{2x}{\sqrt{1 - x^4}}

Keep the two roles of u separate

In Example 2 the inside $x^2$ shows up twice and differently: squared under the root it becomes $x^4$ , differentiated on top it becomes $2x$ . Mixing those two roles (writing $x^2$ under the root, or $2x^2$ on top) is the usual slip.

Check yourself, not just the answer

Type derivatives and get graded on mathematical equivalence, with rule-level hints when you miss, in the Derivative Practice Checker.

Frequently asked questions

Why is the derivative of arcsin x equal to 1/sqrt(1-x^2)?

Because $\arcsin x$ is the inverse of sine on $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ . Rewrite $y = \arcsin x$ as $\sin y = x$ , differentiate to get $\cos y \cdot \frac{dy}{dx} = 1$ , then use $\cos y = \sqrt{1-x^2}$ (positive on that range) to get $\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}$ .

What is the derivative of arccos x?

It is $-\frac{1}{\sqrt{1-x^2}}$ , the negative of the arcsin derivative. Since $\arcsin x + \arccos x = \frac{\pi}{2}$ is a constant, differentiating both sides shows the two derivatives must be exact opposites.

What is the derivative of arcsin(u)?

Use the chain rule: $\frac{d}{dx}[\arcsin(u)] = \frac{u'}{\sqrt{1-u^2}}$ , the derivative of the inside over the square root of one minus the inside squared. For example, $\frac{d}{dx}[\arcsin(3x)] = \frac{3}{\sqrt{1-9x^2}}$ .

Where is the derivative of arcsin x undefined?

At $x = \pm 1$ . The denominator $\sqrt{1-x^2}$ is 0 there, which matches the vertical tangents on the graph of $\arcsin$ , so the derivative exists only on the open interval $-1 < x < 1$ even though $\arcsin x$ itself is defined on $[-1, 1]$ .