AP Calculus AB and BC

Derivative of arcsin 2x: Answer, Proof, Mistakes

Q: What is the derivative of \arcsin 2x?

It is \frac{2}{\sqrt{1-4x^2}}, valid for |x| < \frac{1}{2}.

Q: Why is the radicand 1 - 4x^2 and not 1 - 2x^2?

The formula uses 1 - u^2 with u = 2x, and (2x)^2 = 4x^2, so the inside is 1 - 4x^2.

Q: Where does the 2 in the numerator come from?

It is the chain rule factor, the derivative of the inner function 2x.

The derivative of arcsin 2x is 2 divided by the square root of (1 minus 4x^2). In prime notation, if f(x) = arcsin 2x then f'(x) = 2/sqrt(1-4x^2). The inverse-sine derivative 1/sqrt(1-u^2) is evaluated at u = 2x and multiplied by the inner derivative 2.

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

How to differentiate arcsin 2x

The outer function is $\arcsin$ and the inner is $2x$ . The inverse-sine derivative is $\frac{d}{dx}\arcsin u = \frac{u'}{\sqrt{1-u^2}}$ , so substitute $u = 2x$ and $u' = 2$ .

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

Squaring the inside gives $(2x)^2 = 4x^2$ , which is why the radicand is $1 - 4x^2$ rather than $1 - 2x^2$ .

Watch the domain

The radicand must be positive, so $1 - 4x^2 > 0$ , that is $|x| < \frac{1}{2}$ . The derivative is stated on that open interval.

Where the derivative of arcsin 2x shows up on the AP exam

Inverse trig derivatives combined with the chain rule are Unit 3 material (Topics 3.1 and 3.4). The key move is to keep the inner function squared correctly inside the radical and to carry the inner derivative to the numerator.

The same pattern drives the reverse direction on BC: an integrand shaped like $\frac{1}{\sqrt{1-4x^2}}$ is an inverse-sine antiderivative in disguise.

Common mistakes with the derivative of arcsin 2x

Writing $\frac{1}{\sqrt{1-4x^2}}$ and dropping the inner factor of $2$ in the numerator.
Writing $\frac{2}{\sqrt{1-2x^2}}$ , forgetting to square the whole inside so that $(2x)^2 = 4x^2$ .
Using the arctangent form $\frac{1}{1+u^2}$ by mistake. Arcsine uses the square root of $1 - u^2$ .

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

What is the derivative of $\arcsin 2x$ ?

It is $\frac{2}{\sqrt{1-4x^2}}$ , valid for $|x| < \frac{1}{2}$ .

Why is the radicand $1 - 4x^2$ and not $1 - 2x^2$ ?

The formula uses $1 - u^2$ with $u = 2x$ , and $(2x)^2 = 4x^2$ , so the inside is $1 - 4x^2$ .

Where does the 2 in the numerator come from?

It is the chain rule factor, the derivative of the inner function $2x$ .