AP Calculus AB and BC

Derivative of arccos(2x): Answer, Proof, and Domain

Differentiating arccos(2x) gives negative 2 divided by the square root of 1 - 4x^2. Two things decide it: the minus sign, which is the only difference between the arccosine and arcsine formulas, and the chain rule factor of 2 from the inner derivative. That same inner 2x narrows the domain to x between -1/2 and 1/2.

ddx[arccos⁡(2x)]=−21−4x2\frac{d}{dx}\left[\arccos\left(2x\right)\right] = -\frac{2}{\sqrt{1-4x^{2}}}

Where the minus sign and the 2 come from

The arccosine rule is dduarccos⁡u=−11−u2\frac{d}{du}\arccos u = -\frac{1}{\sqrt{1-u^{2}}}, identical to the arcsine rule apart from the leading minus. Here u=2xu = 2x, so u′=2u' = 2 and u2=4x2u^{2} = 4x^{2}.

ddxarccos⁡(2x)=−11−(2x)2⋅2=−21−4x2\frac{d}{dx}\arccos\left(2x\right) = -\frac{1}{\sqrt{1-\left(2x\right)^{2}}}\cdot 2 = -\frac{2}{\sqrt{1-4x^{2}}}

The minus sign is structural, not an accident of this problem. arccos⁡u\arccos u and arcsin⁡u\arcsin u add to π2\frac{\pi}{2} for every uu in the domain, so their derivatives must be exact negatives of each other.

The domain shrinks when the input is doubled

arccos⁡\arccos accepts inputs from −1-1 to 11, so arccos⁡(2x)\arccos\left(2x\right) needs −1≤2x≤1-1\le 2x\le 1. Dividing by 22 gives −12≤x≤12-\frac{1}{2}\le x\le\frac{1}{2}, half the usual width.

1−4x2>0⟹x2<14⟹−12<x<121-4x^{2} > 0 \quad\Longrightarrow\quad x^{2} < \frac{1}{4} \quad\Longrightarrow\quad -\frac{1}{2} < x < \frac{1}{2}

The derivative is negative across that whole interval, so arccos⁡(2x)\arccos\left(2x\right) is strictly decreasing, dropping from π\pi at x=−12x = -\frac{1}{2} to 00 at x=12x = \frac{1}{2}. It steepens toward both ends, with vertical tangents where the radical vanishes.

The mistakes students make

Each of the three errors below loses exactly one piece of the answer: the sign, the squared coefficient, or the chain rule factor.

  • Dropping the minus sign and writing 21−4x2\frac{2}{\sqrt{1-4x^{2}}}. That is the derivative of arcsin⁡(2x)\arcsin\left(2x\right), a function that increases where this one decreases.
  • Squaring only the xx to get −21−2x2-\frac{2}{\sqrt{1-2x^{2}}}, leaving the coefficient alone. Squaring 2x2x squares both parts, giving 4x24x^{2}.
  • Finishing with −11−4x2-\frac{1}{\sqrt{1-4x^{2}}}, handling the inner function inside the radical but forgetting to multiply by the inner derivative 22.

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 arccos(2x)?

It is −21−4x2-\frac{2}{\sqrt{1-4x^{2}}}, valid for −12<x<12-\frac{1}{2}<x<\frac{1}{2}. The chain rule contributes the factor 22 and the arccosine rule contributes the minus sign.

Why is the derivative of arccos negative?

Because arccos⁡u+arcsin⁡u=π2\arccos u+\arcsin u = \frac{\pi}{2} is constant, so the two derivatives must cancel. Whatever arcsin⁡\arcsin gives, arccos⁡\arccos gives the negative of it.

What is the domain of arccos(2x)?

It runs from x=−12x = -\frac{1}{2} to x=12x = \frac{1}{2}, because the input 2x2x has to stay between −1-1 and 11. The derivative is defined on the open version of that interval.