# Derivative of cos²x: Formula, Proof and Examples

> The derivative of cos²x is -2 sin x cos x, which equals -sin 2x. See the chain rule proof, an identity check and examples like cos²(4x).

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

$$ \frac{d}{dx}\,\cos^2 x = -2\sin x\cos x = -\sin 2x $$

## At a glance

- **Derivative:** $-2\sin x\cos x$
- **Simplified:** $-\sin 2x$
- **Chain rule form:** $\dfrac{d}{dx}\cos^2 u = -2\cos u\sin u\cdot u'$
- **Not the same as:** $\cos(x^2)$, whose derivative is $-2x\sin(x^2)$
- **Integral of cos²x:** $\int\cos^2 x\,dx = \frac{x}{2} + \frac{\sin 2x}{4} + C$

## Key takeaways

- The derivative of $\cos^2 x$ is $-2\sin x\cos x = -\sin 2x$.
- Read $\cos^2 x$ as $(\cos x)^2$, then use the chain rule.
- It is exactly the negative of the derivative of $\sin^2 x$, because $\sin^2 x + \cos^2 x = 1$.
- The identity $\cos^2 x = \frac{1 + \cos 2x}{2}$ gives the same answer.

The **derivative of $\cos^2 x$ is $-2\sin x\cos x$**, which is the same as $-\sin 2x$.

$$\frac{d}{dx}\,\cos^2 x = -2\sin x\cos x = -\sin 2x$$

## Graph of cos²x and its derivative

```graph
title: 'Graph of y = cos²x and its derivative y = -sin 2x'
caption: '$\cos^2 x$ starts at its peak of $1$ when $x = 0$ and falls. Its derivative $-\sin 2x$ is negative while it falls and positive while it rises.'
x: [-0.5, 6.5]
y: [-1.5, 1.6]
functions:
  - expr: cos(x)^2
    label: y = cos²x
  - expr: -sin(2x)
    label: y = -sin 2x (derivative)
    dashed: true
points:
  - [0, 1, 'peak, slope 0']
  - [0.7854, 0.5, 'slope -1']
```

## Proof with the chain rule

$\cos^2 x$ means $(\cos x)^2$. The outside is the square and the inside is $\cos x$. With the [chain rule](/rules/chain-rule) and the [derivative of cos x](/derivatives/cos-x):

$$\frac{d}{dx}(\cos x)^2 = 2\cos x\cdot(-\sin x) = -2\sin x\cos x$$

Since $2\sin x\cos x = \sin 2x$, this is $-\sin 2x$.

## A check with a power-reducing identity

Use $\cos^2 x = \frac{1 + \cos 2x}{2}$:

$$\frac{d}{dx}\left(\frac{1}{2} + \frac{1}{2}\cos 2x\right) = \frac{1}{2}\cdot(-2\sin 2x) = -\sin 2x$$

## A check with sin²x

Because $\cos^2 x = 1 - \sin^2 x$, its derivative must be $0 - \sin 2x$, the negative of the [derivative of sin²x](/derivatives/sin-squared-x). The two graphs are mirror images across $y = \frac{1}{2}$, so their slopes are always opposite.

## Chain rule variants

$$\frac{d}{dx}\cos^2 u = -2\cos u\sin u\cdot u' = -\sin(2u)\cdot u'$$

| Function | Derivative | Simplified |
| --- | --- | --- |
| $\cos^2(4x)$ | $-8\cos 4x\sin 4x$ | $-4\sin 8x$ |
| $\cos^2(x^2)$ | $-4x\cos(x^2)\sin(x^2)$ | $-2x\sin(2x^2)$ |
| $\cos^2(\pi x)$ | $-2\pi\cos(\pi x)\sin(\pi x)$ | $-\pi\sin(2\pi x)$ |

## Worked examples

**Example 1: a value.** At $x = \frac{\pi}{6}$, the slope is $-\sin\frac{\pi}{3} = -\frac{\sqrt{3}}{2} \approx -0.866$.

**Example 2: a higher power.** $\frac{d}{dx}\cos^3 x = 3\cos^2 x\cdot(-\sin x) = -3\cos^2 x\sin x$.

**Example 3: a product.** $\frac{d}{dx}\left(x\cos^2 x\right) = \cos^2 x - x\sin 2x$.

**Example 4: where is it flat?** $-\sin 2x = 0$ when $x = \frac{k\pi}{2}$. At even multiples of $\frac{\pi}{2}$, $\cos^2 x = 1$ (peaks). At odd multiples, $\cos^2 x = 0$ (valleys).

## Where you will meet it

The function $\cos^2 x$ shows up in physics and engineering whenever something is squared after a cosine wave. The power of an alternating current is proportional to $\cos^2(\omega t)$, and the intensity of polarized light through a filter follows $\cos^2\theta$. In each case the derivative tells you how fast that quantity changes. Because the answer simplifies to $-\sin 2x$, the rate of change is itself a simple wave, twice as fast as the original cosine. Writing the answer in that short form also makes it easier to find where the slope is zero, positive or negative.

## Common mistakes

1. **Losing the minus sign.** The inner derivative $-\sin x$ carries it.
2. **Answering $2\cos x$ or $-\sin^2 x$.** Neither applies the chain rule correctly.
3. **Confusing $\cos^2 x$ with $\cos(x^2)$.** $\frac{d}{dx}\cos(x^2) = -2x\sin(x^2)$.

The integral of $\cos^2 x$ uses the same identity, see the [integral of cos²x](/integrals/cos-squared-x).

## Frequently asked questions

### What is the derivative of cos²x?

It is $-2\sin x\cos x$, which simplifies to $-\sin 2x$.

### What is the derivative of cos²(2x)?

By the chain rule, $2\cos(2x)\cdot(-\sin(2x))\cdot 2 = -4\sin(2x)\cos(2x) = -2\sin(4x)$.

### Why is there a minus sign?

The inner derivative is $\frac{d}{dx}\cos x = -\sin x$. The minus comes from there.

### What is the derivative of cos²x - sin²x?

Since $\cos^2 x - \sin^2 x = \cos 2x$, the derivative is $-2\sin 2x$. Term by term, $-\sin 2x - \sin 2x$ gives the same.