# Derivative of sin 2x: Formula, Proof and Examples

> The derivative of sin 2x is 2cos 2x. See the chain rule proof, a check with 2 sin x cos x, the general sin(ax) rule and worked examples.

Source: https://practicecalculus.com/derivatives/sin-2x
Updated: 2026-11-06

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

## At a glance

- **Derivative:** $2\cos 2x$
- **General rule:** $\dfrac{d}{dx}\sin(ax) = a\cos(ax)$
- **Second derivative:** $-4\sin 2x$
- **Period:** $\pi$, half the period of $\sin x$
- **Integral of sin 2x:** $\int\sin 2x\,dx = -\frac{1}{2}\cos 2x + C$

## Key takeaways

- The derivative of $\sin 2x$ is $2\cos 2x$. The 2 inside comes out front.
- More generally, $\frac{d}{dx}\sin(ax) = a\cos(ax)$.
- The inside $2x$ stays the same in the answer. Do not change it to $x$.
- Rewriting $\sin 2x = 2\sin x\cos x$ and using the product rule gives the same result.

The **derivative of $\sin 2x$ is $2\cos 2x$**. The chain rule pulls the inner factor $2$ out front, and the inside $2x$ stays the same.

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

## Graph of sin 2x and its derivative

```graph
title: 'Graph of y = sin 2x and its derivative y = 2cos 2x'
caption: '$\sin 2x$ oscillates twice as fast as $\sin x$, so its slopes are twice as steep. The derivative reaches $\pm 2$.'
x: [-3.3, 3.3]
y: [-2.5, 2.5]
functions:
  - expr: sin(2x)
    label: y = sin 2x
  - expr: 2cos(2x)
    label: y = 2cos 2x (derivative)
    dashed: true
points:
  - [0, 0, 'slope 2']
  - [0.7854, 1, 'peak, slope 0']
```

## Proof with the chain rule

Outer function $\sin u$, inner function $u = 2x$, so $u' = 2$. With the [chain rule](/rules/chain-rule) and the [derivative of sin x](/derivatives/sin-x):

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

## A check with the double angle formula

Rewrite $\sin 2x = 2\sin x\cos x$ and use the [product rule](/rules/product-rule):

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

The double angle formula $\cos^2 x - \sin^2 x = \cos 2x$ turns this into $2\cos 2x$. Two methods, one answer.

## Why the 2 comes out

Squeezing the graph horizontally by a factor of 2 makes it twice as steep everywhere. That is what the factor $2$ in front of $\cos 2x$ records. In general:

$$\frac{d}{dx}\sin(ax) = a\cos(ax), \qquad \frac{d}{dx}\sin(ax + b) = a\cos(ax + b)$$

| Function | Derivative |
| --- | --- |
| $\sin 3x$ | $3\cos 3x$ |
| $\sin\frac{x}{2}$ | $\frac{1}{2}\cos\frac{x}{2}$ |
| $\sin(2x + 1)$ | $2\cos(2x + 1)$ |
| $\sin(\pi x)$ | $\pi\cos(\pi x)$ |
| $5\sin 2x$ | $10\cos 2x$ |

## Worked examples

**Example 1: a value.** At $x = \frac{\pi}{6}$, the slope is $2\cos\frac{\pi}{3} = 2\cdot\frac{1}{2} = 1$.

**Example 2: a product.** $\frac{d}{dx}(x\sin 2x) = \sin 2x + 2x\cos 2x$.

**Example 3: higher derivatives.** Each derivative brings out another $2$: $\frac{d^2}{dx^2}\sin 2x = -4\sin 2x$, $\frac{d^3}{dx^3}\sin 2x = -8\cos 2x$, and $\frac{d^4}{dx^4}\sin 2x = 16\sin 2x$.

**Example 4: a link to sin²x.** The [derivative of sin²x](/derivatives/sin-squared-x) is $\sin 2x$. So $\sin^2 x$ is an antiderivative of $\sin 2x$, and so is $-\frac{1}{2}\cos 2x$. They differ by a constant, since $\sin^2 x = \frac{1}{2} - \frac{1}{2}\cos 2x$.

## Where you will meet it

The function $\sin 2x$ appears whenever a wave repeats twice as often as $\sin x$. It also shows up as a simplified answer, for example as the derivative of $\sin^2 x$, so knowing its own derivative saves time on second derivatives. In physics, $\sin 2\theta$ appears in the range of a projectile, and its derivative tells you how the range changes as you tilt the launch angle.

## Common mistakes

1. **Forgetting the 2.** $\frac{d}{dx}\sin 2x$ is not $\cos 2x$.
2. **Changing the inside.** The answer is $2\cos 2x$, not $2\cos x$.
3. **Dividing instead of multiplying.** $\frac{1}{2}$ shows up when you integrate, not when you differentiate. See the [integral of sin 2x](/integrals/sin-2x).
4. **Confusing $\sin 2x$ with $2\sin x$.** $\frac{d}{dx}(2\sin x) = 2\cos x$. The graphs are different.

For more chain rule drills, try the [chain rule practice problems](/practice-problems/chain-rule).

## Frequently asked questions

### What is the derivative of sin 2x?

It is $2\cos 2x$, by the chain rule.

### Why is the derivative of sin 2x not cos 2x?

Because of the chain rule. The inside $2x$ has derivative $2$, and you must multiply by it.

### What is the derivative of 2 sin x cos x?

It is $2\cos 2x$, since $2\sin x\cos x = \sin 2x$. The product rule gives $2(\cos^2 x - \sin^2 x)$, which is the same.

### What is the derivative of sin(2x + 1)?

It is $2\cos(2x + 1)$. The constant $1$ does not change the inner derivative.