# Derivative of sin²x: Formula, Proof and Examples

> The derivative of sin²x is 2 sin x cos x, which equals sin 2x. See the chain rule proof, a check with a double angle identity and examples.

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

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

## At a glance

- **Derivative:** $2\sin x\cos x$
- **Simplified:** $\sin 2x$
- **Chain rule form:** $\dfrac{d}{dx}\sin^2 u = 2\sin u\cos u\cdot u'$
- **Not the same as:** $\sin(x^2)$, whose derivative is $2x\cos(x^2)$
- **Integral of sin²x:** $\int\sin^2 x\,dx = \frac{x}{2} - \frac{\sin 2x}{4} + C$

## Key takeaways

- The derivative of $\sin^2 x$ is $2\sin x\cos x$, which simplifies to $\sin 2x$.
- Read $\sin^2 x$ as $(\sin x)^2$: the square is the outside function, $\sin x$ is the inside.
- The identity $\sin^2 x = \frac{1 - \cos 2x}{2}$ gives the same answer, a useful check.
- $\sin^2 x$ and $\sin(x^2)$ are different functions with different derivatives.

The **derivative of $\sin^2 x$ is $2\sin x\cos x$**, which the double angle formula shortens to $\sin 2x$.

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

## Graph of sin²x and its derivative

```graph
title: 'Graph of y = sin²x and its derivative y = sin 2x'
caption: '$\sin^2 x$ wiggles between $0$ and $1$ with period $\pi$. Its slope is $0$ at every peak and every touch of the axis, which is where $\sin 2x = 0$.'
x: [-0.5, 6.5]
y: [-1.5, 1.6]
functions:
  - expr: sin(x)^2
    label: y = sin²x
  - expr: sin(2x)
    label: y = sin 2x (derivative)
    dashed: true
points:
  - [0.7854, 0.5, 'steepest, slope 1']
  - [1.5708, 1, 'peak, slope 0']
```

## Proof with the chain rule

The notation $\sin^2 x$ means $(\sin x)^2$. Outside is the square, inside is $\sin x$. The [chain rule](/rules/chain-rule) gives

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

The double angle identity $\sin 2x = 2\sin x\cos x$ tidies it up to $\sin 2x$.

## A check with a double angle identity

Rewrite first with $\sin^2 x = \frac{1 - \cos 2x}{2}$:

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

Same answer, no chain rule on a square needed. This power-reducing identity is also how you integrate it, see the [integral of sin²x](/integrals/sin-squared-x).

## A check with the product rule

Write $\sin^2 x = \sin x\cdot\sin x$ and use the [product rule](/rules/product-rule): $\cos x\sin x + \sin x\cos x = 2\sin x\cos x$.

## Chain rule variants

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

| Function | Derivative | Simplified |
| --- | --- | --- |
| $\sin^2(3x)$ | $6\sin 3x\cos 3x$ | $3\sin 6x$ |
| $\sin^2(x^2)$ | $4x\sin(x^2)\cos(x^2)$ | $2x\sin(2x^2)$ |
| $\sin^2\left(\frac{x}{2}\right)$ | $\sin\frac{x}{2}\cos\frac{x}{2}$ | $\frac{1}{2}\sin x$ |

## Worked examples

**Example 1: a higher power.** $\frac{d}{dx}\sin^3 x = 3\sin^2 x\cos x$.

**Example 2: a value.** At $x = \frac{\pi}{4}$, the slope is $\sin\frac{\pi}{2} = 1$. That is the steepest point of $\sin^2 x$.

**Example 3: maximum slope.** The derivative $\sin 2x$ is at most $1$, so $\sin^2 x$ never climbs faster than slope $1$, the same top speed as $\sin x$.

**Example 4: second derivative.** $\frac{d}{dx}\sin 2x = 2\cos 2x$, so $\sin^2 x$ is concave up where $\cos 2x > 0$, for example on $\left(-\frac{\pi}{4}, \frac{\pi}{4}\right)$.

## Why the answer has period π

The function $\sin^2 x$ never goes negative, so its graph repeats every $\pi$ instead of every $2\pi$. The derivative $\sin 2x$ also has period $\pi$, which is a good consistency check: a function and its derivative should repeat together. If your answer had period $2\pi$, such as $2\cos x$, something went wrong. Matching the period is a quick way to catch errors without redoing the algebra.

## Common mistakes

1. **Answering $\cos^2 x$.** Squaring the derivative is not the chain rule.
2. **Answering $2\cos x$.** The square's derivative is $2\sin x$, and then you multiply by $\cos x$.
3. **Mixing up $\sin^2 x$ and $\sin(x^2)$.** See the [derivative of sin(x²)](/derivatives/sin-x-squared): $2x\cos(x^2)$.

The cosine version is on the [derivative of cos²x](/derivatives/cos-squared-x) page, and the result $\sin 2x$ has its own [derivative of sin 2x](/derivatives/sin-2x) page.

## Frequently asked questions

### What is the derivative of sin²x?

It is $2\sin x\cos x$, which is the same as $\sin 2x$.

### What is the derivative of sin²(3x)?

By the chain rule twice, $2\sin(3x)\cdot\cos(3x)\cdot 3 = 6\sin(3x)\cos(3x) = 3\sin(6x)$.

### Is the derivative of sin²x equal to 2 cos x?

No. That skips the chain rule. The square gives $2\sin x$, then you multiply by the derivative of $\sin x$, which is $\cos x$.

### Why is the derivative of sin²x plus the derivative of cos²x zero?

Because $\sin^2 x + \cos^2 x = 1$, a constant. Indeed $\sin 2x + (-\sin 2x) = 0$.