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Trigonometric derivatives

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.

Trigonometric derivatives Updated 2 min read

At a glance

\frac{d}{dx}\,\sin^2 x = 2\sin x\cos x = \sin 2x
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
On this page
  1. Graph of sin²x and its derivative
  2. Proof with the chain rule
  3. A check with a double angle identity
  4. A check with the product rule
  5. Chain rule variants
  6. Worked examples
  7. Why the answer has period π
  8. Common mistakes

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 of y = sin²x and its derivative y = sin 2x
\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.

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 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.

A check with the product rule

Write \sin^2 x = \sin x\cdot\sin x and use the 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): 2x\cos(x^2).

The cosine version is on the derivative of cos²x page, and the result \sin 2x has its own derivative of sin 2x page.

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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.

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