Derivative of sin²(x)
The derivative of $\sin^2 x = (\sin x)^2$ is $\dfrac{d}{dx}\sin^2 x = 2\sin x\cos x = \sin(2x)$. It is a good example of how three different methods lead to the same result.
See the derivative of sin²(x) step by step in the calculator →
📈 Graph of sin²(x) and its derivative
🔍 Why is the derivative of sin²(x) equal to $\sin(2x)$?
1. Chain rule: outer function $u^2$, inner function $\sin x$: $\;2\sin x\cdot\cos x$.
2. Product rule: $(\sin x\cdot\sin x)' = \cos x\sin x + \sin x\cos x = 2\sin x\cos x$.
3. Identity: $\sin^2 x = \frac{1-\cos 2x}{2}$, so the derivative is $\frac{0 + 2\sin 2x}{2} = \sin 2x$.
🔗 Derivative of sin²(x) with an inner function
If the argument is a function $u(x)$, the chain rule adds the inner derivative:
✏️ Worked examples
| Function | Derivative | Note | |
|---|---|---|---|
| $\sin^2(3x)$ | $3\sin(6x)$ | Inner derivative $3$, then the double-angle formula. | steps → |
| $\cos^2 x$ | $-\sin(2x)$ | The mirror image: $2\cos x\cdot(-\sin x)$. | steps → |
| $\sin^2 x + \cos^2 x$ | $0$ | The sum is the constant $1$. | steps → |
| $\sin(x^2)$ | $2x\cos(x^2)$ | Not the same function! Here the argument is squared. | steps → |
⚠️ Common mistakes
Mixing up $\sin^2 x$ and $\sin(x^2)$ – they have different derivatives.
Writing $2\sin x$ and forgetting the inner derivative $\cos x$.
❓ Frequently asked questions
What is the derivative of sin^2(x)?
$2\sin x\cos x$, which equals $\sin(2x)$.
What is the derivative of cos^2(x)?
$-2\cos x\sin x = -\sin(2x)$.
Why do sin²(x) and −cos²(x) have the same derivative?
Because $\sin^2 x = 1 - \cos^2 x$ – they differ only by the constant $1$.