Find y' for y = sin² x
Find $y'$ for $y = \sin^{2} x$.
Explanation of Method
Differentiate a squared sine using the product rule and a double-angle identity.
Step 1: Apply the product rule
Write the square as a product, $y = \sin x\cdot\sin x$, and use $y' = u'v + uv'$ with $u = v = \sin x$, $u' = v' = \cos x$: $y' = \cos x\cdot\sin x + \sin x\cdot\cos x = 2\sin x\cos x$.
Step 2: Apply the double-angle identity
Using $\sin 2x = 2\sin x\cos x$: $y' = \sin 2x$.
Step 3: Verification
An independent route uses the power-reduction identity $\sin^{2}x = \dfrac{1-\cos 2x}{2}$: \[y' = \frac{d}{dx}\!\left(\frac{1}{2} - \frac{\cos 2x}{2}\right) = -\frac{1}{2}\cdot(-\sin 2x)\cdot 2 = \sin 2x \;\;\checkmark\]
\[x = 0:\quad y'(0) = \sin 0 = 0, \text{ matching the minimum } y(0) = 0 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = 2\sin x\cos x = \sin 2x\,}\]
Remark
Remark. The derivative has period $\pi$, not $2\pi$, because $\sin^{2}x$ itself has period $\pi$. Note also that $\bigl(\cos^{2}x\bigr)' = -\sin 2x$, so the two derivatives cancel — exactly as they must, since $\sin^{2}x + \cos^{2}x \equiv 1$ is constant.