Derivative Sin Squared F90Fcf

1. **State the problem:** We need to find the derivative of $$y = \sin^2(3x)$$ with respect to $$x$$ and then evaluate it at $$x = \frac{\pi}{12}$$. 2. **Formula and rules:** Use the chain rule for derivatives. If $$y = [f(x)]^2$$, then $$\frac{dy}{dx} = 2f(x) \cdot f'(x)$$. 3. **Apply the chain rule:** Here, $$f(x) = \sin(3x)$$, so $$\frac{dy}{dx} = 2 \sin(3x) \cdot \frac{d}{dx}[\sin(3x)]$$. 4. **Differentiate inner function:** $$\frac{d}{dx}[\sin(3x)] = \cos(3x) \cdot \frac{d}{dx}[3x] = \cos(3x) \cdot 3 = 3\cos(3x)$$. 5. **Substitute back:** $$\frac{dy}{dx} = 2 \sin(3x) \cdot 3 \cos(3x) = 6 \sin(3x) \cos(3x)$$. 6. **Simplify using double-angle identity:** Recall $$\sin(2\theta) = 2 \sin \theta \cos \theta$$, so $$6 \sin(3x) \cos(3x) = 3 \cdot 2 \sin(3x) \cos(3x) = 3 \sin(6x)$$. 7. **Evaluate at $$x = \frac{\pi}{12}$$:** $$\frac{dy}{dx} \bigg|_{x=\frac{\pi}{12}} = 3 \sin\left(6 \cdot \frac{\pi}{12}\right) = 3 \sin\left(\frac{6\pi}{12}\right) = 3 \sin\left(\frac{\pi}{2}\right)$$. 8. **Calculate final value:** $$\sin\left(\frac{\pi}{2}\right) = 1$$, so $$\frac{dy}{dx} \bigg|_{x=\frac{\pi}{12}} = 3 \times 1 = 3$$. **Final answer:** $$\boxed{3}$$