Let $$y = f x = \sin^3\dfrac{\pi}{3}\cos\dfrac{\pi}{3\sqrt{2}} - 4x^3 + 5x^2 + 1^{\frac{3}{2}}$$. Then, at $$x = 1$$,

A

$$2y' + \sqrt{3}\pi^2 y = 0$$

B

$$2y' + 3\pi^2 y = 0$$

C

$$\sqrt{2}y' - 3\pi^2 y = 0$$

D

$$y' + 3\pi^2 y = 0$$