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Cho hàm số \(y = \sin 2x\). Tính \({y^{(n)}}\)

A.

\({y^{(n)}} = {2^n}\sin (2x + n\frac{\pi }{3})\)

B.

\({y^{(n)}} = {2^n}\sin (2x + \frac{\pi }{2})\)

C.

\({y^{(n)}} = {2^n}\sin (x + \frac{\pi }{2})\)

D.

\({y^{(n)}} = {2^n}\sin (2x + n\frac{\pi }{2})\)

Trả lời:

Đáp án đúng: D


Ta có \(y = \sin 2x\) \(y' = 2\cos 2x = 2\sin (2x + \frac{\pi}{2})\) \(y'' = -4 \sin 2x = 2^2 \sin (2x + 2.\frac{\pi}{2})\) \(y''' = -8 \cos 2x = 2^3 \sin (2x + 3.\frac{\pi}{2})\) Suy ra \({y^{(n)}} = {2^n}\sin (2x + n\frac{\pi }{2})\).

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