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Câu hỏi: Hàm số $y={{e}^{x}}.\sin 2x$ có đạo hàm là
A. ${y}'={{e}^{x}}.\cos 2x$.
B. ${y}'={{e}^{x}}.\left( \sin 2x-\cos 2x \right)$.
C. ${y}'={{e}^{x}}.\left( \sin 2x+\cos 2x \right)$.
D. ${y}'={{e}^{x}}.\left( \sin 2x+2\cos 2x \right)$.
A. ${y}'={{e}^{x}}.\cos 2x$.
B. ${y}'={{e}^{x}}.\left( \sin 2x-\cos 2x \right)$.
C. ${y}'={{e}^{x}}.\left( \sin 2x+\cos 2x \right)$.
D. ${y}'={{e}^{x}}.\left( \sin 2x+2\cos 2x \right)$.
Ta có: ${y}'={{\left( {{e}^{x}} \right)}^{\prime }}.\sin 2x+{{e}^{x}}.{{\left( \sin 2x \right)}^{\prime }}={{e}^{x}}.\sin 2x+2.{{e}^{x}}.\cos 2x={{e}^{x}}.\left( \sin 2x+2\cos 2x \right)$.
Đáp án D.