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Câu hỏi: Hàm số $y=2\sqrt{\sin x}-2\sqrt{\cos x}$ có đạo hàm là:
A. $y'=\dfrac{1}{\sqrt{\sin x}}-\dfrac{1}{\sqrt{\cos x}}$
B. $y'=\dfrac{1}{\sqrt{\sin x}}+\dfrac{1}{\sqrt{\cos x}}$
C. $y'=\dfrac{\cos x}{\sqrt{\sin x}}-\dfrac{\sin x}{\sqrt{\cos x}}$
D. $y'=\dfrac{\cos x}{\sqrt{\sin x}}+\dfrac{\sin x}{\sqrt{\cos x}}$
A. $y'=\dfrac{1}{\sqrt{\sin x}}-\dfrac{1}{\sqrt{\cos x}}$
B. $y'=\dfrac{1}{\sqrt{\sin x}}+\dfrac{1}{\sqrt{\cos x}}$
C. $y'=\dfrac{\cos x}{\sqrt{\sin x}}-\dfrac{\sin x}{\sqrt{\cos x}}$
D. $y'=\dfrac{\cos x}{\sqrt{\sin x}}+\dfrac{\sin x}{\sqrt{\cos x}}$
$y'=2{{\left( \sqrt{\sin x} \right)}^{\prime }}-2{{\left( \sqrt{\cos x} \right)}^{\prime }}=2.\dfrac{{{\left( \sin x \right)}^{\prime }}}{2\sqrt{\sin x}}-2.\dfrac{{{\left( \cos x \right)}^{\prime }}}{2\sqrt{\cos x}}=\dfrac{\cos x}{\sqrt{\sin x}}+\dfrac{\sin x}{\sqrt{\cos x}}$
Đáp án D.