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Câu hỏi: Tính đạo hàm của hàm số $y=\ln \left( 1+\sqrt{2x+1} \right).$
A. ${y}'=\dfrac{1}{2x+1+\sqrt{2x+1}}.$
B. ${y}'=\dfrac{2}{2x+1+\sqrt{2x+1}}.$
C. ${y}'=\dfrac{\sqrt{2x+1}}{2x+1+\sqrt{2x+1}}.$
D. ${y}'=\dfrac{2\sqrt{2x+1}}{2x+1+\sqrt{2x+1}}.$
A. ${y}'=\dfrac{1}{2x+1+\sqrt{2x+1}}.$
B. ${y}'=\dfrac{2}{2x+1+\sqrt{2x+1}}.$
C. ${y}'=\dfrac{\sqrt{2x+1}}{2x+1+\sqrt{2x+1}}.$
D. ${y}'=\dfrac{2\sqrt{2x+1}}{2x+1+\sqrt{2x+1}}.$
Ta có ${y}'=\dfrac{1}{1+\sqrt{2x+1}}.{{\left( 1+\sqrt{2x+1} \right)}^{\prime }}=\dfrac{1}{1+\sqrt{2x+1}}.\dfrac{2}{2\sqrt{2x+1}}=\dfrac{1}{2x+1+\sqrt{2x+1}}.$
Đáp án A.