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Câu hỏi: Đạo hàm của hàm số $y=\log \left( 1-\sqrt{x+1} \right)$ là
A. ${y}'=-\dfrac{1}{2\sqrt{x+1}\left( 1-\sqrt{x+1} \right)\ln 10}.$
B. ${y}'=\dfrac{1}{\left( 1-\sqrt{x+1} \right)\ln 10}.$
C. ${y}'=\dfrac{\ln 10}{2\sqrt{x+1}\left( 1-\sqrt{x+1} \right)}.$
D. ${y}'=-\dfrac{1}{2\sqrt{x+1}\left( 1-\sqrt{x+1} \right)}.$
A. ${y}'=-\dfrac{1}{2\sqrt{x+1}\left( 1-\sqrt{x+1} \right)\ln 10}.$
B. ${y}'=\dfrac{1}{\left( 1-\sqrt{x+1} \right)\ln 10}.$
C. ${y}'=\dfrac{\ln 10}{2\sqrt{x+1}\left( 1-\sqrt{x+1} \right)}.$
D. ${y}'=-\dfrac{1}{2\sqrt{x+1}\left( 1-\sqrt{x+1} \right)}.$
Ta có ${y}'=\dfrac{{{\left( 1-\sqrt{x+1} \right)}^{\prime }}}{\left( 1-\sqrt{x+1} \right)\ln 10}=-\dfrac{1}{2\sqrt{x+1}\left( 1-\sqrt{x+1} \right)\ln 10}.$
Đáp án A.