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Câu hỏi: Tính đạo hàm của hàm số $y=\ln \dfrac{x-1}{x+2}$.
A. ${y}'=\dfrac{-3}{\left( x-1 \right)\left( x+2 \right)}.$
B. ${y}'=\dfrac{-3}{\left( x-1 \right){{\left( x+2 \right)}^{2}}}.$
C. ${y}'=\dfrac{3}{\left( x-1 \right)\left( x+2 \right)}.$
D. ${y}'=\dfrac{3}{\left( x-1 \right){{\left( x+2 \right)}^{2}}}.$
A. ${y}'=\dfrac{-3}{\left( x-1 \right)\left( x+2 \right)}.$
B. ${y}'=\dfrac{-3}{\left( x-1 \right){{\left( x+2 \right)}^{2}}}.$
C. ${y}'=\dfrac{3}{\left( x-1 \right)\left( x+2 \right)}.$
D. ${y}'=\dfrac{3}{\left( x-1 \right){{\left( x+2 \right)}^{2}}}.$
Ta có ${y}'={{\left( \ln \dfrac{x-1}{x+2} \right)}^{\prime }}={{\left( \ln (x-1)-\ln (x+2) \right)}^{\prime }}=\dfrac{1}{x-1}-\dfrac{1}{x+2}=\dfrac{3}{\left( x-1 \right)\left( x+2 \right)}$.
Đáp án C.