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Câu hỏi: Tính đạo hàm của hàm số $y=\dfrac{x+1}{\ln \text{x}},\left( x>0;x\ne 1 \right)$.
A. ${y}'=\dfrac{\ln \text{x}-x-1}{x{{\left( \ln \text{x} \right)}^{2}}}$
B. ${y}'=\dfrac{x\ln \text{x}-x-1}{x{{\left( \ln \text{x} \right)}^{2}}}$
C. ${y}'=\dfrac{\ln \text{x}-x-1}{{{\left( \ln \text{x} \right)}^{2}}}$
D. ${y}'=\dfrac{\ln \text{x}-x-1}{x\ln \text{x}}$
A. ${y}'=\dfrac{\ln \text{x}-x-1}{x{{\left( \ln \text{x} \right)}^{2}}}$
B. ${y}'=\dfrac{x\ln \text{x}-x-1}{x{{\left( \ln \text{x} \right)}^{2}}}$
C. ${y}'=\dfrac{\ln \text{x}-x-1}{{{\left( \ln \text{x} \right)}^{2}}}$
D. ${y}'=\dfrac{\ln \text{x}-x-1}{x\ln \text{x}}$
${y}'=\dfrac{{{\left( x+1 \right)}^{\prime }}\ln \text{x}-{{\left( \ln \text{x} \right)}^{\prime }}\left( x+1 \right)}{{{\left( \ln \text{x} \right)}^{2}}}=\dfrac{\ln \text{x}-\dfrac{1}{x}\left( x+1 \right)}{{{\left( \ln \text{x} \right)}^{2}}}=\dfrac{x\ln \text{x}-x-1}{x{{\left( \ln \text{x} \right)}^{2}}}$
Đáp án B.