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