Cho hàm số $y = \dfrac{1}{{x + 1 + \ln x}}$ với $x > 0$. Khi đó $- \dfrac{{y'}}{{{y^2}}}$ bằng:

$\left( {0;1} \right)$

$R$

$R\backslash \left\{ 0 \right\}$

$\left( {0; + \infty } \right)$

$y = {\log _2}\left( {x - 1} \right)$

$y = {\log _2}\left( {{x^2} - 1} \right)$

$y = {\log _2}\left( {{x^2} + 1} \right)$

$y = {\log _2}\left( {\sqrt x } \right)$

$y' = \dfrac{{\ln 2017}}{x}.$

$y' = \dfrac{{{{\log }_{2017}}e}}{x}.$

$y' = \dfrac{1}{{x.\log 2017}}.$

$y' = \dfrac{{2017}}{{x.\ln 2017}}.$

$y' = \dfrac{1}{{x\ln 2}}$.

$y' = \dfrac{1}{{x\ln 10}}$.

$y' = \dfrac{1}{{2x\ln 10}}$.

$y' = \dfrac{{\ln 10}}{x}$.

$y' = \dfrac{2}{{2x + 1}}.$

$y' = \dfrac{1}{{2x + 1}}.$

$y' = \dfrac{2}{{\left( {2x + 1} \right)\ln 2}}.$

$y' = \dfrac{1}{{\left( {2x + 1} \right)\ln 2}}.$

$10$

$9$

$3$

$37$

$y = {\log _{\frac{{\sqrt 2 }}{2}}}x$.

$y = {\log _{\frac{e}{3}}}x$.

$y = {\log _{\frac{e}{2}}}x$.

$y = {\log _{\frac{\pi }{4}}}x$.