Câu hỏi 42

Ta có: \(y' = 1 + \dfrac{{2x}}{{2\sqrt {{x^2} + 1} }} = 1 + \dfrac{x}{{\sqrt {{x^2} + 1} }} = \dfrac{{\sqrt {{x^2} + 1}  + x}}{{\sqrt {{x^2} + 1} }}\).

\(\begin{array}{l} \Rightarrow \dfrac{{y'}}{y} = \dfrac{{\sqrt {{x^2} + 1}  + x}}{{\sqrt {{x^2} + 1} }}:\left( {x + \sqrt {{x^2} + 1} } \right)\\\,\,\,\,\,\,\,\,\,\,\,\, = \dfrac{{\sqrt {{x^2} + 1}  + x}}{{\sqrt {{x^2} + 1} }}.\dfrac{1}{{x + \sqrt {{x^2} + 1} }} = \dfrac{1}{{\sqrt {{x^2} + 1} }}\end{array}\)

Chọn A.