2 câu trả lời
\[\begin{array}{l} y = \sqrt {{x^2} + 1} \\ \Rightarrow y' = \left( {\sqrt {{x^2} + 1} } \right)' = \frac{{2x}}{{2\sqrt {{x^2} + 1} }} = \frac{x}{{\sqrt {{x^2} + 1} }}. \end{array}\]
\[\begin{array}{l} y = \sqrt {{x^2} + 1} \\ \Rightarrow y' = \left( {\sqrt {{x^2} + 1} } \right)' = \frac{{2x}}{{2\sqrt {{x^2} + 1} }} = \frac{x}{{\sqrt {{x^2} + 1} }}. \end{array}\]
