2 câu trả lời
\[\begin{array}{l} y = \frac{{\sqrt x }}{x}\,\,\,\,\left( {x \ge 0} \right)\\ \Rightarrow y = \frac{1}{{\sqrt x }}\\ \Rightarrow y' = \left( {\frac{1}{{\sqrt x }}} \right)' = \frac{{ - \left( {\sqrt x } \right)'}}{x} = \frac{{ - \frac{1}{{2\sqrt x }}}}{x} = - \frac{1}{{2x\sqrt x }}. \end{array}\]
$y=\dfrac{\sqrt{x}}{x}=\dfrac{1}{\sqrt{x}}$
$\to y'=\dfrac{-(\sqrt{x})'}{(\sqrt{x})^2}=\dfrac{-1}{2x\sqrt{x}}$
