2 câu trả lời
$y=\dfrac{1}{\sqrt{x}}$
$y'=\dfrac{-(\sqrt{x})'}{x}$
$=\dfrac{-1}{2x\sqrt{x}}$
$=\dfrac{-1}{2\sqrt{x^3}}$
$$\eqalign{ & y = {1 \over {\sqrt x }} = {1 \over {{x^{{1 \over 2}}}}} = {x^{ - {1 \over 2}}} \cr & y' = - {1 \over 2}{x^{ - {1 \over 2} - 1}} = - {1 \over 2}{x^{ - {3 \over 2}}} \cr & = - {1 \over 2}.{1 \over {{x^{{3 \over 2}}}}} = - {1 \over 2}.{1 \over {\sqrt {{x^3}} }} = - {1 \over {2\sqrt {{x^3}} }} \cr} $$
