$y=\dfrac{\sqrt[]{1-2x}}{x}$
$y'=\dfrac{(\sqrt[]{1-2x})'.x-(\sqrt[]{1-2x}).x'}{x^2}$
$=\dfrac{\dfrac{1}{2\sqrt[]{1-2x}}.(1-2x)'.x-(\sqrt[]{1-2x})}{x^2}$
$=(\dfrac{-2x}{2\sqrt[]{1-2x}}-\sqrt[]{1-2x}):x^2$
$=\dfrac{\dfrac{-x}{\sqrt[]{1-2x}}-\sqrt[]{1-2x}}{x^2}$
$=\dfrac{-x-(1-2x)}{\sqrt[]{1-2x}}.$ $\dfrac{1}{x^2}$
$=\dfrac{x-1}{x^2(\sqrt[]{1-2x})}$
BẠN THAM KHẢO.
