Ta có
$y = (\sqrt{x} + 1) \left( \dfrac{1}{\sqrt{x}} - 1 \right)$
$= 1 - \sqrt{x} + \dfrac{1}{\sqrt{x}} - 1$
$= x^{-\frac{1}{2}} - x^{\frac{1}{2}}$
Suy ra
$y' = -\dfrac{1}{2} x^{-\frac{3}{2}} - \dfrac{1}{2} x^{-\frac{1}{2}}$
$= -\dfrac{1}{2} \dfrac{1}{x\sqrt{x}} - \dfrac{1}{2} \dfrac{1}{\sqrt{x}}$
$= -\dfrac{1}{2x\sqrt{x}} - \dfrac{1}{2\sqrt{x}}$
Vậy
$y' = -\dfrac{1}{2x\sqrt{x}} - \dfrac{1}{2\sqrt{x}}$
