Đáp án: $y'=\dfrac{-x-1}{2x\sqrt{x}}$
Giải thích các bước giải:
Ta có:
$y=(\sqrt{x}+1)(\dfrac{1}{\sqrt{x}}-1)$
$\to y=(\sqrt{x}+1)\cdot \dfrac{1-\sqrt{x}}{\sqrt{x}}$
$\to y= \dfrac{(1+\sqrt{x})(1-\sqrt{x})}{\sqrt{x}}$
$\to y= \dfrac{1-x}{\sqrt{x}}$
$\to y'= (\dfrac{1-x}{\sqrt{x}})'$
$\to y'=\dfrac{\left(1-x\right)'\sqrt{x}-\left(\sqrt{x}\right)'\left(1-x\right)}{\left(\sqrt{x}\right)^2}$
$\to y'=\dfrac{\left(-1\right)\sqrt{x}-\dfrac{1}{2\sqrt{x}}\left(1-x\right)}{\left(\sqrt{x}\right)^2}$
$\to y'=\dfrac{-x-1}{2x\sqrt{x}}$
