giúp mik vs m.n ơi tính đạo hàm y=(√x +1)(1/√x -1) 16/11/2021 Bởi Nevaeh giúp mik vs m.n ơi tính đạo hàm y=(√x +1)(1/√x -1)
$y=\Big(\sqrt{x}+1\Big)\Big(\dfrac{1}{\sqrt{x}}-1\Big)$ $=1-\sqrt{x}+\dfrac{1}{\sqrt{x}}-1$ $=\dfrac{1}{\sqrt{x}}-\sqrt{x}$ $y’=\dfrac{-(\sqrt{x})’}{x}-\dfrac{1}{2\sqrt{x}}$ $=\dfrac{-1}{2x\sqrt{x}}-\dfrac{1}{2\sqrt{x}}$ $=\dfrac{-x-1}{2x\sqrt{x}}$ Bình luận
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 $y’ = -\dfrac{1}{2x\sqrt{x}} – \dfrac{1}{2\sqrt{x}}$ Bình luận
$y=\Big(\sqrt{x}+1\Big)\Big(\dfrac{1}{\sqrt{x}}-1\Big)$
$=1-\sqrt{x}+\dfrac{1}{\sqrt{x}}-1$
$=\dfrac{1}{\sqrt{x}}-\sqrt{x}$
$y’=\dfrac{-(\sqrt{x})’}{x}-\dfrac{1}{2\sqrt{x}}$
$=\dfrac{-1}{2x\sqrt{x}}-\dfrac{1}{2\sqrt{x}}$
$=\dfrac{-x-1}{2x\sqrt{x}}$
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
$y’ = -\dfrac{1}{2x\sqrt{x}} – \dfrac{1}{2\sqrt{x}}$