Toán Tính đạo hàm của hamf số Y= √x +1 ( 1/√x -1 ) 09/07/2021 By Allison Tính đạo hàm của hamf số Y= √x +1 ( 1/√x -1 )
$y=\sqrt{x+1}.\Big(\dfrac{1}{\sqrt{x}}-1\Big)$ $=\dfrac{ \sqrt{x+1} }{\sqrt{x}} – \sqrt{x+1}$ $y’= \dfrac{(\sqrt{x+1})’\sqrt{x}-\sqrt{x+1}(\sqrt{x})’ }{x} – \dfrac{1}{2\sqrt{x+1}}$ $=\dfrac{ \dfrac{1}{2\sqrt{x+1}}.\sqrt{x} – \sqrt{x+1}.\dfrac{1}{2\sqrt{x}} }{x} -\dfrac{1}{2\sqrt{x+1}}$ $=\dfrac{ \sqrt{x}.2\sqrt{x}-\sqrt{x+1}.2\sqrt{x+1} }{x.4\sqrt{x}.\sqrt{x+1}} -\dfrac{1}{2\sqrt{x+1}}$ $=\dfrac{ 2x-2x-2 }{4x\sqrt{x}\sqrt{x+1}} -\dfrac{1}{2\sqrt{x+1}}$ $=\dfrac{-1}{2x\sqrt{x}\sqrt{x+1}} -\dfrac{1}{2\sqrt{x+1}}$ $=\dfrac{-x\sqrt{x}-1 }{2\sqrt{x}\sqrt{x+1}}$ Trả lời
Giải thích các bước giải: $y=\sqrt{x+1}(\dfrac{1}{\sqrt{x}}-1)$ $\to y=\dfrac{\sqrt{x+1}}{\sqrt{x}}-\sqrt{x+1}$ $\to y=\sqrt{1+\dfrac{1}{x}}-\sqrt{x+1}$ $\to y=(1+\dfrac{1}{x})^{\frac 12}-\sqrt{x+1}$ $\to y’=\dfrac 12(1+\dfrac{1}{x})’.(1+\dfrac{1}{x})^{\frac 12-1}-\dfrac{1}{2\sqrt{x+1}} $ $\to y’=\dfrac{-1}{2x^2}.(1+\dfrac{1}{x})^{-\frac 12}-\dfrac{1}{2\sqrt{x+1}} $ Trả lời
$y=\sqrt{x+1}.\Big(\dfrac{1}{\sqrt{x}}-1\Big)$
$=\dfrac{ \sqrt{x+1} }{\sqrt{x}} – \sqrt{x+1}$
$y’= \dfrac{(\sqrt{x+1})’\sqrt{x}-\sqrt{x+1}(\sqrt{x})’ }{x} – \dfrac{1}{2\sqrt{x+1}}$
$=\dfrac{ \dfrac{1}{2\sqrt{x+1}}.\sqrt{x} – \sqrt{x+1}.\dfrac{1}{2\sqrt{x}} }{x} -\dfrac{1}{2\sqrt{x+1}}$
$=\dfrac{ \sqrt{x}.2\sqrt{x}-\sqrt{x+1}.2\sqrt{x+1} }{x.4\sqrt{x}.\sqrt{x+1}} -\dfrac{1}{2\sqrt{x+1}}$
$=\dfrac{ 2x-2x-2 }{4x\sqrt{x}\sqrt{x+1}} -\dfrac{1}{2\sqrt{x+1}}$
$=\dfrac{-1}{2x\sqrt{x}\sqrt{x+1}} -\dfrac{1}{2\sqrt{x+1}}$
$=\dfrac{-x\sqrt{x}-1 }{2\sqrt{x}\sqrt{x+1}}$
Giải thích các bước giải:
$y=\sqrt{x+1}(\dfrac{1}{\sqrt{x}}-1)$
$\to y=\dfrac{\sqrt{x+1}}{\sqrt{x}}-\sqrt{x+1}$
$\to y=\sqrt{1+\dfrac{1}{x}}-\sqrt{x+1}$
$\to y=(1+\dfrac{1}{x})^{\frac 12}-\sqrt{x+1}$
$\to y’=\dfrac 12(1+\dfrac{1}{x})’.(1+\dfrac{1}{x})^{\frac 12-1}-\dfrac{1}{2\sqrt{x+1}} $
$\to y’=\dfrac{-1}{2x^2}.(1+\dfrac{1}{x})^{-\frac 12}-\dfrac{1}{2\sqrt{x+1}} $