tính đạo hàm của hàm số sau: y=($\sqrt{x}$ +1)($\frac{1}{\sqrt{x}}$ -1) 28/10/2021 Bởi Reese tính đạo hàm của hàm số sau: y=($\sqrt{x}$ +1)($\frac{1}{\sqrt{x}}$ -1)
$y=(\sqrt{x}+1)(\dfrac{1}{\sqrt{x}-1})$ $=1-\sqrt{x}+\dfrac{1}{\sqrt{x}}-1$ $=\dfrac{1}{\sqrt{x}}-\sqrt{x}$ $\to y’=(\dfrac{1}{\sqrt{x}})’-(\sqrt{x})’$ $=\dfrac{-(\sqrt{x})’}{x}-\dfrac{1}{2\sqrt{x}}$ $=\dfrac{-1}{2x\sqrt{x}}-\dfrac{1}{2\sqrt{x}}$ $=\dfrac{-1-x}{2x\sqrt{x}}$ Bình luận
Đáp án: $y’=\dfrac{-x-1}{2x\sqrt{x}}$ Giải thích các bước giải: Ta có:$y=(\sqrt{x}+1)(\dfrac{1}{\sqrt{x}}-1)$ $\to y=(\sqrt{x}+1)\cdot \dfrac{1-\sqrt{x}}{\sqrt{x}}$ $\to y= \dfrac{(1+\sqrt{x})(1-\sqrt{x})}{\sqrt{x}}$ $\to y= \dfrac{1-x}{\sqrt{x}}$ $\to y’= (\dfrac{1-x}{\sqrt{x}})’$ $\to y’=\dfrac{\left(1-x\right)’\sqrt{x}-\left(\sqrt{x}\right)’\left(1-x\right)}{\left(\sqrt{x}\right)^2}$ $\to y’=\dfrac{\left(-1\right)\sqrt{x}-\dfrac{1}{2\sqrt{x}}\left(1-x\right)}{\left(\sqrt{x}\right)^2}$ $\to y’=\dfrac{-x-1}{2x\sqrt{x}}$ Bình luận
$y=(\sqrt{x}+1)(\dfrac{1}{\sqrt{x}-1})$
$=1-\sqrt{x}+\dfrac{1}{\sqrt{x}}-1$
$=\dfrac{1}{\sqrt{x}}-\sqrt{x}$
$\to y’=(\dfrac{1}{\sqrt{x}})’-(\sqrt{x})’$
$=\dfrac{-(\sqrt{x})’}{x}-\dfrac{1}{2\sqrt{x}}$
$=\dfrac{-1}{2x\sqrt{x}}-\dfrac{1}{2\sqrt{x}}$
$=\dfrac{-1-x}{2x\sqrt{x}}$
Đáp án: $y’=\dfrac{-x-1}{2x\sqrt{x}}$
Giải thích các bước giải:
Ta có:
$y=(\sqrt{x}+1)(\dfrac{1}{\sqrt{x}}-1)$
$\to y=(\sqrt{x}+1)\cdot \dfrac{1-\sqrt{x}}{\sqrt{x}}$
$\to y= \dfrac{(1+\sqrt{x})(1-\sqrt{x})}{\sqrt{x}}$
$\to y= \dfrac{1-x}{\sqrt{x}}$
$\to y’= (\dfrac{1-x}{\sqrt{x}})’$
$\to y’=\dfrac{\left(1-x\right)’\sqrt{x}-\left(\sqrt{x}\right)’\left(1-x\right)}{\left(\sqrt{x}\right)^2}$
$\to y’=\dfrac{\left(-1\right)\sqrt{x}-\dfrac{1}{2\sqrt{x}}\left(1-x\right)}{\left(\sqrt{x}\right)^2}$
$\to y’=\dfrac{-x-1}{2x\sqrt{x}}$