Giải thích các bước giải:
1.Ta có: $y'=3x^2-4x+3$
2.Ta có: $y'=4x^3+\dfrac{1}{\sqrt{x}}$
3.Ta có:
$y=\sqrt{x}-2x$
$\to y=x^{\dfrac12}-2x$
$\to y'=\dfrac12x^{\dfrac12-1}-2$
$\to y'=\dfrac12x^{-\dfrac12}-2$
4.Ta có:
$y=\dfrac{2x-1}{x+2}$
$\to y'=\dfrac{(2x-1)'\cdot (x+2)-(2x-1)\cdot (x+2)'}{(x+2)^2}$
$\to y'=\dfrac{2\cdot (x+2)-(2x-1)\cdot 1}{(x+2)^2}$
$\to y'=\dfrac{5}{(x+2)^2}$
5.Ta có:
$y=\dfrac{x^2+3x-1}{x+1}$
$\to y'=\dfrac{\left(x^2+3x-1\right)'\left(x+1\right)-\left(x+1\right)'\left(x^2+3x-1\right)}{\left(x+1\right)^2}$
$\to y'=\dfrac{\left(2x+3\right)\left(x+1\right)-1\cdot \left(x^2+3x-1\right)}{\left(x+1\right)^2}$
$\to y'=\dfrac{x^2+2x+4}{\left(x+1\right)^2}$
6.Ta có:
$y=\dfrac{x^2}{x-1}$
$\to y'=\dfrac{\left(x^2\right)'\left(x-1\right)-\left(x-1\right)'\:x^2}{\left(x-1\right)^2}$
$\to y'=\dfrac{2x\left(x-1\right)-1\cdot \:x^2}{\left(x-1\right)^2}$
$\to y'=\dfrac{x^2-2x}{\left(x-1\right)^2}$
7.Ta có:
$y=\dfrac{x^2+2x+2}{x+1}$
$\to y=\dfrac{(x+1)^2+1}{x+1}$
$\to y=(x+1)+\dfrac{1}{x+1}$
$\to y'=1-\dfrac{1}{(x+1)^2}$
8.Ta có:
$y=\dfrac{x^2+3x+3}{x+1}$
$\to y=\dfrac{(x+1)^2+(x+1)+1}{x+1}$
$\to y=(x+1)+1+\dfrac{1}{x+1}$
$\to y'=1-\dfrac{1}{(x+1)^2}$
