Đạo hàm của hàm số $y={\left(\frac{2x+1}{x-1}\right)}^3$ là

A. $3{\left(\frac{2x+1}{x-1}\right)}^2$

B. $-9\frac{{\left(2x+1\right)}^2}{{\left(x-1\right)}^5}$

C. $-9\frac{{\left(2x+1\right)}^2}{{\left(x-1\right)}^4}$

D. $9\frac{{\left(2x+1\right)}^2}{{\left(x-1\right)}^4}$

Đáp án đúng là: C

Có $y^{\text{'}}={\left[{\left(\frac{2x+1}{x-1}\right)}^3\right]}^{\text{'}}$$=3{\left(\frac{2x+1}{x-1}\right)}^2{\left(\frac{2x+1}{x-1}\right)}^{\text{'}}$

 $=3{\left(\frac{2x+1}{x-1}\right)}^2\frac{{\left(2x+1\right)}^{\text{'}}\left(x-1\right)-\left(2x+1\right){\left(x-1\right)}^{\text{'}}}{{\left(x-1\right)}^2}$

$=3{\left(\frac{2x+1}{x-1}\right)}^2\frac{2\left(x-1\right)-\left(2x+1\right)}{{\left(x-1\right)}^2}$

$=3{\left(\frac{2x+1}{x-1}\right)}^2\frac{2x-2-2x-1}{{\left(x-1\right)}^2}$$=-9\frac{{\left(2x+1\right)}^2}{{\left(x-1\right)}^4}$

Vậy $y^{\text{'}}=-9\frac{{\left(2x+1\right)}^2}{{\left(x-1\right)}^4}$ .