Giải thích các bước giải:
$(x+1)'=1\\$
$\begin{split}(\dfrac{x^2}{x-1})'&=(\dfrac{x^2-1+1}{x-1})'\\&=(x+1+\dfrac{1}{x-1})'\\&=1-\dfrac{1}{(x-1)^2}\end{split}\\$
$\dfrac{2x^2+x}{x-1}=\dfrac{2x^2-2x+3x-3+3}{x-1}=2x+3+\dfrac{3}{x-1}$
$\begin{split}\rightarrow (\dfrac{2x^2+x}{x-1})'&=(2x+3+\dfrac{3}{x-1})'\\&=2+0-\dfrac{3}{(x-1)^2}\\&=2-\dfrac{3}{(x-1)^2}\end{split}$