Cho hàm số f(x) = x(2x – 1)². Tính f'(0) và f'(1).

+ Có f'(0) =  $\underset{x\to 0}{\lim }\frac{f\left(x\right)-f\left(0\right)}{x-0}$$=\underset{x\to 0}{\lim }\frac{x{\left(2x-1\right)}^2}{x}$$=\underset{x\to 0}{\lim }{\left(2x-1\right)}^2=1$

Vậy f'(0) = 1.

+ Có f'(1) = $\underset{x\to 1}{\lim }\frac{f\left(x\right)-f\left(1\right)}{x-1}$$=\underset{x\to 1}{\lim }\frac{x{\left(2x-1\right)}^2-1}{x-1}$

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

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

$=\underset{x\to 1}{\lim }\frac{\left(x-1\right){\left(2x-1\right)}^2+4x\left(x-1\right)}{x-1}$

$=\underset{x\to 1}{\lim }\frac{\left(x-1\right)\left[{\left(2x-1\right)}^2+4x\right]}{x-1}$

$=\underset{x\to 1}{\lim }\left[{\left(2x-1\right)}^2+4x\right]$$=\underset{x\to 1}{\lim }\left(4x^2+1\right)=5$

Vậy f'(1) = 5.