Dùng định nghĩa để tính đạo hàm của các hàm số sau:     

b) f(x) = x² − 2x;             

b) Với bất kì x₀ ∈ ℝ, ta có:

 $f^{\text{'}}\left(x_0\right)=\underset{x\to x_0}{\lim }\frac{\left(x^3-2x\right)-\left(x_0^3-2x_0\right)}{x-x_0}$

  $=\underset{x\to x_0}{\lim }\frac{x^3-2x-x_0^3+2x_0}{x-x_0}=\underset{x\to x_0}{\lim }\frac{\left(x^3-x_0^3\right)-\left(2x-2x_0\right)}{x-x_0}$

 $=\underset{x\to x_0}{\lim }\frac{\left(x-x_0\right)\left(x^2+x.x_0+x_0^2\right)-2\left(x-x_0\right)}{x-x_0}$

 $=\underset{x\to x_0}{\lim }\frac{\left(x-x_0\right)\left(x^2+x.x_0+x_0^2-2\right)}{x-x_0}$

 $=\underset{x\to x_0}{\lim }\left(x^2+x.x_0+x_0^2-2\right)$

 $=x_0^2+x_0.x_0+x_0^2-2=3x_0^2-2$.

Vậy  $f^{\text{'}}\left(x\right)={\left(x^3-2x\right)}^{\text{'}}=3x^2-2$ trên ℝ.