Tính đạo hàm của hàm số:

a) y = ax² (a là hằng số) tại điểm x₀ bất kì.

b) $y=\frac{1}{x-1}$ tại điểm x₀ bất kì, x₀ ≠ 1.

a) Đặt y = f(x) = ax².

Ta có y'(x₀) = $\underset{x\to x_0}{\lim }\frac{f\left(x\right)-f\left(x_0\right)}{x-x_0}$$=\underset{x\to x_0}{\lim }\frac{a{x}^2-a{x}_0^2}{x-x_0}$

$=\underset{x\to x_0}{\lim }\frac{a\left(x^2-x_0^2\right)}{x-x_0}$$=\underset{x\to x_0}{\lim }\frac{a\left(x-x_0\right)\left(x+x_0\right)}{x-x_0}$$=\underset{x\to x_0}{\lim }\left[a\left(x+x_0\right)\right]=2a{x}_0$

Vậy y'(x₀) = 2ax₀.

b) Đặt $y=f\left(x\right)=\frac{1}{x-1}$

Ta có y'(x₀) = $\underset{x\to x_0}{\lim }\frac{f\left(x\right)-f\left(x_0\right)}{x-x_0}$$=\underset{x\to x_0}{\lim }\frac{\frac{1}{x-1}-\frac{1}{x_0-1}}{x-x_0}$

$=\underset{x\to x_0}{\lim }\frac{\frac{x_0-1-\left(x-1\right)}{\left(x-1\right)\left(x_0-1\right)}}{x-x_0}$$=\underset{x\to x_0}{\lim }\frac{\frac{\left(x_0-x\right)}{\left(x-1\right)\left(x_0-1\right)}}{x-x_0}$$=\underset{x\to x_0}{\lim }\frac{-1}{\left(x-1\right)\left(x_0-1\right)}$$=\frac{-1}{{\left(x_0-1\right)}^2}$

Vậy $y^{\text{'}}\left(x_0\right)=\frac{-1}{{\left(x_0-1\right)}^2}$ , x₀ ≠ 1.