Question

Đã trả lời bởi chuyên gia

Answer 1

a) Đặt y = f(x) = ax².

Ta có y'(x₀) = $\lim_{x \to x_{0}} \frac{f (x) - f (x_{0})}{x - x_{0}}$ $= \lim_{x \to x_{0}} \frac{a x^{2} - a x_{0}^{2}}{x - x_{0}}$

$= \lim_{x \to x_{0}} \frac{a (x^{2} - x_{0}^{2})}{x - x_{0}}$ $= \lim_{x \to x_{0}} \frac{a (x - x_{0}) (x + x_{0})}{x - x_{0}}$ $= \lim_{x \to x_{0}} [a (x + x_{0})] = 2 a x_{0}$

Vậy y'(x₀) = 2ax₀.

b) Đặt $y = f (x) = \frac{1}{x - 1}$

Ta có y'(x₀) = $\lim_{x \to x_{0}} \frac{f (x) - f (x_{0})}{x - x_{0}}$ $= \lim_{x \to x_{0}} \frac{\frac{1}{x - 1} - \frac{1}{x_{0} - 1}}{x - x_{0}}$

$= \lim_{x \to x_{0}} \frac{\frac{x_{0} - 1 - (x - 1)}{(x - 1) (x_{0} - 1)}}{x - x_{0}}$ $= \lim_{x \to x_{0}} \frac{\frac{(x_{0} - x)}{(x - 1) (x_{0} - 1)}}{x - x_{0}}$ $= \lim_{x \to x_{0}} \frac{- 1}{(x - 1) (x_{0} - 1)}$ $= \frac{- 1}{{(x_{0} - 1)}^{2}}$

Vậy $y^{'} (x_{0}) = \frac{- 1}{{(x_{0} - 1)}^{2}}$ , x₀ ≠ 1.

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Answer 2