Tính được đạo hàm của mỗi hàm số sau tại các điểm đã chỉ ra. Khi đó:

a) Đúng

b) Sai

c) Đúng

d) Đúng

 

a) \({f^\prime }(1) = \mathop {\lim }\limits_{x \to 1} \frac{{f(x) - f\left( {{x_0}} \right)}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - x - f(1)}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{{x^2} - x}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} x = 1\).

b) \({f^\prime }(1) = \mathop {\lim }\limits_{x \to 1} \frac{{f(x) - f\left( {{x_0}} \right)}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to 1} \frac{{\sqrt x  - f(1)}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{{\sqrt x  - 1}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} \frac{1}{{\sqrt x  + 1}} = \frac{1}{2}\).

c) \({f^\prime }(0) = \mathop {\lim }\limits_{x \to 0} \frac{{f(x) - f\left( {{x_0}} \right)}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{1}{{{x^2} + 1}} - f(0)}}{x} =  = \mathop {\lim }\limits_{x \to 0} \frac{{\frac{1}{{{x^2} + 1}} - 1}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{ - x}}{{{x^2} + 1}} = 0\).

d) \({f^\prime }(2) = \mathop {\lim }\limits_{x \to 2} \frac{{f(x) - f\left( {{x_0}} \right)}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to 2} \frac{{\frac{1}{{x + 1}} - f(2)}}{{x - 2}} = \mathop {\lim }\limits_{x \to 2} \frac{{\frac{1}{{x + 1}} - \frac{1}{3}}}{{x - 2}} =  = \mathop {\lim }\limits_{x \to 2} \frac{{ - 1}}{{3(x + 1)}} =  - \frac{1}{9}\).