Bài 1 trang 41 SGK Toán 11 tập 2 - Chân trời sáng tạo

Đề bài

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

a) $f\left( x \right) =  - {x^2}$;

b) $f\left( x \right) = {x^3} - 2x$;

c) $f\left( x \right) = \frac{4}{x}$.

Lời giải chi tiết

a) Với bất kì ${x_0} \in \mathbb{R}$, ta có:

$f'\left( {{x_0}} \right) = \mathop {\lim }\limits_{x \to {x_0}} \frac{{\left( { - {x^2}} \right) - \left( { - x_0^2} \right)}}{{x - {x_0}}}$

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

$= \mathop {\lim }\limits_{x \to {x_0}} \left( { - x - {x_0}} \right) =  - {x_0} - {x_0} =  - 2{{\rm{x}}_0}$

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

b) Với bất kì ${x_0} \in \mathbb{R}$, ta có:

$f'\left( {{x_0}} \right) = \mathop {\lim }\limits_{x \to {x_0}} \frac{{\left( {{x^3} - 2{\rm{x}}} \right) - \left( {x_0^3 - 2{{\rm{x}}_0}} \right)}}{{x - {x_0}}}$

$= \mathop {\lim }\limits_{x \to {x_0}} \frac{{{x^3} - 2{\rm{x}} - x_0^3 + 2{{\rm{x}}_0}}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} \frac{{\left( {{x^3} - x_0^3} \right) - 2\left( {x - {x_0}} \right)}}{{x - {x_0}}}$

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

$= \mathop {\lim }\limits_{x \to {x_0}} \left( {{x^2} + x.{x_0} + x_0^2 - 2} \right) = x_0^2 + {x_0}.{x_0} + x_0^2 - 2 = 3{\rm{x}}_0^2 - 2$

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

c) Với bất kì ${x_0} \ne 0$, ta có:

$f'\left( {{x_0}} \right) = \mathop {\lim }\limits_{x \to {x_0}} \frac{{\frac{4}{x} - \frac{4}{{{x_0}}}}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} \frac{{\frac{{4{x_0} - 4x}}{{x{x_0}}}}}{{x - {x_0}}}$

$= \mathop {\lim }\limits_{x \to {x_0}} \frac{{4{x_0} - 4x}}{{x{x_0}\left( {x - {x_0}} \right)}} = \mathop {\lim }\limits_{x \to {x_0}} \frac{{ - 4\left( {x - {x_0}} \right)}}{{x{x_0}\left( {x - {x_0}} \right)}}$

$= \mathop {\lim }\limits_{x \to {x_0}} \frac{{ - 4}}{{x{{\rm{x}}_0}}} = \frac{{ - 4}}{{{x_0}.{x_0}}} =  - \frac{4}{{x_0^2}}$

Vậy $f'\left( x \right) = {\left( {\frac{4}{x}} \right)^\prime } =  - \frac{4}{{{x^2}}}$ trên các khoảng $\left( { - \infty ;0} \right)$ và $\left( {0; + \infty } \right)$.