Câu hỏi 1 - Mục Bài tập trang 45

1. Nội dung câu hỏi

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

a) \(f\left( x \right) = \sqrt {4x + 1} \) tại \(x = 2\);

b) \(f\left( x \right) = {x^4}\) tại \(x =  - 1\);

c) \(f\left( x \right) = \frac{1}{{x + 1}}\);

d) \(f\left( x \right) = \sqrt[3]{{{x^2} + 1}}\).

3. Lời giải chi tiết 

a) Với bất kì \({x_0} \ge \frac{{ - 1}}{4}\) ta có: \(f'\left( {{x_0}} \right) = \mathop {\lim }\limits_{x \to {x_0}} \frac{{f\left( x \right) - f\left( {{x_0}} \right)}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} \frac{{\sqrt {4x + 1}  - \sqrt {4{x_0} + 1} }}{{x - {x_0}}}\)

\( = \mathop {\lim }\limits_{x \to {x_0}} \frac{{\left( {\sqrt {4x + 1}  - \sqrt {4{x_0} + 1} } \right)\left( {\sqrt {4x + 1}  + \sqrt {4{x_0} + 1} } \right)}}{{\left( {x - {x_0}} \right)\left( {\sqrt {4x + 1}  + \sqrt {4{x_0} + 1} } \right)}} = \mathop {\lim }\limits_{x \to {x_0}} \frac{{4x + 1 - 4{x_0} - 1}}{{\left( {x - {x_0}} \right)\left( {\sqrt {4x + 1}  + \sqrt {4{x_0} + 1} } \right)}}\)

\( = \mathop {\lim }\limits_{x \to {x_0}} \frac{{4\left( {x - {x_0}} \right)}}{{\left( {x - {x_0}} \right)\left( {\sqrt {4x + 1}  + \sqrt {4{x_0} + 1} } \right)}} = \mathop {\lim }\limits_{x \to {x_0}} \frac{4}{{\left( {\sqrt {4x + 1}  + \sqrt {4{x_0} + 1} } \right)}} = \frac{4}{{2\sqrt {4{x_0} + 1} }} = \frac{2}{{\sqrt {4{x_0} + 1} }}\)

Suy ra: \(f'\left( x \right) = \frac{2}{{\sqrt {4x + 1} }}\). Do đó, \(f'\left( 2 \right) = \frac{2}{{\sqrt {4.2 + 1} }} = \frac{2}{3}\)

b) Với bất kì \({x_0}\) ta có: \(f'\left( {{x_0}} \right) = \mathop {\lim }\limits_{x \to {x_0}} \frac{{f\left( x \right) - f\left( {{x_0}} \right)}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} \frac{{{x^4} - x_0^4}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} \frac{{\left( {{x^2} + x_0^2} \right)\left( {x - {x_0}} \right)\left( {x + {x_0}} \right)}}{{x - {x_0}}}\)

\( = \mathop {\lim }\limits_{x \to {x_0}} \left( {{x^2} + x_0^2} \right)\left( {x + {x_0}} \right) = \left( {x_0^2 + x_0^2} \right)\left( {{x_0} + {x_0}} \right) = 2x_0^2.2{x_0} = 4x_0^3\)

Do đó, \(f'\left( x \right) = 4{x^3}\). Suy ra \(f'\left( { - 1} \right) = 4.{\left( { - 1} \right)^3} =  - 4\)

c) Với bất kì \({x_0} \ne  - 1\) ta có: \(f'\left( {{x_0}} \right) = \mathop {\lim }\limits_{x \to {x_0}} \frac{{f\left( x \right) - f\left( {{x_0}} \right)}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} \frac{{\frac{1}{{x + 1}} - \frac{1}{{{x_0} + 1}}}}{{x - {x_0}}}\)

\( = \mathop {\lim }\limits_{x \to {x_0}} \frac{{{x_0} + 1 - x - 1}}{{\left( {x - {x_0}} \right)\left( {x + 1} \right)\left( {{x_0} + 1} \right)}} = \mathop {\lim }\limits_{x \to {x_0}} \frac{{ - \left( {x - {x_0}} \right)}}{{\left( {x - {x_0}} \right)\left( {x + 1} \right)\left( {{x_0} + 1} \right)}} = \mathop {\lim }\limits_{x \to {x_0}} \frac{{ - 1}}{{\left( {x + 1} \right)\left( {{x_0} + 1} \right)}}\)

\( =  - \frac{1}{{{{\left( {{x_0} + 1} \right)}^2}}}\)

Vậy \(f'\left( x \right) = \frac{{ - 1}}{{{{\left( {x + 1} \right)}^2}}}\)

d) Với bất kì \({x_0}\) ta có:

\(f'\left( {{x_0}} \right) = \mathop {\lim }\limits_{x \to {x_0}} \frac{{f\left( x \right) - f\left( {{x_0}} \right)}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} \frac{{\sqrt[3]{{{x^2} + 1}} - \sqrt[3]{{x_0^2 + 1}}}}{{x - {x_0}}}\)

\( = \mathop {\lim }\limits_{x \to {x_0}} \frac{{\left( {\sqrt[3]{{{x^2} + 1}} - \sqrt[3]{{x_0^2 + 1}}} \right)\left( {\sqrt[3]{{{{\left( {{x^2} + 1} \right)}^2}}} + \sqrt[3]{{\left( {{x^2} + 1} \right)\left( {x_0^2 + 1} \right)}} + \sqrt[3]{{{{\left( {x_0^2 + 1} \right)}^2}}}} \right)}}{{\left( {x - {x_0}} \right)\left( {\sqrt[3]{{{{\left( {{x^2} + 1} \right)}^2}}} + \sqrt[3]{{\left( {{x^2} + 1} \right)\left( {x_0^2 + 1} \right)}} + \sqrt[3]{{{{\left( {x_0^2 + 1} \right)}^2}}}} \right)}}\)

\( = \mathop {\lim }\limits_{x \to {x_0}} \frac{{{x^2} + 1 - x_0^2 - 1}}{{\left( {x - {x_0}} \right)\left( {\sqrt[3]{{{{\left( {{x^2} + 1} \right)}^2}}} + \sqrt[3]{{\left( {{x^2} + 1} \right)\left( {x_0^2 + 1} \right)}} + \sqrt[3]{{{{\left( {x_0^2 + 1} \right)}^2}}}} \right)}}\)

\( = \mathop {\lim }\limits_{x \to {x_0}} \frac{{\left( {x - {x_0}} \right)\left( {x + {x_0}} \right)}}{{\left( {x - {x_0}} \right)\left( {\sqrt[3]{{{{\left( {{x^2} + 1} \right)}^2}}} + \sqrt[3]{{\left( {{x^2} + 1} \right)\left( {x_0^2 + 1} \right)}} + \sqrt[3]{{{{\left( {x_0^2 + 1} \right)}^2}}}} \right)}}\)

\( = \mathop {\lim }\limits_{x \to {x_0}} \frac{{x + {x_0}}}{{\sqrt[3]{{{{\left( {{x^2} + 1} \right)}^2}}} + \sqrt[3]{{\left( {{x^2} + 1} \right)\left( {x_0^2 + 1} \right)}} + \sqrt[3]{{{{\left( {x_0^2 + 1} \right)}^2}}}}} = \frac{{2{x_0}}}{{3\sqrt[3]{{{{\left( {x_0^2 + 1} \right)}^2}}}}}\)

Vậy \(f'\left( x \right) = \frac{{2x}}{{3\sqrt[3]{{{{\left( {{x^2} + 1} \right)}^2}}}}}\).