Bài 1 trang 48 SGK Toán 11 tập 2 – Chân trời sáng tạo

Đề bài

Tính đạo hàm của các hàm số sau:

a) \(y = 2{{\rm{x}}^3} – \frac{{{x^2}}}{2} + 4{\rm{x}} – \frac{1}{3}\);

b) \(y = \frac{{ – 2{\rm{x}} + 3}}{{{\rm{x}} – 4}}\);

c) \(y = \frac{{{x^2} – 2{\rm{x}} + 3}}{{{\rm{x}} – 1}}\);  d) \(y = \sqrt {5{\rm{x}}} \).

Lời giải chi tiết

a) \(y’ = 2.3{{\rm{x}}^2} – \frac{1}{2}.2{\rm{x}} + 4.1 – 0 = 6{{\rm{x}}^2} – x + 4\).

b) \(y’ = \frac{{{{\left( { – 2{\rm{x}} + 3} \right)}^\prime }.\left( {{\rm{x}} – 4} \right) – \left( { – 2{\rm{x}} + 3} \right).{{\left( {{\rm{x}} – 4} \right)}^\prime }}}{{{{\left( {{\rm{x}} – 4} \right)}^2}}}\)

\( = \frac{{ – 2\left( {{\rm{x}} – 4} \right) – \left( { – 2{\rm{x}} + 3} \right).1}}{{{{\left( {{\rm{x}} – 4} \right)}^2}}}\)

\( = \frac{{ – 2{\rm{x}} + 8 + 2{\rm{x}} – 3}}{{{{\left( {{\rm{x}} – 4} \right)}^2}}} = \frac{5}{{{{\left( {{\rm{x}} – 4} \right)}^2}}}\)

c) \(y’ = \frac{{{{\left( {{x^2} – 2{\rm{x}} + 3} \right)}^\prime }\left( {{\rm{x}} – 1} \right) – \left( {{x^2} – 2{\rm{x}} + 3} \right){{\left( {{\rm{x}} – 1} \right)}^\prime }}}{{{{\left( {{\rm{x}} – 1} \right)}^2}}}\)

\( = \frac{{\left( {2{\rm{x}} – 2} \right)\left( {{\rm{x}} – 1} \right) – \left( {{x^2} – 2{\rm{x}} + 3} \right).1}}{{{{\left( {{\rm{x}} – 1} \right)}^2}}}\) \( = \frac{{2{{\rm{x}}^2} – 2{\rm{x}} – 2{\rm{x}} + 2 – {x^2} + 2{\rm{x}} – 3}}{{{{\left( {{\rm{x}} – 1} \right)}^2}}}\)

\( = \frac{{{x^2} – 2{\rm{x}} – 1}}{{{{\left( {{\rm{x}} – 1} \right)}^2}}}\)

d) \(y’ = {\left( {\sqrt 5 .\sqrt x } \right)^\prime } = \sqrt 5 .\frac{1}{{2\sqrt x }} = \frac{{\sqrt 5 }}{{2\sqrt x }} = \frac{5}{{2\sqrt {5x} }}\).