Bài 1 trang 48 SGK Toán 11 tập 2 - Chân trời sáng tạo — Không quảng cáo

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