Cho hàm số f(x) có đạo hàm \[f\prime (x) = 2x + 4\;\] với mọi \[x \in \mathbb{R}\]. Hàm số \[g(x) = 2f(x) + 3x - 1\;\] có đạo hàm là

A.\[y' = \frac{{3x}}{{\sqrt {{x^2} + 1} }}\]

B. \[y' = \frac{{9{x^2} - x + 3}}{{\sqrt {{x^2} + 1} }}\]

C. \[y' = \frac{{9{x^2} - 2x + 3}}{{\sqrt {{x^2} + 1} }}\]

D. \[y' = \frac{{6{x^2} - x + 3}}{{\sqrt {{x^2} + 1} }}\]

A.\[y' = \frac{{ - 13{x^2} - 10x + 1}}{{{{\left( {{x^2} - 5x + 2} \right)}^2}}}\]

B. \[y' = \frac{{ - 13{x^2} + 5x + 11}}{{{{\left( {{x^2} - 5x + 2} \right)}^2}}}\]

C. \[y' = \frac{{ - 13{x^2} + 5x + 1}}{{{{\left( {{x^2} - 5x + 2} \right)}^2}}}\]

D. \[y' = \frac{{ - 13{x^2} + 10x + 1}}{{{{\left( {{x^2} - 5x + 2} \right)}^2}}}\]

A.\[(uv)' = u'v + v'u\]

B. \[(u + v)' = u' + v'\]

C. \[(u - v)' = u' - v'\]

D. \[{\left( {\frac{u}{v}} \right)^\prime } = \frac{{u'v + v'u}}{{{v^2}}}\]

D. \[ - \frac{3}{2}\]

A.\[\frac{1}{6}\]

B. \[\frac{1}{{12}}\]

C. \[ - \frac{1}{6}\]

D. \[ - \frac{1}{{12}}\]

A.12

B.6

C.24

D.4

A.\[ - \sqrt 3 \]

B. 4

C. -3

D. \(\sqrt 3 \)