Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
y = 2{x^3} \Rightarrow y' = 2.3{x^2} = 6{x^2}\\
2,\\
y = 2\sin x + \cos x + 3\tan x - \cot x\\
\Rightarrow y' = 2\cos x - \sin x + \dfrac{3}{{{{\cos }^2}x}} + \dfrac{1}{{{{\sin }^2}x}}\\
3,\\
y = {\left( {2{x^2} + 5x} \right)^3}\\
\Rightarrow y' = 3.\left( {2{x^2} + 5x} \right)'.{\left( {2{x^2} + 5x} \right)^2}\\
= 3.\left( {4x + 5} \right).{\left( {2{x^2} + 5x} \right)^2}\\
4,\\
y = {x^3} - 3{x^2} - 9x + 2\\
\Rightarrow y' = 3{x^2} - 6x - 9\\
y' < 0 \Leftrightarrow 3{x^2} - 6x - 9 < 0\\
\Leftrightarrow {x^2} - 2x - 3 < 0\\
\Leftrightarrow \left( {x + 1} \right)\left( {x - 3} \right) < 0\\
\Leftrightarrow - 1 < x < 3\\
5,\\
y = \left( {{x^2} + 2x} \right).\left( {5 + 2x - 3{x^2}} \right)\\
\Rightarrow y' = \left( {{x^2} + 2x} \right)'.\left( {5 + 2x - 3{x^2}} \right) + \left( {{x^2} + 2x} \right).\left( {5 + 2x - 3{x^2}} \right)'\\
= \left( {2x + 2} \right).\left( {5 + 2x - 3{x^2}} \right) + \left( {{x^2} + 2x} \right).\left( {2 - 6x} \right)
\end{array}\)
