Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
y = 2{x^3} \Rightarrow y' = 2.3{x^2} = 6{x^2}\\
2,\\
y = 2\sin x + \cos x + 3\tan x - \cot x\\
\Rightarrow y' = 2.\cos x - \sin x + \dfrac{3}{{{{\cos }^2}x}} + \dfrac{1}{{{{\sin }^2}x}}\\
3,\\
y = {\left( {2{x^2} + 5x} \right)^3}\\
\Rightarrow y' = 3.\left( {2{x^2} + 5x} \right)'.{\left( {2{x^2} + 5x} \right)^2} = 3.\left( {4x + 5} \right){\left( {2{x^2} + 5x} \right)^2}\\
4,\\
y = \left( {{x^2} + 2x} \right).\left( {5 + 2x - 3{x^2}} \right)\\
\Rightarrow y' = \left( {{x^2} + 2x} \right)'.\left( {5 + 2x - 3{x^2}} \right) + \left( {{x^2} + 2x} \right).\left( {5 + 2x - 3{x^2}} \right)'\\
= \left( {2x + 2} \right)\left( {5 + 2x - 3{x^2}} \right) + \left( {{x^2} + 2x} \right)\left( {2 - 6x} \right)\\
5.\\
y = 2\sin 3x.\cos 2x\\
\Rightarrow y' = 2.\left( {\sin 3x} \right)'.\cos 2x + 2\sin 3x.\left( {\cos 2x} \right)'\\
= 2.\left( {3x} \right)'.cos3x.\cos 2x + 2.\sin 3x.\left( {2x} \right)'\left( { - \sin 2x} \right)\\
= 6\cos 3x.\cos 2x - 4\sin 3x.\sin 2x\\
6,\\
y = \sqrt {{x^2} - 2x} \\
\Rightarrow y' = \dfrac{{\left( {{x^2} - 2x} \right)'}}{{2\sqrt {{x^2} - 2x} }} = \dfrac{{2x - 2}}{{2\sqrt {{x^2} - 2x} }} = \dfrac{{x - 1}}{{\sqrt {{x^2} - 2x} }}\\
8,\\
y = {x^2} \Rightarrow y' = 2x\\
9,\\
y = {\tan ^3}3x\\
\Rightarrow y' = 3.\left( {\tan 3x} \right)'.{\tan ^2}3x = 3.\dfrac{{\left( {3x} \right)'}}{{{{\cos }^2}3x}}.{\tan ^2}3x = \dfrac{{9{{\sin }^2}3x}}{{{{\cos }^4}3x}}\\
10,\\
\lim \sqrt {{n^2} + 3n + 1} = \lim \left( {n\sqrt {1 + \dfrac{3}{n} + \dfrac{1}{{{n^2}}}} } \right) = + \infty \\
\left( \begin{array}{l}
\lim n = + \infty \\
\lim \sqrt {1 + \dfrac{3}{n} + \dfrac{1}{{{n^2}}}} = \sqrt 1 = 1
\end{array} \right)
\end{array}\)
