Đáp án:
$\begin{array}{l}
1)y = \dfrac{1}{{\sin x}} + \dfrac{1}{{\cos x}}\\
\Leftrightarrow y' = \dfrac{{ - \cos x}}{{{{\sin }^2}x}} + \dfrac{{\sin x}}{{{{\cos }^2}x}}\\
2)y = \sqrt {2x + 1} .\left( {{x^2} + 2} \right)\\
\Leftrightarrow y' = \dfrac{2}{{2\sqrt {2x + 1} }}.\left( {{x^2} + 2} \right) + 2x.\sqrt {2x + 1} \\
= \dfrac{{{x^2} + 2}}{{\sqrt {2x + 1} }} + 2x\sqrt {2x + 1} \\
= \dfrac{{{x^2} + 2 + 2x\left( {2x + 1} \right)}}{{\sqrt {2x + 1} }}\\
= \dfrac{{5{x^2} + 2x + 2}}{{\sqrt {2x + 1} }}\\
3)y = {x^3} - 3{x^2} - 2\\
\Leftrightarrow y' = 3{x^2} - 6x\\
a)Khi:y = - 2\\
\Leftrightarrow {x^3} - 3{x^2} - 2 = - 2\\
\Leftrightarrow {x^2}\left( {x - 3} \right) = 0\\
\Leftrightarrow x = 0/x = 3\\
Khi:{x_0} = 0\\
\Leftrightarrow PTTT:y = {y_0}'.x - 2\\
\Leftrightarrow y = - 2\\
+ Khi:{x_0} = 3\\
\Leftrightarrow PTTT:y = {y_0}'.\left( {x - 3} \right) - 2\\
= \left( {{{3.3}^2} - 6.3} \right).\left( {x - 3} \right) - 2\\
= 9x - 29\\
b){x_0} = 2\\
\Leftrightarrow PTTT:y = {y_0}'\left( {x - 2} \right) + {y_0}\\
= \left( {{{3.2}^2} - 6.2} \right).\left( {x - 2} \right) + {2^3} - {3.2^2} - 2\\
= - 6
\end{array}$
