Đáp án:
$\begin{array}{l}
a)y = \left( {3x - 2} \right)\left( {{x^2} - 1} \right)\\
\Leftrightarrow y' = 3.\left( {{x^2} - 1} \right) + 2x.\left( {3x - 2} \right)\\
= 3{x^2} - 3 + 6{x^2} - 4x\\
= 9{x^2} - 4x - 3\\
b)y = \dfrac{{1 - 2x}}{{\sqrt {2x + 1} }}\\
\Leftrightarrow y' = \dfrac{{ - 2.\sqrt {2x + 1} - \left( {1 - 2x} \right).\dfrac{2}{{2\sqrt {2x + 1} }}}}{{2x + 1}}\\
= \dfrac{{ - 2\sqrt {2x + 1} + \dfrac{{2x - 1}}{{\sqrt {2x + 1} }}}}{{2x + 1}}\\
= \dfrac{{ - 2\left( {2x + 1} \right) + 2x - 1}}{{\left( {2x + 1} \right)\sqrt {2x + 1} }}\\
= \dfrac{{ - 2x - 3}}{{\left( {2x + 1} \right)\sqrt {2x + 1} }}\\
c)y = {\sin ^4}\left( {\dfrac{x}{2}} \right)\\
\Leftrightarrow y' = \dfrac{1}{2}.4.cos\dfrac{x}{2}.{\sin ^3}\dfrac{x}{2}\\
= 2\cos \dfrac{x}{2}.{\sin ^3}\dfrac{x}{2}
\end{array}$
