Đáp án:
$\begin{array}{l}
a)y = {\left( {3x - 1} \right)^4}\\
\Rightarrow y' = 4.\left( {3x - 1} \right)'.{\left( {3x - 1} \right)^3}\\
= 4.3.{\left( {3x - 1} \right)^3}\\
= 12.{\left( {3x - 1} \right)^3}\\
b)y = \dfrac{{\sqrt x }}{{1 - 2x}}\\
\Rightarrow y' = \dfrac{{\left( {\sqrt x } \right)'.\left( {1 - 2x} \right) - \sqrt x .\left( {1 - 2x} \right)'}}{{{{\left( {1 - 2x} \right)}^2}}}\\
= \dfrac{{\dfrac{1}{{2\sqrt x }}.\left( {1 - 2x} \right) - \sqrt x .\left( { - 2} \right)}}{{{{\left( {1 - 2x} \right)}^2}}}\\
= \dfrac{{\dfrac{{1 - 2x + 2\sqrt x .2\sqrt x }}{{2\sqrt x }}}}{{{{\left( {1 - 2x} \right)}^2}}}\\
= \dfrac{1}{{2\sqrt x {{\left( {1 - 2x} \right)}^2}}}
\end{array}$
