Đáp án:
$\begin{array}{l}
32)y = \sqrt { - x} \\
\Leftrightarrow y' = \left( { - x} \right)'.\dfrac{1}{{2\sqrt { - x} }} = - \dfrac{1}{{2\sqrt { - x} }}\\
33)y = \dfrac{{2x - 1}}{{3x + 2}}\\
\Leftrightarrow y' = \dfrac{{\left( {2x - 1} \right)'.\left( {3x + 2} \right) - \left( {3x + 2} \right)'.\left( {2x - 1} \right)}}{{{{\left( {3x + 2} \right)}^2}}}\\
= \dfrac{{2\left( {3x + 2} \right) - 3\left( {2x - 1} \right)}}{{{{\left( {3x + 2} \right)}^2}}}\\
= \dfrac{7}{{{{\left( {3x + 2} \right)}^2}}}\\
34)y = \dfrac{{2{x^2} - x + 3}}{{x + 2}}\\
= \dfrac{{\left( {4x - 1} \right)\left( {x + 2} \right) - 2{x^2} + x - 3}}{{{{\left( {x + 2} \right)}^2}}}\\
= \dfrac{{2{x^2} + 8x - 5}}{{{{\left( {x + 2} \right)}^2}}}
\end{array}$
