Đáp án:
$\begin{array}{l}
a)y = \frac{{ - 2x + 3}}{{x - 2}}\\
\Rightarrow y' = \frac{{ - 2.\left( {x - 2} \right) - \left( { - 2x + 3} \right)}}{{{{\left( {x - 2} \right)}^2}}}\\
= \frac{{ - 2x + 4 + 2x - 3}}{{{{\left( {x - 2} \right)}^2}}}\\
= \frac{1}{{{{\left( {x - 2} \right)}^2}}}\\
b)y = \left( {x - 2} \right).\sqrt {{x^2} + 1} \\
\Rightarrow y' = \sqrt {{x^2} + 1} + \left( {x - 2} \right).\frac{{2x}}{{2\sqrt {{x^2} + 1} }}\\
= \sqrt {{x^2} + 1} + \left( {x - 2} \right).\frac{x}{{\sqrt {{x^2} + 1} }}\\
= \frac{{{x^2} + 1 + {x^2} - 2x}}{{\sqrt {{x^2} + 1} }}\\
= \frac{{2{x^2} - 2x + 1}}{{\sqrt {{x^2} + 1} }}\\
c)y = co{s^5}\left( {2x + \frac{{4\pi }}{3}} \right)\\
\Rightarrow y' = 5.2.\left( { - \sin \left( {2x + \frac{{4\pi }}{3}} \right)} \right).co{s^4}\left( {2x + \frac{{4\pi }}{3}} \right)\\
= - 10\sin \left( {2x + \frac{{4\pi }}{3}} \right).c{\rm{o}}{{\rm{s}}^4}\left( {2x + \frac{{4\pi }}{3}} \right)
\end{array}$
