Đáp án:
$\begin{array}{l}
a)f\left( x \right) = \left( {{x^{10}} - 16\sqrt x + 16} \right)\left( {{x^2} - 16x + 16} \right)\\
\Rightarrow f'\left( x \right) = \left( {10{x^9} - \frac{8}{{\sqrt x }}} \right).\left( {{x^2} - 16x + 16} \right)\\
+ \left( {{x^{10}} - 16\sqrt x + 16} \right).\left( {2x - 16} \right)\\
= 12{x^{11}} - 176{x^{10}} + 160{x^9} - 40x\sqrt x + 32x + \\
244\sqrt x - \frac{{128}}{{\sqrt x }} - 256\\
b)y = \frac{{16x - 10}}{{16x + 20}} = 1 - \frac{{30}}{{16x + 20}}\\
\Rightarrow y' = \frac{{30.16}}{{{{\left( {16x + 20} \right)}^2}}} = \frac{{30}}{{{{\left( {4x + 5} \right)}^2}}}\\
c)y = \frac{{10x + 16}}{{20x - 16}} = \frac{1}{2} + \frac{{24}}{{20x - 16}} = \frac{1}{2} + \frac{6}{{5x - 4}}\\
\Rightarrow y' = \frac{{ - 6.5}}{{{{\left( {5x - 4} \right)}^2}}} = \frac{{ - 30}}{{{{\left( {5x - 4} \right)}^2}}}\\
d)y = {\left( {16{x^{20}} - 16{x^7} + 16x - 16} \right)^{10}}\\
\Rightarrow y' = \left( {320{x^{19}} - 112{x^6} + 16} \right).{\left( {16{x^{20}} - 16{x^7} + 16x - 16} \right)^9}\\
e)y = \sin \sqrt {16{x^2} - 16x + 16} \\
\Rightarrow y' = \frac{{32x - 16}}{{2\sqrt {16{x^2} - 16x + 16} }}.cos\sqrt {16{x^2} - 16x + 16} \\
= \frac{{4x - 2}}{{\sqrt {{x^2} - x + 1} }}.cos\sqrt {16{x^2} - 16x + 16}
\end{array}$
