Đáp án:
$\begin{array}{l}
y = \sin \left( {\dfrac{\pi }{3} - 2x} \right)\\
\Rightarrow y' = \left( {\dfrac{\pi }{3} - 2x} \right)'.\cos \left( {\dfrac{\pi }{3} - 2x} \right)\\
\Rightarrow y' = - 2.\cos \left( {\dfrac{\pi }{3} - 2x} \right)\\
\Rightarrow y'\left( {\dfrac{\pi }{{12}}} \right) = - 2.\cos \left( {\dfrac{\pi }{3} - \dfrac{\pi }{6}} \right)\\
= - 2.\cos \dfrac{\pi }{6}\\
= - \sqrt 3 \\
\Rightarrow A\\
30)\\
y = \cos {\left( {\dfrac{\pi }{3} - \dfrac{x}{2}} \right)^2}\\
\Rightarrow y' = - \left[ {{{\left( {\dfrac{\pi }{3} - \dfrac{x}{2}} \right)}^2}} \right]'.sin{\left( {\dfrac{\pi }{3} - \dfrac{x}{2}} \right)^2}\\
= - 2.\left( { - \dfrac{1}{2}} \right).\left( {\dfrac{\pi }{3} - \dfrac{x}{2}} \right).sin{\left( {\dfrac{\pi }{3} - \dfrac{x}{2}} \right)^2}\\
= \left( {\dfrac{\pi }{3} - \dfrac{x}{2}} \right).sin{\left( {\dfrac{\pi }{3} - \dfrac{x}{2}} \right)^2}\\
\Rightarrow A
\end{array}$
