Đáp án:
$\begin{array}{l}
y = \ln \left| {\sin 2x} \right|\\
y' = \dfrac{{\left| {\sin 2x} \right|'}}{{\left| {\sin 2x} \right|}}\\
= \dfrac{{2\cos 2x}}{{\left| {\sin 2x} \right|}}\\
\Rightarrow y'\left( {\dfrac{\pi }{8}} \right) = \dfrac{{2.\cos \dfrac{\pi }{4}}}{{\left| {\sin \dfrac{\pi }{4}} \right|}} = \dfrac{{2.\dfrac{{\sqrt 2 }}{2}}}{{\left| {\dfrac{{\sqrt 2 }}{2}} \right|}} = 2
\end{array}$
