Đáp án:
$\begin{array}{l}
a)\\
y = {\log _2}\left( {\sin x + 2} \right)\\
\Rightarrow y' = \left( {\sin x + 2} \right)'.\dfrac{1}{{\ln 2.\left( {\sin x + 2} \right)}}\\
= \dfrac{{ - \cos x}}{{\ln 2.\left( {\sin x + 2} \right)}}\\
b)y = \sqrt {\sin x} = {\left( {\sin x} \right)^{\dfrac{1}{2}}}\\
\Rightarrow y' = \dfrac{1}{2}.\left( {\sin x} \right)'.{\left( {\sin x} \right)^{1 - \dfrac{1}{2}}}\\
\Rightarrow y' = \dfrac{{ - \cos x}}{2}.{\left( {\sin x} \right)^{ - \dfrac{1}{2}}}\\
\Rightarrow y' = \dfrac{{ - \cos x}}{{2\sqrt {\sin x} }}
\end{array}$
