Đáp án:
$\begin{array}{l}
a)y = {\tan ^2}\left( {\sqrt x + 3} \right)\\
\Rightarrow y' = 2.\left( {\sqrt x + 3} \right)'.\left( {\tan \left( {\sqrt x + 3} \right)'} \right).tan\left( {\sqrt x + 3} \right)\\
= 2.\frac{1}{{2\sqrt x }}.\frac{1}{{{{\cos }^2}\left( {\sqrt x + 3} \right)}}.\tan \left( {\sqrt x + 3} \right)\\
= \frac{{\tan \left( {\sqrt x + 3} \right)}}{{\sqrt x .{{\cos }^2}\left( {\sqrt x + 3} \right)}}\\
b)y = \sqrt {{{\sin }^2}x + \cos 3x} \\
y'= \left( {{{\sin }^2}x + \cos 3x} \right)'.\frac{1}{{2\sqrt {{{\sin }^2}x + \cos 3x} }}\\
= \left( {2.cosx.sinx - 3sin3x} \right).\frac{1}{{2\sqrt {{{\sin }^2}x + \cos 3x} }}\\
= \left( {\sin 2x - 3\sin 3x} \right).\frac{1}{{2\sqrt {{{\sin }^2}x + \cos 3x} }}
\end{array}$
