Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
f\left( x \right) = \cos 3x\\
\Rightarrow f'\left( x \right) = \left( {3x} \right)'.\left( { - \sin 3x} \right) = - 3\sin 3x\\
\Rightarrow f''\left( x \right) = - 3.\left( {3x} \right)'.\left( {\cos 3x} \right) = - 9\cos 3x\\
\Rightarrow f''\left( { - \dfrac{\pi }{2}} \right) = 0;\,\,\,\,f''\left( 0 \right) = - 9;\,\,\,\,f''\left( {\dfrac{\pi }{9}} \right) = - \dfrac{{9\sqrt 3 }}{2}\\
b,\\
g\left( x \right) = x.\sin 2x\\
\Rightarrow g'\left( x \right) = x'.\sin 2x + x.\left( {\sin 2x} \right)'\\
= 1.\sin 2x + x.\left( {2x} \right)'.\cos 2x = \sin 2x + 2x\cos 2x\\
\Rightarrow g''\left( x \right) = \left( {\sin 2x + 2x.\cos 2x} \right)'\\
= \left( {2x} \right)'.cos2x + \left( {2x} \right)'.cos2x + 2x.\left( {2x} \right)'.\left( { - \sin 2x} \right)\\
= 2.\cos 2x + 2\cos 2x - 4x.\sin 2x\\
= 4\cos 2x - 4x.\sin 2x\\
\Rightarrow g''\left( 0 \right) = 4.\cos 0 - 4.0.sin0 = 4
\end{array}\)
