Đáp án:
$y' = -6\sin3x.\sin(\sin3x).\cos(\cos3x)$
Giải thích các bước giải:
$\quad y = \sin^2(\cos3x)$
$\to y' = 2\sin(\cos 3x).\left[\sin(\cos3x)\right]'$
$\to y' = 2\sin(\cos3x).(\cos3x)'.\cos(\cos3x)$
$\to y' = 2\sin(\cos3x).(3x)'.(-\sin3x).\cos(\cos3x)$
$\to y' = -6\sin3x.\sin(\sin3x).\cos(\cos3x)$