Ta có
$f'(x) = 3 (x+1)' \cos x + 3 (x+1) (\cos x)'$
$= 3\cos x - 3 (x+1) \sin x$
Suy ra
$f''(x) = [f'(x)]'$
$= 3 (\cos x)' - 3 [(x+1) \sin x]'$
$= -3\sin x - 3[(x+1)' \sin x + (x+1)(\sin x)']$
$= -3\sin x - 3[\sin x + (x+1)\cos x]$
$= -6\sin x - 3(x+1)\cos x$
Vậy
$f'(x) = 3\cos x - 3 (x+1) \sin x$, $f''(x) =-6\sin x - 3(x+1)\cos x$.