Найдите производную функции f (x)=sin (4-3x)tg (4-3x)

Решите задачу:

$f'_x (x)=[sin (4-3x)tg (4-3x)]'_x=$
$=[sin(4-3x)]'_x*tg(4-3x)+sin(4-3x)*[tg(4-3x)]'_x=$
$=cos(4-3x)*[4-3x]'_x*tg(4-3x)+$
$+sin(4-3x)* \frac{1}{cos^2(4-3x)}*[4-3x]'_x =$
$=[cos(4-3x)*tg(4-3x)+ \frac{sin(4-3x)}{cos^2(4-3x)}]*[4-3x]'_x =$
$=[cos(4-3x)* \frac{sin(4-3x)}{cos(4-3x)} + \frac{sin(4-3x)}{cos^2(4-3x)}]*[0-3]=$
$=-3[sin(4-3x) + \frac{sin(4-3x)}{cos^2(4-3x)}] =-3sin(4-3x)[ 1+ \frac{1}{cos^2(4-3x)}]$