Question

2sin2x*cos5x+sin3x найдите производную функции

Answer 1

$y=2sin(2x)*cos(5x)+sin(3x)$
$(2sin(2x)*cos(5x)+sin(3x))' = (2sin(2x)*cos(5x))' + (sin(3x))' = \ =(-10sin(2x)*sin(5x)+4cos(2x)*cos(5x)) + 3cos(3x) = \ =-10sin(2x)*sin(5x)+4cos(2x)*cos(5x)+3cos(3x) \ \ (2sin(2x)*cos(5x))' = 2((cos(5x))'sin(2x)+cos(5x)*(sin(2x))') = \ \ =2((-5sin(5x))*sin(2x)+cos(5x)*2cos(2x)) \ \ (cos(5x))' = (cos(5x))'(5x)' = -5sin(5x) \ (5x)' = 5 \ (sin(2x))' = (sin(2x))'(2x)' = 2cos(2x) \ (2x)' = 2 \ (sin(3x))' = (sin(3x))'(3x)' = 3cos(3x) \ (3x)' = 3 \ -10sin(2x)*sin(5x)+4cos(2x)*cos(5x)+3cos(3x)$
$-10sin(2x)*sin(5x)+4cos(2x)*cos(5x)+3cos(3x)=7(cos)7x$