Question

Найдите:
$y=cos5xcos8x; y'( \frac{ \pi }{6} )=?$

Answer 1

$y=cos5x*cos8x\\y'=-sin5x*(5x)'*cos8x-sin(8x)*(8x)'*cos5x=\\=-(5sin5x*cos8x+8cos5x*sin8x)\\\\y'(\frac{\pi}6)=-(5sin(5*\frac{\pi}6)*cos(8*\frac{\pi}6)+8cos(5*\frac{\pi}6)*sin(8*\frac{\pi}6))=\\\\=-(5sin(\frac{5\pi}6)*cos(\frac{4\pi}3)+8cos(\frac{5\pi}6)*sin(\frac{4\pi}3))=\\\\=-(5*\frac{1}2*(-\frac{1}2)+8*(-\frac{\sqrt3}2)*(-\frac{\sqrt3}2))=\frac{5}4-\frac{8*3}4=-\frac{19}4$