Question

найти произведение в точке x0 f(x)=4cos^2*2x ,x0= п/6

Answer 1

$f`(x)=(4cos ^{2}2x)`=8cos2x\cdot (cos2x)`= 8cos2x\cdot (-sin2x)\cdot (2x)`= \\ =8cos2x\cdot (-sin2x)\cdot 2=-8\cdot (2sin2x\cdot cos 2x)=-8 sin 4x \\ f`( \frac{ \pi }{6})=-8sin \frac{4 \pi }{6} =-8sin \frac{2 \pi }{3}=-8sin \frac{ \pi }{3}=-8\cdot \frac{ \sqrt{3} }{2}=-4 \sqrt{3}$