Question

ПОМОГИТЕ,СРОЧНО
с производной плиз

Answer 1

$\displaystyle f(x)=3cos2x-sin2x\\\\f`(x)=3*(cos2x)`-(sin2x)`=3*(-sin2x)*2-(cos2x)*2=\\\\=-2(3sin2x+cos2x)\\\\f`(\frac{\pi}{8})=-2(3sin(2*\frac{\pi}{8})+cos(2*\frac{\pi}{8}))=-2(3sin\frac{\pi}{4}+cos(\frac{\pi}{4}))=\\\\=-2(3\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2})=-2*4\frac{\sqrt{2}}{2}=-4\sqrt{2}$

Answer 2

$f(x)=3Cos2x-Sin2x\\\\f'(x)=3(Cos2x)'-(Sin2x)' =3*(-Sin2x)*(2x)'-Cos2x*(2x)'=-6Sin2x-2Cos2x\\\\f'(\frac{\pi }{8})=-6*Sin(2*\frac{\pi }{8})-2Cos(2*\frac{\pi }{8} )=-6Sin\frac{\pi }{4}-2Cos\frac{\pi }{4}=-6*\frac{\sqrt{2} }{2}-2*\frac{\sqrt{2} }{2}=-3\sqrt{2}-\sqrt{2} =-4\sqrt{2}$