Question

ПОМОГИТЕ,СРОЧНО
с производной плиз

Answer 1

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

Answer 2

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