Tìm đạo hàm của hàm số y=cos^6x+2sin^4xcos^2x
Giải thích
y=sin4x1+2cos2x+cos4x3sin2x+cos2x
=sin4x1+2cos2x+cos4x1+2sin2x
=sin4x+cos4x+2sin4xcos2x+2sin2xcos4x
=cos2x+sin2x2−2sin2xcos2x+2sin2xcos2xcos2x+sin2x
=1.
y=sin4x1+2cos2x+cos4x3sin2x+cos2x
=sin4x1+2cos2x+cos4x1+2sin2x
=sin4x+cos4x+2sin4xcos2x+2sin2xcos4x
=cos2x+sin2x2−2sin2xcos2x+2sin2xcos2xcos2x+sin2x
=1.