Verified answer

Ответ:

f'(x)=4cos 2x ×(sin 2x+1)

Объяснение:

f'(x)=(sin 2x+1)^2)'=2× (sin 2x+1) × (sin 2x+1)'=2×(sin 2x+1)×(cos 2x ×(2x)' +0)=2(sin 2x+1)×cos 2x ×2=4cos 2x ×(sin 2x+1)