Cho \(y=\cos^2x\) . Tính \(f=\left(y'''+y"\right)+16\left(y'+y\right)\).
\(0\).\(8\).\(16\cos4x\).\(-16\cos4x\).Hướng dẫn giải:\(y=\cos^22x=\dfrac{1+\cos4x}{2}\Rightarrow y'=-2\sin4x\Rightarrow y"=-8\cos4x\Rightarrow y'''=32\sin4x\)
Suy ra \(f=\left(y'''+y"\right)+16\left(y'+y\right)=\left(32\sin4x-8\cos4x\right)+16\left(-2\sin4x+\dfrac{1+\cos x}{2}\right)=8\)