Ta có : \(f\left(x\right)=\frac{1-\cos2ax}{2}.\cos bx=\frac{1}{2}\cos bx-\frac{1}{2}\cos2ax.\cos bx\)
\(=\frac{1}{2}\cos bx-\frac{\cos\left(2a+b\right)x+\cos\left(2a-b\right)x}{4}\)
\(=\frac{1}{2}\cos bx-\frac{1}{4}\cos\left(2a+b\right)x-\frac{1}{4}\cos\left(2a-b\right)x\)
\(f^{\left(n\right)}\left(x\right)=\frac{1}{2}.b^n\cos\left(bx+\frac{b\pi}{2}\right)-\frac{1}{4}\left(2a+b\right)^n\cos\left[\left(2a+b\right)x+\frac{n\pi}{2}\right]-\frac{1}{4}\left(2a-b\right)^n\cos\left[\left(2a-b\right)x+\frac{n\pi}{2}\right]\)
Áp dụng : Khi a=1,b=2 tức là nếu \(f\left(x\right)=\sin^2x\cos2x\) ta có :
\(f^{\left(n\right)}\left(x\right)=\frac{1}{2}.2^n\cos\left(2x+\frac{n\pi}{2}\right)-\frac{1}{4}.4^n\cos\left(4x+\frac{n\pi}{2}\right)\)
\(=2^{n-1}\cos\left(2x+\frac{n\pi}{2}\right)-4^{n-1}\cos\left(4x+\frac{n\pi}{2}\right)\)
