\(y'=-2.sin2x\)
\(y''=-2^2.cos2x\)
\(y^{\left(3\right)}=2^3.sin2x\)
\(y^{\left(4\right)}=2^4.cos2x\)
\(\Rightarrow\) \(y^{\left(n\right)}=\left[{}\begin{matrix}-2^n.sin2x\text{ nếu }x=4k+1\\-2^n.cos2x\text{ nếu }x=4k+2\\2^n.sin2x\text{ nếu }x=4k+3\\2^n.cos2x\text{ nếu }x=4k\end{matrix}\right.\)
Do \(2019=2016+3=4.504+3\) có dạng 4k+3
\(\Rightarrow y^{\left(2019\right)}=2^{2019}.sin2x\)