Let \(u= e^{-3x}\), \(v=cos 3x\).
Since the derivative of y=uv is y'=udv+vdu, the derivatives of u and v are taken. Where \(du=-3 e^{-3x} dx\) and \(dv=-3 sin3x dx\).
The derivative of the equation is
\(y'=-3 e^{-3x} sin 3x dx -3 e^{-3x} cos 3x dx \\ y'=-3 e^{-3x}dx (sin 3x+cos 3x)\)
