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