Question

FInd the nth derivative of y=e^-3x×cos^3x

Answer 1

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)`