Question

If `g(x)=x/(e^x)`, what is `g^(n) (x)`?

Answer 1

`g^((n))(x) = (-1)^n e^-x (x-n)`

Explanation:

Rewrite as `g(x)=xe^-x`

We can then use the product rule; `d/dx(uv)=u(dv)/dx+v(du)/dx`

`g'(x) = { (x)(d/dxe^-x) + (e^-x)(d/dxx) }`
`:. g'(x) = { (x)(-e^-x) + (e^-x)(1) }`
`:. g'(x) = e^-x - xe^-x`

We can now write down the second derivative:
`g''(x) = -e^-x - ( e^-x - xe^-x )`
`:. g''(x) = -2e^-x + xe^-x`

Similarly the third derivative:
`g^((3))(x) = 2e^-x + e^-x - xe^-x`
`:. g^((3))(x) = 3e^-x - xe^-x`

So it looks like clear pattern is forming, but let us just check by looking at the fourth derivative;
`:g^((4))(x) = -3e^-x - (e^-x - xe^-x)`
`:. g^((4))(x) = -4e^-x + xe^-x`

In summary;
`g^((0))(x) = -0e^-x + xe^-x`
`g^((1))(x) = +e^-x - xe^-x`
`g^((2))(x) = -2e^-x + xe^-x`
`g^((3))(x) = +3e^-x - xe^-x`
`g^((4))(x) = -4e^-x + xe^-x`

And this would lead s to conclude that
`g^((n))(x) = -(-1)^n(n)e^-x + (-1)^n xe^-x`
`:. g^((n))(x) = (-1)^n e^-x (x-n)`

NOTE:
This is NOT a vigorous proof! In order to prove this result is valid we would need to start with the result and use proof by Induction.

Answer 2

`g^((n))(x)=(-1)^n(g(x)- nx), n=1, 2, 3,...`

Explanation:

`=xe^(-x)`

`g'=-xe^(-x)+e^(-x)=-g+e^(-x)`

`g''=-g'-e^(-x)=g+2e^(-x)`

And so, `g^((3))=-g-3e^(-x)`, and so on.

Thus, `g^((n))=(-1)^n(g- n e^(-x)) , n=1, 2, 3, .`..