If #g(x)=x/(e^x)#, what is #g^(n) (x)#?
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# g^((n))(x) = (-1)^n e^-x (x-n) #
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.
#g^((n))(x)=(-1)^n(g(x)- nx), n=1, 2, 3,...#
#=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, .#..
