find the 1800th derivative of f(x)=xe^-x
i thought it would work out to be e^-x(1800-x)
but thats not right.... and i tried everything i could think of and i'm not getting this right..
pleaseee help!!
2007-10-25 12:23:00 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
f(x)=xe^-x
f' = e^-x -xe^-x = e^-x -f
f" = -e^-x - {e^-x -xe^-x} = -2e^-x +xe^-x
f"' = +2e^-x +e^-x -xe^-x = 3e^-x -xe^-x
f"" = -3e^-x -e^-x +xe^-x = -4e^-x +xe^-x
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Notice the sign change from - to + when the order of the derivative is even or odd
the 1800th derivative is =
-1800e^-x +xe^-x = (x - 1800)e^-x
2007-10-25 12:39:24 · answer #1 · answered by Anonymous · 1⤊ 0⤋
I'm getting e^(-x) (x - 1800).
f(x) = xe^(-x)
f'(x) = e^(-x) - xe^(-x) = (1-x)e^(-x)
f''(x) = -e^(-x) - (1-x)e^(-x) = (x-2)e^(-x)
f'''(x) = e^(-x) - (x-2)e^(-x) = (3-x)e^(-x)
etc..
1800 being even gives the answer above.
2007-10-25 12:39:09 · answer #2 · answered by ♣ K-Dub ♣ 6 · 0⤊ 0⤋