derivative of e?

Question

Find the 1800th derivative of f(x) = xe^-x.

Answer 1

f(x)=xe^(-x)
f'(x)=e^(-x)-xe^(-x)
f"(x)=-e^(-x)-f
f'''(x)=e^(-x)-e^(-x)+x^e^(-x)=f
The pattern repeats itself, my guess is the 1800th derivative is = f(x)

Answer 2

I took the first, second, and third derivatives. If you do this and simplify the answer every time, then each derivative consists of two terms, with a very nice pattern. It is then easy to just write down the 1800th derivative, or any other derivative. (Hint: the pattern involves an alternating sign. 1800 is an even number, and thus the signs in the 1800th derivative will follow the pattern for the 2nd, 4th, etc.)

Answer 3

f(x) = xe^(-x)
=> f'(x) = e^(-x) - xe^(-x)
=> f"(x) = - e^(-x) - e^(-x) + xe^(-x) = f(x) - 2e^(-x)
=> f"'(x) = f'(x) + 2e^(-x)
=> f""(x) = f"(x) - 2e^(-x) = f(x) - 4e^(-x)
Pattern keeps repeating
=> 1800th derivative is f(1800th derivative)(x) = f(x) - 1800e^(-x).
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