f(x)=xe^-x
Any steps would be appreciated, thank you.
Any steps would be appreciated, thank you.
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f'(x) = e^-x + x(-e^-x) {Found by product rule}
f'(x) = e^-x - xe^-x
f''(x) = -e^-x -(e^-x - xe^-x) {as we already know the derivative of xe^-x}
f''(x) = -2e^-x + xe^-x
I'm sure you can see the pattern here. All you have to notice are the sign switches at odd and even, and the leading coefficient IS the derivative number.
f(500)(x) = -500e^-x + xe^-x
f'(x) = e^-x - xe^-x
f''(x) = -e^-x -(e^-x - xe^-x) {as we already know the derivative of xe^-x}
f''(x) = -2e^-x + xe^-x
I'm sure you can see the pattern here. All you have to notice are the sign switches at odd and even, and the leading coefficient IS the derivative number.
f(500)(x) = -500e^-x + xe^-x