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what is the derivative and second derivative of xe^-x?
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product rule: if f = u.v, then f' = u.v' + u'.v where u = x & v = e^-x
I got it! Thanks!
let A = xe^-x take log both side, ln(A) = ln( xe^-x) ln(A) = lnx - xln(e), since lne =1 lnA = lnx - x now differentiate both side (1/A)A' = 1/x - 1 A' = ((1-x)/x)A A' = (1-x)e^-x
That's what I got too! :)
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