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Derivatives of x / (x^x)?
Does anyone know how to find the first 5 or 6 derivatives of x / (x^x)?
I'm trying to find the Taylor Series centered at a = 1 of this function, if you're curious.
Thanks for your help.
2 Answers
- Anonymous1 decade agoFavorite Answer
I would use implicit differentiation.
For example, let y = x/ (x^x)
Then taking the logs of both sides,
lny = lnx - xlnx
Then 1/y (dy/dx) = 1/x - lnx-1
Hence if y = x/(x^x), then dy/dx = x/(x^x)[1/x-lnx-1].
You can similarly obtain the next derivatives by doing the same thing.
- Scott VLv 51 decade ago
f'(x) = x(-ln(x - 1)*x^(-x) + x^(-x)
f''(x) = -x^(-x) + ((x/(x^x))*(-ln(x) - 1)^2 + 2(-ln(x) - 1)*x^(-x)
f'''(x) = -3(-ln(x) - 1)*x^(-x) - ((2x^(-x) / x)) + (x / (x^x)*(-ln(x) - 1)^3 + 3x^(-x)*(-ln(x) - 1)^2
f''''(x) = (-(8(-ln(x) - 1)*x^(-x)) / x) - 6x^(-x)*(-ln(x) - 1)^2 + (x / (x^x))*(-ln(x) - 1)^4 + (2x^(-x)) / x^2) + (3x^(-x) / (x)) + 4x^(-x)*(-ln(x) - 1)^3
f'''''(x) = (-(20*x^(-x)*(-ln(x) - 1)^2 / (x)) - 10*x^(-x)*(-ln(x) - 1)^3 - (4x^(-x) / (x^3)) + (x / (x^x))*(-ln(x) - 1)^5 + (5x^(-x) / x^2)) + 5x^(-x)*(-ln(x) - 1)^4 + ((10*(-ln(x) - 1)^(-x) / x^2)) + ((15*(-ln(x) - 1)*x^(-x) / (x))
