Can anyone find the derivative of y=x^1/x -1
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Can anyone find the derivative of y=x^1/x -1

[From: ] [author: ] [Date: 11-11-12] [Hit: ]
If the numerator is x^(1/(x-1)), you can find its derivative using the same procedure.......
I'm trying to solve the limit as x approaches infinity of x(x^1/x-1) which is an indeterminate form of infinity times x. I changed the problem to (x^1/x-1)/(1/x). From there I need to do L Hospitals Rule, but I'm having trouble finding the derivative of the numerator.

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y = x^(1/x) -1

Derivative of a constant is zero, so you need to find the derivative of x^(1/x)

u= x^(1/x)

lnu = (1/x)( lnx)

Take derivative of both sides

u'/u = (-1/x^2)lnx + (1/x)(1/x) = (1/x^2)(1-lnx)

==>y'=u' = (1/x^2)(1-lnx)u = (1/x^2)(1-lnx)[x^(1/x)]

If the numerator is x^(1/(x-1)), you can find its derivative using the same procedure.
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