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.
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.