Question

How do I find the derivative of
x^ (1/(x-1))

is it
(1/(x-1)) x^[1-(1/(x-1))] * (-1/ (x-1)^2)

??
If not, can you please write out what rule I should use?

Thank you very much!

Answer 1

Use the chain rule. Let u = 1/(x-1)
u-1 = (1-x +1)/x-1 = -x/(x-1)

F(x) = x^u
dF/dx = dF/du * du/dx
= u*x^(u-1) * [-1/(x-1)^2]
= [1/(x-1)]*x^[-x/(x-1)]*[-1/(x-1)^2]

Our middle terms seem to differ.

Answer 2

y = x^(1/(x-1))
take ln of both sides
ln y = 1/(x-1) ( lnx)
lny = (x-1)^-1 (lnx)
now take derivative
(dy/dx)/y = -1(x-1)^-2 (lnx) + (x-1)^-1 (1/x)
dy/dx = y[-1(x-1)^-2 (lnx) + (x-1)^-1 (1/x)]
or
dy/dx = x^(1/(x-1))[-1(x-1)^-2 (lnx) + (x-1)^-1 (1/x)]