differentiate x^(1/x)?

Answer 1

For (x^(1/x))'....
Let y=(x^(1/x))
ln y = ln (x^(1/x))
ln y = (1/x)lnx
(ln y)' = [(1/x)lnx]'
(1/y) y' = (1/x)(1/x) + (lnx)(-1/x^2)
y' = [(1/x^2) - (lnx)/(x^2)][x^(1/x)]

Answer 2

First the formula: d/dx(a^x) = (a^x)(lna)(dx/dx)
ln = Natural Log,
(d/dx) = Derivative with respect to x,
(dx/dx) = Derivative of x with respect to x

So...

d/dx(x^(1/x)) = (x^(1/x))(lnx)(-x^-2)
(1/x can be rewritten x^-1)
=(x^(1/x))(lnx)(1/x^2)

I think this is right. I hope I used the right formula. Hope this helps.

Answer 3

you can use logarithmic differentiation.
x^1/x = y
ln x^1/x = ln y
1/x ln x = ln y
1/y dy/dx = -ln x/x^2 + 1/x^2
dy/dx = y (1/x^2 - lnx/x^2)

now y = x^1/x so sub that in
dy/dx = x^1/x (1/x^2 - lnx/x^2)

and then u can simplify or whatever if u want.

Answer 4

x^(1-2x)

There is no natural log crap going on there. Why do you guys above make it harder than it really is? Doing all that unnessary stuff just leads to errors.

Answer 5

x^(1/x)=e^(ln x/x)
d(e^(ln x/x)/dx = x^(1/x) * (1/x² - ln x/x²) = (1-ln x)x^(1/x - 2)