Could someone help me determine the derivative of this? PTS WILL bE GIVEN IMMEDIATLEY!?

Question

how do you find the derivative of this logarithmic function?

y= x ^ (1/ lnx)

Answer 1

I'd agree with Matt's answer, but when you get to
ln y = 1
it's easier to write as the next step
y = e (constant function), so dy/dx = 0.

Note that you can also see this by letting x = e^t. Then we get
y = (e^t)^(1/t) = e^(t.(1/t)) = e^1 = e, so dy/dx = 0.

Answer 2

0

because x^(1/ln(x)) = e

therefore the derivative is = 0 because the derivative of any constant is 0

Answer 3

take the log of both sides

ln y = ln(x^1/lnx)
ln y = (1/ln(x))*ln(x)
clean that up
ln y = 1
take the derivative knowing that d/dx(ln y) = y'/y
y'/y = 0
y' = 0

Answer 4

y=x^(1/ln(x))
ln (y)= (1/ln(x)) ln(x)
(1/y)(dy/dx)= (1/ln(x))(1/x) + ln(x) (-1/ln^2(x))(1/x)
(1/y)(dy/dx)=1/(x ln(x)) -1/(xln(x))
dy/dx=y[0]
=0