what is the derivitive of y=x^(1/lnx)?

Question

what is the derivitive of y=x^(1/lnx)
it equals 0
but i need to know how to get it
(muy dificil mi amigos!!!)

Answer 1

The solution is simple

y= x^(1/lnx)

now take the ln of both sides

lny = ln[x^(1/lnx)]

This becomes

lny = (1/lnx)*(lnx)

because ln(a^b) = b*lna

So lny = 1

now take the exponent of both sides

y = e, which is a constant

dy/dx = 0

Answer 2

Take ln of both sides.
ln y = ln x ^ 1/lnx
ln y = 1/ln x (ln x) bring exponent down
ln y = 1 cancellation
y = e
y' = 0.

Answer 3

e^(6) is purely a variety (approx 403, do it on your calculator). so which you will think of of the question as being something like: e^(7x) - 403 The by-manufactured from a variety is 0, as a result the by-manufactured from the question is: 7e^(7x) i'm a severe college maths instructor and the respond is 7e^(7x) you're precise: the rule of thumb of derivitives is that der.e^x=e^x. notwithstanding, it is purely for e^x. listed right here are some examples of derivatives e^2x ? 2e^2x e^x² ? 2xe^x² e^(x² - 3x) ? (2x -3)e^(x² - 3x) so which you ought to multiply e with tips from the differential of the flexibility.

Answer 4

y = x^(1/lnx) ----(i)
=> let us put lnx = t
=> x = e^t
substituting these values in equation (i)
=> y = e^t(1/t)
=> y = e (cuz t X 1/t = 1)
Differentiating above equation, we get
=> dy/dt = 0

Answer 5

take ln on both sides of equation. you get
lny=ln(x^(1/lnx))=(1/lnx)*lnx=1
therefore
lny=1
taking derivative,
d(lny)/dx=0
or, 1/y * dy/dx = 0
since y not equal to 0, dy/dx = 0

Answer 6

take ln on both sides
lny=1*lnx/lnx
lny=1
differentiate now
(1/y)*dy/dx=0
dy/dx=0

Answer 7

You have to use a differentiation technique called "logarithmic differention." Basically, you take the natural logarithm of both sides of the equation and then use implicit differention. It is important to know that the derivative of ln(x) is 1/x:

y = x^(1/ln(x))
ln(y) = ln(x)^(1/ln(x)) [took natural log of both sides]
ln(y) = (1/ln(x))ln(x) [simplified using laws of logarithms*]
ln(y) = 1 [simplified]
(1/y)(dy/dx) = 0 [differentiatedd]
(dy/dx) = 0

*ln(A)^B = B(ln(A)) (property of logarithms)

Good luck.

Answer 8

let e^z = x^(1/lnx)
z = (1/lnx)lnx = 1
d/dx(x^(1/lnx) = d/dx(e^1) = 0