# Thread: ln function...finding value of f'(x)

1. ## ln function...finding value of f'(x)

If $f(x)=ln(ln\:x)$, find the value of $f'(\frac{1}{e})$

Choices:

a) $-\frac{1}{e}$
b)undefined
c) $\frac{1}{e}$
d) $2e$
e) $\sqrt{e}$

alright, heres where im at right now:

1)It could be undefined because $\frac{1}{e}$ is not in the domain of f(x)?
2)the answer is $-e$ and my prof made a mistake(seems unlikely since he's been doing this for 20 years)?
3)my friend says its $\frac{1}{e}$?

Can anyone confirm which is correct with justification?

Thanks!

2. Hello, polymerase!

If $f(x)\:=\:\ln(\ln x)$, find the value of $f'\left(\frac{1}{e}\right)$

Choices: . $a)\;\text{-}\frac{1}{e}\qquad b)\;\text{unde{f}ined}\qquad c)\;\frac{1}{e}\qquad d) \;2e \qquad e) \;\sqrt{e}$

I agree with your answer . . . ${\color{blue}\text{b) unde{f}ined}}$

$f\left(\frac{1}{e}\right) \:=\:\ln\left[\ln\left(\frac{1}{e}\right)\right] \;=\;\ln\left[\ln\left(e^{-1}\right)\right] \;=\;\ln(-1)$ .??

The function does not exist at that point, so neither does its derivative.

3. Originally Posted by Soroban
Hello, polymerase!

I agree with your answer . . . ${\color{blue}\text{b) unde{f}ined}}$

$f\left(\frac{1}{e}\right) \:=\:\ln\left[\ln\left(\frac{1}{e}\right)\right] \;=\;\ln\left[\ln\left(e^{-1}\right)\right] \;=\;\ln(-1)$ .??

The function does not exist at that point, so neither does its derivative.
Hello, I did this and used Maple to confirm. Finding the derivative of this double ln first:
ln(ln(x)) = 1/ln(x) * 1/x
now you sub in (1/e)
1/ln(1/e) = 1/ln(e^-1) noting that.. ln (e^x) = x
so = 1/ln(e^-1) = 1/-1 * 1/(1/e)