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

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

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

Choices:

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

alright, heres where im at right now:

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

Can anyone confirm which is correct with justification?

Thanks!

2. Hello, polymerase!

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

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

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

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

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 . . . $\displaystyle {\color{blue}\text{b) unde{f}ined}}$

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

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)