Math Help Forum - View Single Post

lmao6 · #3 December 7th, 2007, 12:32 AM

Quote:

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)
so multiply the reciprical since your dividing by a fraction and your final answer is -e!
Im pretty sure im right, and im matching the answer of your 20 years experience prof :P