allery$\dfrac{d}{dx}\left[\log(\log x)\right]$ at $x = e^e$ equals:
A$\dfrac{1}{(\log x) \cdot x}$ evaluated; = $1/e^{e+1}$
B$\dfrac{1}{e \cdot e^e}$
C$\dfrac{1}{e^{e+1}}$
D$\dfrac{1}{e^e}$
Answer & Solution
Correct answer: A. $\dfrac{1}{(\log x) \cdot x}$ evaluated; = $1/e^{e+1}$
$d/dx[\log(\log x)] = (1/\log x) \cdot (1/x) = 1/(x \log x)$.
At $x = e^e$: $\log x = e$.
**So derivative = $1/(e^e \cdot e) = 1/e^{e+1}$.**
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