The interval in which \(y=\ln (\ln (x)), x>1\) is decreasing is
The interval in which \(y=\ln (\ln (x)), x>1\) is decreasing is
\((-\infty, 0) \cup(2, \infty)\)
\((0,2)\)
\((0,1)\)
None of the above
Solution
Given function \(y=l(\ln (x)), x > 1\)
\(\therefore \quad \frac{d y}{d x}=\frac{1}{x \ln x}, x > 1\)
\(\because y\) is a decreasing, then
\(\frac{d y}{d x} < 0 \Rightarrow \frac{1}{x \ln x} < 0\)
\(\because x > 1 \Rightarrow \ln x < 0 \Rightarrow x < 1\), which is not possible as \(x > 1\)
\(\therefore x \in \phi\), for decreasing