 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2011/paper-1/9/solution

For a real-valued expression,

ln x > 0

requires

x > 1

. Applying the chain rule twice gives

\frac{d}{d x} ln (ln x) = \frac{1}{ln x} \frac{1}{x} = \frac{1}{x ln x}, x > 1.

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