Question
Differentiate the log (log x), x > 1 w.r.t. x.
✓
Answer
Let y = log(log x)
So, by using chain rule, we get
$\frac{d y}{d x}=\frac{d}{d x}(\log (\log x))$
$=\frac{1}{\log x} \cdot \frac{d}{d x}(\log x)$
$=\frac{1}{\log x} \cdot \frac{1}{x}$
$=\frac{1}{\mathrm{xlogx}}$
So, by using chain rule, we get
$\frac{d y}{d x}=\frac{d}{d x}(\log (\log x))$
$=\frac{1}{\log x} \cdot \frac{d}{d x}(\log x)$
$=\frac{1}{\log x} \cdot \frac{1}{x}$
$=\frac{1}{\mathrm{xlogx}}$
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