Question:
Differentiate the following w.r.t. x:
$\log (\log x), x>1$
Solution:
Let $y=\log (\log x)$
By using the chain rule, we obtain
$\frac{d y}{d x}=\frac{d}{d x}[\log (\log x)]$
$=\frac{1}{\log x} \cdot \frac{d}{d x}(\log x)$
$=\frac{1}{\log x} \cdot \frac{1}{x}$
$=\frac{1}{x \log x}, x>1$
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