$\log (\log x),x > 1$
$\log (\log x),x > 1$
Official Solution
VVidaara Team
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NCERT & Exemplar
Let y $=$ $\log (\log x),x > 1$
therefore, $\cfrac{{dy}}{{dx}} = \cfrac{d}{{dx}}\log (\log x) = \cfrac{1}{{(\log x)}} \cdot \cfrac{d}{{dx}}(\log x)$
$= \cfrac{1}{{\log x}} \cdot \cfrac{1}{x} = \cfrac{1}{{x\log x}},x > 1$
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