Continuity and Differentiability — Class 12 Maths Solution

Let$y = \log (\log x)$
$\Rightarrow$ $\cfrac{{dy}}{{dx}} = \cfrac{1}{{\log x}}\cfrac{1}{x}$ therefore,

$\cfrac{{{d^2}y}}{{d{x^2}}} = \cfrac{1}{{\log x}}\left( { - \cfrac{1}{{{x^2}}}} \right) + \cfrac{1}{x}\cfrac{d}{{dx}}\left( {\cfrac{1}{{\log x}}} \right)$

$= \cfrac{{ - 1}}{{{x^2}\log x}} + \cfrac{1}{x}\left[ {\cfrac{{\log x \cdot 0 - 1 \cdot \cfrac{1}{x}}}{{{{(\log x)}^2}}}} \right] = \cfrac{{ - 1}}{{{x^2}\log x}} + \cfrac{1}{x}\left[ {\cfrac{{ - \cfrac{1}{x}}}{{{{(\log x)}^2}}}} \right]$

$= \cfrac{{ - 1}}{{{x^2}\log x}} - \cfrac{1}{{{x^2}{{(\log x)}^2}}} = \cfrac{{ - 1}}{{{x^2}\log x}}\left[ {1 + \cfrac{1}{{\log x}}} \right] = \cfrac{{ - (1 + \log x)}}{{{{(x\log x)}^2}}}$