Continuity and Differentiability — Maths STD 12 Science — Question

$\therefore \frac{{dy}}{{dx}} = \frac{1}{{\log x}}\frac{d}{{dx}}\log x\left[ {\therefore \frac{d}{{dx}}\log f\left( x \right) = \frac{1}{{f\left( x \right)}}\frac{d}{{dx}}f\left( x \right)} \right]$

$ = \frac{1}{{\log x}}.\frac{1}{x} = \frac{1}{{x\log x}}$

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

$= \frac{{\left( {x\log x} \right)\left( 0 \right) - \left[ {x\frac{d}{{dx}}\log x + \log x\frac{d}{{dx}}x} \right]}}{{{{\left( {x\log x} \right)}^2}}}$

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

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