What is \(\frac{{{d^2}y}}{{d{x^2}}}\) at x = 1 equal to?

Concept:

\(\frac{d}{{dx}}\left( {\frac{u}{v}} \right) = \frac{{v \times \frac{{du}}{{dx}} - u \times \frac{{dv}}{{dx}}}}{{{v^2}}}\)

Calculation:

From previous question we got, \(\frac{{dy}}{{dx}} = \frac{{\ln x}}{{{{\left( {1 + \ln x} \right)}^2}}}\)

\( \Rightarrow \frac{{{d^2}y}}{{d{x^2}}} = \frac{d}{{dx}}\left( {\frac{{\ln x}}{{{{\left( {1 + \ln x} \right)}^2}}}} \right) = \frac{{\left( {1 + \ln x} \right) \times \frac{1}{x} - 2 \times \ln x \times \left( {1 + \ln x} \right) \times \frac{1}{x}}}{{{{\left( {1 + \ln x} \right)}^4}}}\)

So, \(\frac{{{d^2}y}}{{d{x^2}}}\;at\;x = 1\;is\;1\)