Question

second derivative of y=ln(x2+1)

Answer 1

Just derive the equation twice.

Answer 2

\[y' = \frac{ 2x }{ x^2+1 }\] --> 1st derivative
\[y'' = \frac{2(x^2+1) - 2x(2x) }{ (x^2+1)^2 }\]
Hence
\[y'' = \frac{2 (1-x^2) }{ (x^2+1)^2 }\]