While reading a paper, I came across the term for linearized curvature operator \begin{eqnarray} \kappa_1 = -\frac{1}{{(1+x^2)^{\frac{3}{2}}}}\frac{\,d }{\,d x^2} + \frac{3x}{{(1+x^2)^{\frac{5}{2}}}}\frac{\,d}{\, dx} \end{eqnarray}
the usual definition of curvature I know is
\begin{eqnarray} \kappa = \frac{y''}{({1+y'^2})^{\frac{3}{2}}} \end{eqnarray}
to make it linearlized, \begin{eqnarray} \kappa &=& y''\left[({1+y'^2})^{-\frac{3}{2}} \right] = y''\left[({1-\frac{3} {2}y'^2}) \right] = y'' -\frac{3}{2}y'^2y''\\ &=& \frac{\,d }{\,d x^2}y -\frac{3}{2}\frac{\,d }{\,d x}\left(\frac{1}{3}y'^2\right) \end{eqnarray}
which is not close to the form given above. how can I get the correct form?