# How to calculate the second functional derivative of the action of a one-particle system?

Given the Lagrangian $$L(q,\dot{q})=m\dot{q}^2/2-V(q)$$ and the corresponding action $$S[q]\equiv\int_0^t dt' (m\dot{q}^2/2-V(q)),$$ I need to be able to evaluate the second functional derivative $\frac{\delta^2S[q]}{\delta q(t) \delta q(t')}|_{q=q_{cl}}$. According to Altland/Simon's CMFT, \begin{equation*} \int_0^t dt \int dt' r(t) \frac{\delta^2 S[q]}{\delta q(t) \delta q(t')}|_{q=q_{cl}} r(t') = -\frac{1}{2} \int dt r(t) [m\partial_t^2 + V''(q_{cl}(t))]r(t). \end{equation*} My problem is that I can't figure out how to get the $\partial_t^2$ part when evaluating the function derivative. I would suspect that it's coming from taking the functional derivative of the kinetic energy part of the action. However, when using the definition $$\frac{\delta F[f]}{\delta f(x)} \equiv \lim_{\epsilon\rightarrow0}\frac{1}{\epsilon}(F[f+\epsilon \delta_x]-F[f]),$$ I'm left with some really odd first- and second-derivatives of the Dirac delta function instead.

Could someone please show me how to evaluate these functional derivatives?