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?  
 A: I know how to do this by simply expanding the action as a Taylor series in the deviation $r(t)$.
Expand the action
\begin{align}
\delta S =& S[q_{\mathrm{cl}} + r(t)] - S[q_{\mathrm{cl}}(t)] \\
   =& \int_0^t\mathrm{d}t' \cdot\left\{ \frac{m}{2}\left[\partial_{t'}( q_{\mathrm{cl}} + r)\right]^2 - V(q_{\mathrm{cl}} + r) 
    -\frac{m}{2}\left[\partial_{t'} q_{\mathrm{cl}} \right]^2 + V(q_{\mathrm{cl}}) \right\} \\
   = & \int_0^t\mathrm{d}t' \cdot\left\{O(r) + \frac{m}{2} (\partial_{t'}r)^2 - \frac{1}{2}V''(q_{\mathrm{cl}})r^2
    + O(r^{3}) + \cdots\right\}
\end{align}
where $O(r^n)$ means the $n$th order term. Then use integration by parts
\begin{align}
  \int_0^t \mathrm{d}t'\cdot \left[\partial_{t'}r(t')\right] \left[\partial_{t'}r(t')\right]
  =& \left.r(t')\partial_{t'}r(t')\right|_0^t - \int_0^t \mathrm{d}t'\cdot r(t')\partial_{t'}^{2}r(t') \\
  = & - \int_0^t \mathrm{d}t'\cdot r(t')\partial_{t'}^{2}r(t')
\end{align}
where we have used the fact that $r(0)=r(t)=0$ . So we get the second order term of $S$
\begin{align}
\int_0^t \mathrm{d}t \int_0^t \mathrm{d}t'\cdot r(t) \frac{\delta^2 S[q]}{\delta q(t) \delta q(t')}|_{q=q_{cl}} r(t') = -\frac{1}{2} \int \mathrm{d}t \cdot r(t) [m\partial_t^2 + V''(q_{cl}(t))]r(t).
\end{align}
Reference:
Appendix A of Density Functional Theory An Advanced Course by Eberhard Engel, Reiner M. Dreizler may be helpful to understand the relation between the Taylor expansion and functional derivative.
