Q:

Explicit calculation of the two-point function by path integrals

I need help carrying out the following calculation:
We have the generating functional of free theory:
$$Z[f] = \exp\left(\frac{i}{2} \int d^4xd^4y f(x)f(y)\Delta(x-y)\right) $$
where $f$ is an external field and $\Delta(x-y)$ the Green function of the Klein-Gordon equation. 
The two-point function $\langle{0}| T\phi(x) \phi(y) |0\rangle$ is then calculated by the rule
$$\frac{1}{i^2} \frac{\delta Z[f]}{\delta f (x_1) \delta f (x_2)}\bigg|_{f=0}$$
The result should of course be $i\Delta(x-y)$.
Can someone show me the explicit steps?  I just obtain gibberish. 

A:

Omitting space-time indices (cf.  DeWitt notation):
$$
Z[f]=\mathrm e^{\frac12f\cdot\Delta\cdot f}
$$
Therefore,
$$
Z'[f]=\Delta\cdot f\ \mathrm e^{\frac12f\cdot\Delta\cdot f}
$$
and
$$
Z''[f]=\Delta\ \mathrm e^{\frac12f\cdot\Delta\cdot f}+(\Delta\cdot f)^2\ \mathrm e^{\frac12f\cdot\Delta\cdot f}
$$
Therefore, setting $f=0$, we get
$$
Z''[0]=\Delta
$$
as required.