Expanding an interacting field theory in its coupling parameter

Question

Simple question about some mathematical syntax in Zee's QFT book, page 49. I have the generating functional

$$ Z=\int\!D\varphi\,\exp\left\{ i\int\!d^4x\,\frac{1}{2}\big[ (\partial\varphi)^2-m^2\varphi^2 \big] -\frac{\lambda}{4!}\varphi^4+J\varphi \right\},$$

which I want to expand in powers of $\lambda$. Obviously, $\varphi$ and $J$ are functions of a continuous variable $\varphi(x^\mu)$ and $J(x^\mu)$. Therefore, I use the separability of the integral as

$$ Z=\int\!D\varphi\,\exp\!\left\{ i\int\!d^4x\,\frac{1}{2}\big[ (\partial\varphi)^2-m^2\varphi^2 \big] +J\varphi \right\}\exp\!\left\{ -\frac{i\lambda}{4!}\int\!d^4w\, \varphi^4\right\}. $$

Since I have created two separate integrals, I have relabeled the integration variable in on case $x^\mu\to w^\mu$. The point of this expansion is bring rewrite $\varphi^4$ as $\partial^4_J$ so that we can move it outside of the path integral over $D\varphi$. We do that as

$$ Z=\exp\!\left\{ -\frac{i\lambda}{4!}\int\!d^4w\, \left[ \frac{\delta}{\delta \big(iJ(w)\big)} \right]^{\!4}\right\}\int\!D\varphi\,\exp\!\left\{ i\int\!d^4x\,\frac{1}{2}\big[ (\partial\varphi)^2-m^2\varphi^2 \big] +J\varphi \right\}. $$

Q1: Why did Zee use the functional derivative $\delta$ instead of the partial derivative $\partial$? This same question appeared here a couple of years ago. The answer skips straight to some stuff about the Dirac $\delta$ which I do not follow and which I do not see on the wiki for functional derivatives. Mainly, since an integral of a function is only one type of functional, I do not think the Dirac $\delta$ is hard-coded into the definition of the functional derivative like the answer to the previous question suggests. Any clarifications?

Q2: What is the meaning of having the variational derivative with respect to $J(w)$ when we have $J(x)$ in the operand expression? This is asked in the other question too but it didn't get answered. I think it tells you that after you do the integral over $d^4x$ and then operate on the integrated expression with $\delta/\delta(iJ(w))$, you need to convert the dummy spacetime variable in the result to $w$ so it gets caught by the outstanding integral over $d^4w$. Is that right?

Answer 1

Answer to Q1
The functional derivative (or variational partial derivative) relates a change in a functional to a change in a function on which the functional depends. Instead the partial derivative relates a change in a function to a change in one of its arguments.

Let us introduce the functional derivative.
Since $J(y) = \int d^4x \delta^4(x - y) J(x)$, we may define $\frac{\partial J(x)}{\partial J(y)} = \delta^4(x - y)$.
That implies
$\frac{\partial}{\partial J(x_1)} \int d^4x J(x) \phi(x) = \phi(x_1)$

Note: To avoid confusion with the Dirac $\delta$ function, I used the symbol $\partial$, which is commonly used for the partial derivative of a function; however here it refers to a functional.

Answer to Q2
That is exactly the meaning of the functional derivative. It varies the value of $J(x)$, where $x$ is the integration variable, at $x = w$, holding all other values of $J(x)$ fixed.