Notation question in calculus of variations -- QFT these two integrals below are equal, but I am not understanding where the $x'$ variable comes from.
\begin{align}
I_0&=e^{ i\int d^4x \left\{ \frac{1}{2}\left[  \left( \partial\varphi(x)  \right)^2-\varphi(x)^2  \right]  -\big[\varphi(x)\big]^4  +J(x)\varphi(x)\right\}}\\
&=e^{  -i\int d^4x'\big[ \frac{\delta}{\delta J(x')} \big]^4 }    e^{ i\int d^4x \left\{ \frac{1}{2}\left[  \left( \partial\varphi(x)  \right)^2-\varphi(x)^2  \right]  +J(x)\varphi(x)\right\}}
\end{align}
The gist of what I have been doing is to write
\begin{align}
I_0&=e^{ i\int d^4x \left\{ \frac{1}{2}\left[  \left( \partial\varphi(x)  \right)^2-\varphi(x)^2  \right]    +J(x)\varphi(x)\right\}}e^{ -i\int d^4x  \big[\varphi(x)\big]^4}  \\
&=e^{ i\int d^4x \left\{ \frac{1}{2}\left[  \left( \partial\varphi(x)  \right)^2-\varphi(x)^2  \right]    +J(x)\varphi(x)\right\}} \left[ 1+\left( -i\int d^4x \big[  \varphi(x)  \big]^4 \right) +...\right]\\
&=\left[ 1+\left( -i\int d^4x \left[\dfrac{d}{dJ} \right]^4\right)+... \right]e^{ i\int d^4x \left\{ \frac{1}{2}\left[  \left( \partial\varphi(x)  \right)^2-\varphi(x)^2  \right]    +J(x)\varphi(x)\right\}} 
\end{align}
The point is to pull the $\varphi^4$ term out of the integral by writing $\varphi$ as $d/dJ$ except in this case I have to use the variation $\delta$ instead and I don't see why.  I used the $d/dJ$ trick in simpler examples without an integral in the exponent, and I am not seeing the connection to the variational notation $\delta$ which appears in the second of the first two equations above.  Obviously, I can recombine the prefactor sum in my final equation into $\text{exp}$, but I do not see the point of the $x'$ variable.  I hope it is clear that if I had $\delta/\delta J(x')$ in my last equation instead of $d/dJ$ then I would get the form of the second equation which is the correct form.  Why can't I just write it as $d/dJ$ like I did before?  Please give me a tip, thanks.
 A: The issue is really what it means to to compute one functional derivative $\delta/\delta f$.  Once we get that part, we can raise it to the nth power and get the result.
How to compute one variation?
\begin{align}
\dfrac{\delta}{\delta J(x')} K&=\dfrac{\delta}{\delta J(x')} e^{ i\int d^4x \left\{ \frac{1}{2}\left[  \left( \partial\varphi(x)  \right)^2-\varphi(x)^2  \right]    +J(x)\varphi(x)\right\}} \\
&=\dfrac{\delta}{\delta J(x')} e^{ i\int d^4x J(x)\varphi(x)}e^{ i\int d^4x \left\{ \frac{1}{2}\left[  \left( \partial\varphi(x)  \right)^2-\varphi(x)^2  \right] \right\}} \\
&=A\dfrac{\delta}{\delta J(x')} e^{ i\int d^4x J(x)\varphi(x)} \\
\end{align}
Let $F=e^{ i\int d^4x J(x)\varphi(x)} $. The next step is what follows.  Instead of using $\Delta$, you need to use $\varepsilon$ times a Dirac delta.  This is part of the definition of the functional derivative I guess, If anyone want to say a little more about that would be nice, but I think it's just part of the definition.  However, wikipedia calls the Dirac delta a "test function" so maybe it is an ansatz of some type.  Anyway, the answer follows:
\begin{align}
\dfrac{\delta}{\delta J(x')}F&=\lim\limits_{\varepsilon\to0}\dfrac{F[J(x)+\varepsilon\delta(x-x')]-F[J(x)]}{\varepsilon}\\
&=\lim\limits_{\varepsilon\to0}\dfrac{  e^{ i\int d^4x [J(x)+\varepsilon\delta(x-x')]\varphi(x)}-e^{ i\int d^4x J(x)\varphi(x)} }{ \varepsilon  }\\
&=\lim\limits_{\varepsilon\to0}\dfrac{  e^{ i\int d^4x [J(x)+\varepsilon\delta(x-x')]\varphi(x)}-e^{ i\int d^4x J(x)\varphi(x)} }{ \varepsilon  }\\
&=F\,\lim\limits_{\varepsilon\to0}\dfrac{  e^{ i\int d^4x \varepsilon\delta(x-x')\varphi(x)}-1 }{ \varepsilon  }\\
&=F\,\lim\limits_{\varepsilon\to0}\dfrac{  e^{ i\varepsilon\varphi(x')}-1 }{ \varepsilon  }
\end{align}
use l'Hopital's rule
\begin{align}
\lim\limits_{\varepsilon\to0}\dfrac{  e^{ i\varepsilon\varphi(x')}-1 }{ \varepsilon  }\stackrel{*}{=}\lim\limits_{\varepsilon\to0}\dfrac{i\varphi(x')  e^{ i\varepsilon\varphi(x')}}{ 1  }=i\varphi(x')
\end{align}
therefore
\begin{align}
\left[\dfrac{\delta}{\delta J}\right]^4F=\left[\varphi(y)\right]^4 F
\end{align}
then 
\begin{align}
AF+\left(-i\int d^4x \left[\dfrac{d}{dJ}\right]^4 \right)AF+...&=AF+\left(-i\int d^4x' \left[\varphi(x')\right]^4\right)AF+...\\
&=e^{ -i\int d^4x'  \big[\varphi(x')\big]^4}AF \\
&=e^{ -i\int d^4x  \big[\varphi(x)\big]^4} AF
\end{align}
Plugging in F and A, we get the expected result.
