I have found two different ways of doing this and I am seeking commentary on the fine nuance.  Suppose there is a Hamiltonian

$$ H=\frac{1}{2}\int\!d^3x \left[   \pi^2+(\nabla\varphi)^{\!2}+m^2\varphi^2  \right] ~~. \tag{1}$$

Hamilton's equations are

$$  \dot\pi=-\dfrac{\delta H}{\delta \varphi} \qquad\text{and}\qquad \dot\varphi=\dfrac{\delta H}{\delta \pi}~~. \tag{2} $$

It follows that 

$$  \dot \varphi=\int\!d^3x\,\pi~~, \tag{3}$$

but I have a question about how to compute $\dot\pi$.  If I trivially take the derivative with respect to $\varphi$ as $$\frac{\delta}{\delta\varphi}\equiv\frac{\partial}{\partial\varphi},\tag{4}$$ skipping over the gradient term, I get the wrong answer $\dot\pi=\int d^4x\, m^2\varphi$.  I have found two different ways to get the correct answer.  The first is to use the identity

$$ (\nabla\varphi)^2=\nabla(\varphi\nabla\varphi)-\varphi\nabla^2\varphi ~~,\tag{5}$$

to rewrite the Hamiltonian as 

$$ H=\frac{1}{2}\int\!d^3x \left[   \pi^2+\nabla(\varphi\nabla\varphi)-\varphi\nabla^2\varphi +m^2\varphi^2  \right]  ~~.\tag{6}$$

Now when I take the partial with respect to $\varphi$ I get the correct answer

$$ \dot \pi=\int\!d^3x\, \left[\nabla^2\varphi-m^2\varphi\right]~~.  \tag{7} $$

Mainly my question is this: What rule is it that dictates that $\frac{\delta}{\delta\varphi}$ has to hit the $\nabla\varphi$ term?  Something about the linearity of the operators, I am sure, but I am not certain exactly what my reasoning is.   I found another way to compute the correct answer and that is also what my question is about because ***the other method suggests the derivative does not have to hit that term***.  I saw something on the internet that says Hamilton's equations for fields use the "functional derivative"

$$  \dfrac{\delta}{\delta \varphi} \equiv\dfrac{\partial}{\partial \varphi}-\nabla\dfrac{\partial}{\partial (\nabla \varphi)} ~~.\tag{8}$$

I can apply this to the original Hamiltonian as 

\begin{align}
 \dot\pi=-\dfrac{\delta}{\delta \varphi} H&=-\left(\dfrac{\partial}{\partial \varphi}-\nabla\dfrac{\partial}{\partial (\nabla \varphi)}   \right)\frac{1}{2}\int\!d^3x \left[   \pi^2+(\nabla\varphi)^{\!2}+m^2\varphi^2  \right]\\
&=-\int\!d^3x \,m^2\varphi+\int\!d^3x\, \nabla\nabla\varphi\\
&=\int\!d^3x\,\nabla^2\varphi-m^2\varphi\tag{9}
\end{align}

Once I have to correct $\dot\varphi,\dot\pi$, I can easily proceed to derive the KG equations of motion.  I am seeking input regarding the relative merits and uses of these two procedures for computing $\dot\pi$.  Initially, I got the wrong answer because I thought $\frac{\delta}{\delta\varphi}(\nabla\varphi)=0$ and I want to better understand why I was wrong.