velocity - Functional derivative in Lagrangian field theory

The following functional derivative holds:

$$\begin{align} \frac{\delta q(t)}{\delta q(t')} ~=~ \delta(t-t') \end{align}$$ and $$\begin{align} \frac{\delta \dot{q}(t)}{\delta q(t')} ~=~ \delta'(t-t') \end{align}$$ where $'$ is $d/dt$.

Question: What is

$$\begin{align} \frac{\delta q(t)}{\delta \dot{q}(t')}? \end{align}$$ I'm asking this because in QFT, the lecturer defined the canonical momentum field to $\phi$ by $$\begin{align} \pi(x,t) ~:=~ \frac{\delta L(t)}{\delta \dot{\phi}(x,t)}, \end{align}$$ where $L$ is the Lagrangian, a functional of the field: $L[\phi,\partial_\mu \phi] = \int d^d x \mathcal{L}(\phi,\partial_\mu \phi)$.

I know I should get

$$\begin{align} \pi(x,t) ~=~ \frac{\partial \mathcal{L(x,t)}}{\partial \dot{\phi}(x,t)}. \end{align}$$ (Note it's now a partial derivative with respect to the Lagrangian density.) But doing it I get: $$\begin{align} \delta L = \int d^dx \frac{\partial \mathcal{L}}{\partial \phi} \delta \phi + \frac{\partial \mathcal{L}}{\partial \partial_\mu \phi} \delta\partial_\mu \phi. \end{align}$$ So somehow we ignore the first term $\int d^dx \frac{\partial \mathcal{L}}{\partial \phi} \delta \phi$! Why is that?

It can't be that we are treating $\delta \phi$ and $\delta \dot{\phi}$ as independent because if I were to take the functional derivative w.r.t. $\phi(x')$, I would have to move the dot from $\delta \dot{\phi}$ over to $\frac{\partial \mathcal{L}}{\partial \dot{\phi}}$ which will give me

$$\int d^dx (\frac{\partial \mathcal{L}}{\partial \phi} - \partial_\mu\frac{\partial \mathcal{L}}{\partial \partial_\mu \phi}) \delta \phi$$

i.e. the functional derivative gives the Euler-Lagrange equations.

So how to I take the functional derivative of a functional w.r.t to the derivative of a function?