Pointwise functional derivative as partial derivative Suppose $x_{1},...,x_{n} \in \mathbb{R}^{d}$ are fixed and $f: \mathcal{S}(\mathbb{R}^{d}) \to \mathbb{C}$ is given by: 
$$ f(\phi) = e^{\sum_{j=1}^{n}\alpha_{j}\phi(x_{j})}$$
with $\alpha_{1},...,\alpha_{n} \in \mathbb{C}$. I'd like to know if there is any kind of derivative whose derivative of $f$ would be:
$$ \sum_{j=1}^{n}\alpha_{j}e^{\sum_{j=1}^{n}\alpha_{j}\phi(x_{j})}$$
In other words, it would be as if we're considering partial derivatives $\partial/\partial \phi(x_{j})$, $j=1,...,n$, such that
$$\sum_{j=1}^{n}\frac{\partial}{\partial \phi(x_{j})} e^{\sum_{j=1}^{n}\alpha_{j}\phi(x_{j})} = \sum_{j=1}^{n}\alpha_{j}e^{\sum_{j=1}^{n}\alpha_{j}\phi(x_{j})}.$$
Is there a way to define such objects in a precise/rigorous way? 
 A: A functional derivative is what you need:
$$
\begin{split}
\frac{\partial f}{\partial \phi} &\triangleq \left.\frac{\mathrm{d}}{\mathrm{d}\varepsilon} f(\phi+\varepsilon\varphi)\right|_{\varepsilon=0} = \left.\frac{\mathrm{d}}{\mathrm{d}\varepsilon}e^{\sum_{j=1}^{n}\alpha_{j}[\phi(x_{j})+\varepsilon\varphi(x_i)]}\right|_{\varepsilon=0}\\
& =\sum_{j=1}^{n}\alpha_{j}\varphi(x_j)e^{\sum_{j=1}^{n}\alpha_{j}\phi(x_{j})}
\end{split}
$$
and choosing $\varphi\in\mathcal{S}(\Bbb R^d)$ such that $\varphi(x_j)=1$ for all $j=1,\ldots,n$ does the Job.
Notes


*

*The choice of the variation $\varphi$. Since $\{x_j: j=1,\ldots,n\}\subsetneq\Bbb R^d$, the choice of a suitable $\varphi(x)$ is always possible and moreover, you could extend the above formula even to the case  $j\in\Bbb N_+=\Bbb N\setminus \{0\}$, provided the indexed (numerable) family of points $\{x_j\}_{j\in\Bbb N_+}$ is contained in a suitable bounded set $M\subsetneq\Bbb R^d$: in all these cases, you can choose a sufficiently large compact set $K\Supset M$ and mollify its characteristic function by the heat kernel (if you do not want a compactly supported function), i.e.
$$
\varphi(x)=\varphi(x,t)=\int\limits_K \mathscr{K}(x-y, t)\,\mathrm{d}y
$$
where
$$
\mathscr{K}(x, t)=\frac{1}{\sqrt{(4\pi t)^d}}e^{-\dfrac{\langle x,x\rangle}{4t}}.
$$

*On the domain of definition of $f$(and thus again on $\varphi$). As user131781 noted in the comments, $f$ is defined on a much wider space than the "sole" $\mathcal{S}(\Bbb R^d)$. Its domain is at least the whole $\mathcal{E}(\Bbb R^d)\equiv C^\infty(\Bbb R^d)$, thus also the functional derivative $\frac{\partial{f}}{\partial{\phi}}$ is defined and meaningful  for every $\varphi\in C^\infty(\Bbb R^d)$ and the choice $\varphi(x) \equiv 1$ is perfectly licit. However, the choices of $\varphi$ described above show that the very same solution holds in more stricter spaces where, perhaps by unstated context consideration, the functional $\varphi$ is bounded to act.  

