I see summation of a sequence written as a function like this:
$f(x) = \sum_{i=1}^{n} x_i $
for $ x_i \in [0,1]$
However $x_i$ are a sequence and yet no such sequence is defined and the function only takes a single $x$ as input. How would you differentiate this function?
As noted by @Eff the function, as stated in your problem appears to be the following $$ f:[0,1]^n \rightarrow \mathbb{R},\quad f(\vec{x}) = \sum_{i=1}^n x_i$$ i.e., this is a real valued function from the $n$-dimensional unit cube into the reals. What seems to have confused you (and others at the beginning of their studies) is the lac of a distinction between a vector $\vec{x}$, which is used as the argument of the function and a scalar $x$. There has been a discussion about that in this MSE post. In short, vectors are often typeset in the same way as scalars and it is assumed that it is obvious from the context which one is meant.
Differentiation of the above function, for example, will be done element-by-element.
In this case, $f(\mathbf x)$ is a function defined as $$f(\mathbf x):\Bbb R^n \rightarrow \Bbb R, f(\mathbf x) = \sum_{i=1}^{n}x_i$$ Where $\mathbf x = (x_1,x_2,....,x_{n-1},x_n)$
Now we would have to consider a notion of gradient ,which is the generalization of our classic differential for single variable calculus. We first define partial derivatives of $f$ in each $x_i$ direction, as $$\frac{\partial f(\mathbf x)}{\partial x_i} = \lim_{h \to 0}\frac{f(\mathbf x + h\mathbf u_i) - f(\mathbf x)}{h}$$ Here, $\mathbf u_i$ represents the unit vector of the $x_i$ direction, i.e. $\mathbf u_i = (0,0,...,1_i,0,..,0)$
Now, we define the gradient as a scalar to vector function, as $$\nabla f(\mathbf x) = \sum_i\frac{\partial f(\mathbf x)}{\partial x_i}\mathbf u_i$$
One can easily check this for single variate case, it matches our notion of differential
