Confusion about second partial derivative term in Ito's lemma with a constraint on the variables Currently I am getting stuck with the stochastic process followed by forward stock prices:
Here's the rough background:
If we assume $S$ is a ito process satisfies that
$$ dS = \mu S dt + \sigma SdW_{t}$$
where $S$  stands for the stock price and $W_t$ is the wiener process. 
And we define $f$ with
$$f(S,t)=Se^{r(T-t)}$$
According to the Ito's lemma:
$$df=\frac{\partial f}{\partial t} dt+ \frac{\partial f}{\partial S} dS+\frac{1}{2}\frac{\partial^2 f}{\partial S^2} {(dS)}^2$$
So to get the $df$, we need 
$$\frac{\partial f}{\partial t} , \quad
 \frac{\partial f}{\partial S} \quad and\quad \frac{\partial^2 f}{\partial S^2}$$
Let us consider the term  $$\frac{\partial f}{\partial S} $$
According to the content on page-23 in textbook (thanks to @m_gnacik),
$$\frac{\partial f}{\partial S} = e^{r(T-t)}$$  
which implies that $S$ is independent to $t$ and that textbook solves this partial derivative with considering $S$ as a constant number relative to $t$.
But in my opinion, due to the assumption that
$$ dS = \mu S dt + \sigma SdW_{t}$$
there's obvious some connection between $S$ and $t$, so there should be a function to describe this relation. So at least 
$$S \quad is \quad independent \quad of \quad t$$
is not that intuitive to me...
I am so confused... Am I missing something here?
LAST UPDATE:
This question is similar to this one
 A: I will try to explain what have you missed. 
Let me start with a notation given a differentiable function of two variables, e.g. $f(x,t)$, the notation $$\frac{\partial f}{\partial t}(x_0, t_0) \ \ \mbox{ for some } x_0 \mbox{ and } t_0$$ stands for the partial derivative of $f$ with respect to $t$ evaluated at the point $(x_0, t_0)$.
Given your Ito process 
$$ dS_t = \mu S_t dt + \sigma S_t dW_t,$$
you consider a nice smooth function of two variables $f(x,t) = xe^{r(T-t)}$, where $T \geq 0$ and $r \in \mathbb{R}$.
In order to apply Ito's lemma you need to find
$$\frac{\partial f}{\partial t}, \ \frac{\partial f}{\partial x}, \ \frac{\partial^2 f}{\partial x^2}. $$
After you find those then you apply Ito's lemma, that is, 
$$ df(S_t, t) = \frac{\partial f}{\partial t}(S_t, t)dt + \frac{\partial f}{\partial x}(S_t,t)dS_t + \frac{1}{2}\frac{\partial^2 f}{\partial x^2}(S_t, t)(dS_t)^2.$$
Since, we know the term $dS_t$ and from Ito's table we have that $(dS_t)^2 = \sigma^2 S_t^2 dt$ we obtain that
$$\begin{align*} df(S_t, t) =& \frac{\partial f}{\partial t}(S_t, t)dt + \frac{\partial f}{\partial x}(S_t,t)\left( \mu S_t dt + \sigma S_t dW_t\right) + \frac{1}{2}\frac{\partial^2 f}{\partial x^2}(S_t, t)\sigma^2 S_t^2 dt, \\
df(S_t, t) =& \left(\frac{\partial f}{\partial t}(S_t, t) + \mu S_t\frac{\partial f}{\partial x}(S_t,t)+ \frac{1}{2}\sigma^2S_t^2\frac{\partial^2 f}{\partial x^2}(S_t, t) \right) dt + \sigma S_t \frac{\partial f}{\partial x}(S_t,t)dW_t.
\end{align*}$$
Therefore, because you evaluate partial derivatives at fixed point $(S_t(\omega), t)$ (for our convenience we often abuse the notation, that is, $S_t$ stands for $S_t(\omega)$ ), then there is no dependence on $t$ while you differentiate with respect to $x$. Sometimes you can find the expression $\frac{\partial f}{\partial S}$ which is the notation for $\frac{\partial f}{\partial x}(S_t,t)$, but as you noticed it may be confusing.   
The last thing:
$$ \frac{\partial f}{\partial t} (x,t)= -rxe^{r(T-t)}$$
not $e^{r(T-t)}$.
Update.
It appeared that the confusion of the author of the question came from lack of understanding partial derivatives rather than stochastic tools used in Ito's calculus. There are some examples in the below comments in which I tried to explain @AoSun doubts about partial differentiation.  
