Suppose that $f(x,t) = x^2$. Clearly, $\frac{\partial f}{\partial t} = 0$.
However, let us now consider $f(X(t),t) = X(t)^2$. The book I am reading claims that $\frac{\partial f}{\partial t}(X(t),t) = 0$. I am wondering why this is so since $X(t)$ is a function of $t$ as well.
To put some context into the above problem, I have been reading a stochastic calculus book which uses the above fact for Ito's formula for $f(X(t),t)$. The book makes a distinction for the special case of $f(X(t))$ which is what I have in the example above since $X(t)^2$ does not depend on $t$ explicitly but implicitly through $X(t)$.
Hoping for someone to clarify my confusions on the matter. Thanks!
The notation $\frac{\partial f}{\partial t}f(X(t),t)$ is somewhat misleading but sadly commonly used. It means that you have to differentiate $f$ with respect to the second variable and evaluate the result at $(X(t),t)$. Since $f$ does not depend on the second variable, the result is 0. It's the same thing with the definition $f(X(t),t)=X(t)^2$. This means that $f(x,y)=x^2$, so $f(X(t),t)=X(t)^2$. I would probably define $f$ by $f(x,y)=x^2$ and then write $\frac{\partial}{\partial y}f(x,y)|_{x=X(t),y=t}=0$ to make all things clear.