partial derivative question with a context from a stochastic process

the simple question is
s ∂f(x,t)/∂t the same with ∂f(x(t),t)/∂t?

given that x is dependent on t, namely x=x(t)

for example, say x(t)=2t, f(x,t)=x+t

is ∂f(x,t)/∂t= 1 or 3?

the context of this question is from a stochastic calculus class where my professor was able to do maneuver a stochastic process f, and say like f(t,w(t))=e^(-1/2(σ^2)t+σw(t)) and my professor would replace w(t) with x, (w(t) is a brownian motion) so it becomes f(t,x)=e^(-1/2(σ^2)t+σx) and he would treat x as a variable that is independent of t. and calculate partial of f(x,t) respect to t, and treat x as a constant. I guess this kinda puzzles me a little because i thought that w(t) is dependent on t, but when he change the notation of w(t) to x, x is treated as a constant when calculating the partial derivative. Is my professor able to do this kinda of maneuver only because the nature of w(t)? or is this something you can freely do for partial derivatives?

 

yes, you can treat all other variables as if they are constants
here are some examples:
https://www.mathsisfun.com/calculus/derivatives-partial.html

 

You know Calculus 3? I'm very impressed. I will surely need your help when I get there. I expect to be there in 2023.

 

If x= 2t and f= x+ t then partial f/partial t= 1. Of course, since x= 2t, f(t)= 3t and df/dt= 3 but that is not the partial derivative.

By the chain rule, df/dt= partial f/partial t+ (partial f/partial x)(dx/dt)= 1+ 1(2)= 3. In order for that to be true we must have both df/dt and df/dx equal to 1.