The partial derivative of a call option with respect to $t$ [closed]

In Black-Scholes related computations, why do we not treat the stock price $$S$$ as a function of $$t$$ when taking partial derivatives with respect to $$t$$? For example, if $$c(t,T)=SN(d_1)-Ke^{-r(T-t)}N(d_2)$$ is the price of a call option and we want to find $$\partial C/\partial t$$, we never include the term $$\partial S/\partial t$$ and don't consider $$S$$ as a function of $$t$$, but as a separate variable. How can this be justified?

• S is just the current observed spot price.
– user70573
Feb 28 at 3:57
• So S doesn't change over time? Feb 28 at 4:01
• What's the partial derivative of x*y with respect to y? By definition, a partial derivative is holding all else equal.
– user70573
Feb 28 at 4:07
• I get your point; however, S = S(t), so how can we ignore its partial derivative. What if, in your example, x is a function of y? x(y)*y Feb 28 at 4:08
• It's not a function x(y) though, it's just S. What about IV? Surely S is dependent on VOL? Yet, it's not in the closed form solution, which looks at current S. That's why S is a constant in computing theta or vega. You also hold S constant when you compute a bump and reprice theta. Feb 28 at 4:15

But regarding your question : It depends on what you mean by "justified" and how you are going to use the partial derivative. Yes, $$S(t)$$ depends on t. However, the process $$S(t)$$ here is random and almost nowhere differentiable so taking the derivative would not make much sense in the first place.
The reason we don't take the time derivative of $$S(t)$$ here though is that we consider formal partial derivatives and not full derivatives. As a simple example, let us say $$S(t) = t^2$$ and we consider the function $$z(t) = S(t) \cdot t = S \cdot t$$ We now want to calculate $$\partial{z} / \partial{t}$$. What would you say this is? It actually depends on how you represent $$z(t)$$. If you use the above representation you get $$\partial{z} / \partial{t} = S = S(t) = t^2$$, but if you represent $$z(t)$$ as $$z(t) = t^3$$ you instead get $$\partial{z} / \partial{t} = 3 \cdot t^2.$$ When calculating the formal partial derivatives, you see the variables as placeholders. There is no contradiction here. And as I mentioned in the beginning, the justification depends on how you are using those partial derivatives later.
So in the Black-Scholes formula above, we see the value of $$S = S(t)$$ as such a placeholder for the value of the process $$S(t)$$ at a fixed $$t$$.