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If I have a function $f(x,t)=x$, and I also happen to know that $x=t^2$, then the partial derivative

$$\frac{\partial f}{\partial t} = 0$$

Now, I get that the idea is 'hold $x$ constant, take the derivative with respect to $t$', but I still keep making the mistake of thinking the chain rule should apply, since I know $x$ can't be held constant while varying $t$.

When I find myself thinking that the chain rule should apply, it's because I'm thinking $x$ is a function of $t$, and so $f(x(t),t)=g(t)=t^2$.

Should I not be thinking of it this way? Because $f(x,t)$ and $g(t)$ are completely different beasts, even if they take the same value for all $t$?

I think I'm getting confused about the difference between functions and variables as well. In the above, $x$ isn't a function of $t$, correct? $x$ and $t$ are both just variables and I happen to have a relation/constraint between them? If $x$ were a function, so we had $f(x(t),t)=x(t)$ and I were asked to calculate $\partial f /\partial t$, then would the chain rule apply, because f is a composed function of $t$, or would it be ambiguous whether we were thinking of f as a function of one variable, or as a function of two variables with a relation between them?

Andrew
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1 Answers1

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but I still keep making the mistake of thinking the chain rule should apply, since I know x can't be held constant while varying t.

You shouldn't be thinking of it as $x=t^2$ is being held constant, but the first coordinate of the function $f(x,t)$ is being held constant. When you take the partial derivative with respect to a specific coordinate, you treat all other coordinates as fixed, regardless of their dependence on the others. So in your example $f(x,t)=x$, even though $x=t^2$, when you take the partial with respect to the second argument you treat the first argument, $x$, as a constant. It may seem weird, but your asking how does the function change as $t$ changes as compared to how does it change when $t^2$ changes.

With $f(x,t)=xt=t^3$, where $x=t^2$, you would have $\frac{\partial f}{\partial x}=t$ while $\frac{\partial f}{\partial t}=t^2$.

user162520
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