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Consider the function $f(x(s,t),y(s,t))$. Then, the partial derivative with respect to t is defined as:

$\frac{\partial f}{\partial t} = \frac{\partial f}{\partial x}*\frac{\partial x}{\partial t} + \frac{\partial f}{\partial y}*\frac{\partial y}{\partial t}$

Taken as a whole, this formula makes sense to me as you are considering how the function changes when you change $x$ (holding $y$ constant) by changing $t$. Then, you look at how the function changes when you change $y$ (holding $x$ constant) by changing $t$. After doing so, you add both these changes together.

What I don't understand are the terms $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ taken individually since it doesn't make sense to me that we can change $x$ holding $y$ constant or vice versa when $x$ and $y$ are not independent variables.

Are we pretending they are independent for the sake of calculating those partial derivatives? Why can we do so?

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    $\begingroup$ I will make it more visual. If $x$ and $y$ have increments $\Delta x$ and $\Delta y$ separately, then the increment of $f$ is $\dfrac{\partial f}{\partial x}\Delta x+\dfrac{\partial f}{\partial y}\Delta y+o(\sqrt{{\Delta x}^2+{\Delta y}^2})$. It does not matter whether $\Delta x$ and $\Delta y$ are dependent. $\endgroup$ Commented Nov 6 at 15:36

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Actually, the pretence that is going on is a pretence in the opposite direction: by taking lots of notational shortcuts, we get into a kind of pretend version of the chain rule.

Let me explain the standard formal notation of the chain rule, which avoids all of the confusions.

Suppose we are given three two-variable functions: $f(x,y)$; and $g(s,t)$; and $h(s,t)$. We can then form a composite function $$F(s,t) = f(g(s,t),h(s,t)) $$ In this situation we have the chain rule formula $$\frac{\partial F}{\partial t} = \frac{\partial f}{\partial x} \cdot \frac{\partial g}{\partial t} + \frac{\partial f}{\partial y} \cdot \frac{\partial h}{\partial t} $$ This should be quite straightforward.

One thing to notice: all symbols are either independent variables ($s,t,x,y$) or function symbols ($g,h,f,F$). No dependent variables at all.

How can we make it confusing?

Let's introduce some confusing notational shortcuts!

First, each of those independent variables $x,y$ for the function $f$ could serve a double role: $x$ could also be the dependent variable of the function $g$, and $y$ could also be the dependent variable of the function $h$: $$x = g(s,t) \qquad y = h(s,t) $$ And then, why bother at all with those pesky function symbols $g$ and $h$? Let's just let $x$ and $y$ take on triple roles: independent variables for $f$; dependent variables for those pesky functions; and, heck, let's make them the names for those pesky middle functions themselves!!

And, while we're at it, who cares that $F$ is really a function of $s$ and $t$ whereas $f$ is really a function of $x$ and $y$. They're really the same, so let's confuse them too by replacing $F$ with $f$.

And voila, we get the pretend version of the chain rule in your post: $$\frac{\partial f}{\partial t} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial t} $$ The thing is, this pretend formula is kind of nice, because it's easy to remember. So it does serve a purpose.

But, its purpose is only served if you are able to apply it corectly. Perhaps you can learn to use the shorcut notation correctly. But perhaps, when doing a computation, you may need to undo all of the confusing notational shortcuts in order to actually compute the various partial derivatives correctly. If so then you can always revert to the first, formal version of the chain rule.

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  • $\begingroup$ When you defined F(s,t) = f(g(s,t),h(s,t)), didn't x and y lose independence there since that indirectly implies x = g(s,t) and y = h(s,t)? $\endgroup$ Commented Nov 6 at 18:24
  • $\begingroup$ Never mind, I think you substitute x = g(s,t) and y = h(s,t) after taking the partial derivatives of f w.r.t x or y. Let me know if this makes sense. $\endgroup$ Commented Nov 6 at 18:37
  • $\begingroup$ If that makes sense to you then it's fine. But what I actually did when writing out the formal version of the chain rule was to ignore the unhelpful variable labels "dependent/independent". Instead, I changed the focus to the functions themselves --- which, by the way, is what happens in more advanced calculus classes. $\endgroup$ Commented Nov 6 at 18:50

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