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EDIT : I had earlier asked four doubts together and I thought that maybe I should ask some of them in a separate question.

I have some very basics doubts about partial differentials ....

Doubt #1 : What is the correct mathematical interpretation of the statement "$x$ and $y$ are independent variables" ? Does it mean $\large \frac {\partial x}{\partial y}=\frac{\partial y}{\partial x}=0$ ? According to my understanding of independent variables, if $x$ and $y$ are given to be independent, then any change in $y$ and $y$ only should not cause any change in $x$. So $ \large \frac{\partial x}{\partial y}$ must be zero. Similar logic for $\large \frac {\partial y}{\partial x} =0$. Is it okay to use this logic ?

Doubt #2 : : Under what circumstances is $\large\frac{\partial x }{\partial y} = \Large \frac{1}{\frac{\partial y}{\partial x}}$ valid?

These doubts have become a real hinderance to my understanding of partial derivatives. I need some insightful examples so that I can understand the subtleties that I am currently overlooking.

Rest of the doubts can be found here : Some basic doubts about partial differentiation - PART II

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  • $\begingroup$ This is a rather interesting question, although you seem to be bunching up a lot of sub-questions together, which is prompting people to close this as "too broad". You can ask them separately instead. $\endgroup$ – YiFan Oct 5 '19 at 14:55
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Let me try so answer your questions, which seem to me interesting.

3) This is always true, for any differentiable function $f$.It doesn't even matter if your variables $x_1,x_2,\dots$ are independent or depend on some other variables.

4) The first one is true. If you call, say, $g(x,y)=h(x,y,z(x,y))$ then $$ dg=\frac{\partial g}{\partial x}dx+\frac{\partial g}{\partial y}dy $$ BTW if you replace $dz$ in the expression for $dh$ by $\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}dy$ and you group the terms containing $dx$ and $dy$, you obtain formulas for the partial derivatives of $g$ in terms of the partials of $h$ like $$ \frac{\partial g}{\partial x}=\frac{\partial h}{\partial x}+\frac{\partial h}{\partial z}\frac{\partial z}{\partial x} $$ and a similar expression for $\frac{\partial g}{\partial y}$ (chain rules for partial derivatives).

$h$ is a function of $x,y,z$ and $\frac{\partial h}{\partial z}$ makes perfect sense. You cannot assume any extra dependence $z=z(x,y)$ here. In the chain rule above,$\frac{\partial h}{\partial z}$ is taken at the point $(x,y,z(x,y))$.

2) This is not valid in general. For ex. if you have the relation $$ zx+z^2y=1, $$ Then if you think of $x$ as a function of $y$ and $z$ you get $$ \frac{\partial x}{\partial y}=-z $$ but if you think of $y$ as a function of $x$ and $z$ you get $$ \frac{\partial y}{\partial x}=-1/z $$ Observe that when you write something like $\frac{\partial x}{\partial y}$ you are assuming that $x$ is a function of $y$ and something else, unless your derivative is a total derivative, in which case it would be better to denote it as $\frac{dx}{dy}$. In this case your relation holds, as well as in some special cases where none of the partial derivatives depends on the other variable, like in the relation $$ z(x+y)=1 $$ where both are equal to $-1$.

As for (1), I think that it is better to think of independent variables as those that can be varied independently, whereas the change of the dependent variable becomes fixed once you fix the changes of the independent ones. In any relation between variables, you are free to choose which ones you can consider independent (typically, their number is the total number of variables minus the number of relations (equations) relating them).

In general, it is better to indicate the variables that you fix in the partial derivatives, as it is done in Thermodynamics.

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  • $\begingroup$ Since points 3 and 4 of the original question have been moved to a new question, please consider splitting your answer and posting your replies to 3 and 4 at the other question. $\endgroup$ – Daniel Fischer Oct 10 '19 at 13:04
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To address the first question, I think the answer is that there is no formal definition of what $x$ and $y$ being independent variables means. What one has in mind when one says that $x,y$ are independent is that if we know $x$, we do not necessarily know $y$, and vice versa: in other words $x,y$ can take on different values "independently" of the other. So for instance, say we wanted to find the minima and maxima of $$f(x,y)=\sin\left( x^2\right)+\cos(xy^2)$$ we can say that $x,y$ are independent, but if we were to do so under the additional restriction that $x=y^2$, then they are no longer independent because knowing the value of $x$ tells us the value of $y$ and vice versa.

You also want to take note that it doesn't really make much sense to make a statement like $$\frac{\partial x}{\partial y}=\frac{\partial y}{\partial x}=0.$$ A notation like $\frac{\partial f}{\partial x}$ means taking the partial derivative of $f$ with respect to $x$, but here what are we taking the partial derivatives of? A constraint like the example $x=y^2$ given above? In that case, it is certainly not true: as long as we have such a constraint, then the variables are already not said to be independent.

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