I'm having a hard time understanding why \begin{equation}\frac{\partial x}{\partial y}\frac{\partial y}{\partial z}\frac{\partial z}{\partial x}=-1 \tag{1}\label{1} \end{equation} Wikipedia provides this derivation. I have two problems with it.
The proof starts by stating that there is a function f such that $f(x,y,z)=0$ and that $z$ can be made a function of $x,y$. Furthermore it states that there can be found a curve, along which $dz=0$ and $y$ is a function of $x$, such that we can then write the differential of $z$ in terms of the differential of $x$ as $$dz=\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}\frac{\partial y}{\partial x}dx$$
The rest follows naturally from setting $dz=0$ and multiplying some partial derivatives by their inverses.
I have two problems with this proof

1.Chain rule

The first one is that since \begin{equation} \tag{2} \label{2} \frac{\partial z}{\partial x}=-\frac{\partial z}{\partial y}\frac{\partial y}{\partial x} \end{equation} That would mean that, by chain rule, $$\frac{\partial z}{\partial x}=-\frac{\partial z}{\partial x}$$ which would imply this partial derivative to be zero. However if this is true, \ref{1} yields $0=-1$. Is the chain rule not valid in this case? If so, why?

2.Inverse of the partials
The second is that, while applying the last step, it is implied that we obtain \ref{1} by multiplying by the inverse of the righthand-side in \ref{2}. I thought the relationship $$\frac{\partial y}{\partial x}=\frac{1}{\frac{\partial x}{\partial y}}$$ was in general not true, as pointed out in this post. Is it true in this case? And if so, why is that?
Also, if that really is the case, then using the chain rule again yields, from \ref{1}, $$\frac{\partial x}{\partial z}\frac{\partial z}{\partial x}=-1 \iff 1=-1$$ What am I doing wrong?

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    $\begingroup$ Your main problem is lack of context. $x,y,z$ do not automatically have any relationships, so those partial derivatives do not have any meaning unless you set something up. And if (for example) $z$ is assumed to be a function of $x,y$, then, still, $\partial x/\partial z$ has no meaning, without further assumptions. You are in a situation where the natural symbol heuristics are misleading, unfortunately... Can you give your larger context? $\endgroup$ Jan 9, 2022 at 23:15
  • $\begingroup$ the context is the context of the Wikipedia demonstration that I linked at the top of the question. Maybe I should edit my post to make it clearer? But anyway, the context is, given $f(x,y,z)=0$, consider a path along which, writing $z$ as a function of $x$ and $y$, $dz=0$. They further parametrize $y$ with $x$ and write out the differential of $z$ in the basis of the differential of $x$, from whence everything else follows $\endgroup$ Jan 9, 2022 at 23:21
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    $\begingroup$ Ah, yes, please do insert the context. The style on this site, for better or worse, is to ignore links... $\endgroup$ Jan 9, 2022 at 23:22
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    $\begingroup$ Suppose $x+y+z=0$. Then, informally, ${\partial z\over\partial x}=-1$, ${\partial z\over\partial y}=-1$, and ${\partial y\over\partial x}=-1$, so you see that (2) is correct, and what you are calling the chain rule doesn't apply. $\endgroup$ Jan 9, 2022 at 23:22
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    $\begingroup$ I suggest you read this post and this post instead of Wikipedia. :) $\endgroup$ Jan 9, 2022 at 23:28

2 Answers 2


At least as a place-holder, in line with @GerryMyerson's apt comment:

Yes, there is an appealing heuristic that suggests that ${\partial z\over \partial x}={\partial z\over \partial y}{\partial y\over \partial x}$, ... and such things.

In a different universe, it might not matter that these named variables were related by $f(x,y,z)=0$ or $z=f(x,y)$ or some other relation. But, in our universe, this does have some relevance.

Even in a simpler situation, $f(x,y)=0$, whether or not we rename $f$ to $z$, a person might imagine that (via some sort of implicit function theorem, making $y$ a function of $x$) ${\partial y\over \partial x}={\partial f\over \partial x}/{\partial f\over \partial y}$... but that's off by a sign!?!?! :)

Careful application of the chain rule corrects the sign. :)

EDIT: When $y$ is (locally) defined as a function of $x$ by a relation $f(x,y)=0$, differentiating this with respect to $x$ gives $$ 0 \;=\; f_1(x,y)\cdot {dx\over dx} + f_2(x,y)\cdot {dy\over dx} \;=\; f_1(x,y)+f_2(x,y){dy\over dx} $$ where $f_i$ is the partial derivative of $f$ with respect to the $i$-th argument. This gives $$ {dy\over dx} \;=\; -{f_1(x,y)\over f_2(x,y)} $$ If we somewhat-abuse notation by thinking that $f_1=f_x$ and $f_2=f_y$, then this would be $$ {dy\over dx} \;=\; -{{\partial f\over \partial x}\over {\partial f\over \partial y}} $$ which does not give the expected heuristic outcome, being off by a sign. :)

  • $\begingroup$ Could you clarify how "careful application of the chain rule" corrects the sign? $\endgroup$ Jan 9, 2022 at 23:40
  • $\begingroup$ Thank you :) However I still feel a bit uneasy, as I previously explained in my comments to Ted and Gerry. I feel like thinking "heuristically" should yield the right results, and I'm still struggling on why it isn't. Why is the situation in question different from just having a function $z(x,y)$ and a parametrization $\lambda(x)=y(x)$? I mean, I know what is different, the relationship $f(x,y,z)=0$, but what I mean is: why does having this constraint alter a chain rule which should be general for any two function and parametrization( given they are differentiable)? $\endgroup$ Jan 10, 2022 at 0:13
  • $\begingroup$ @LourencoEntrudo, I think I do understand/sympathize with your question/concern... as this was a fact that disturbed me years ago, as well. Apparently, this is a fairly extreme case of a symbol-oriented heuristic being significantly "off" from the truth. "Only by a sign", ... :) In general, we would indeed want our notation to suggest only what is true, but this is not universally possible... apparently! :) $\endgroup$ Jan 10, 2022 at 0:55
  • $\begingroup$ It is not so much a matter of notation as it is a matter of understanding why the derivative of the composition of two functions is not given by the chain rule $\endgroup$ Jan 10, 2022 at 1:08
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    $\begingroup$ @LourencoEntrudo, the composition of two functions is given by the chain rule, but we can accidentally choose a notation that presents functions misleadingly. $\endgroup$ Jan 10, 2022 at 19:07

It has come to me with all your help that my confusion was only a matter of damned notation and that in fact the chain rule is not broken in the implicit function theorem. It all boils down to what the original wikipedia article calls partial derivative and the way I also thought of it. For me, partial differentiation is always when only one parameter is free to move. It is the natural way of defining it in real analysis. Everything else is just a derivative of the composition of a function with a parametrization (what the physicists like to call the "total derivative"). And that is the derivative that is being used and that eluded me. So as an exercise and to check that I have understood everything, I will answer my own post and "correct" the enunciation of the triple product.
Let $f=f(x,y,z)$, $z=z(x,y)$ and $y=y(x)$. Consider the differential of z, $$dz=\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}dy$$ This is the definition of the differential. But now , when considering the differential of y, we shouldn't write $dy=\frac{\partial y}{\partial x}dx$, but $dy=\frac{d y}{d x}dx$. This is the notation I was used to and a notation which indicates that not only x is being free to vary: we are taking the derivative of y=y(x,z) composed with z=z(x). We can even see the relationship of this derivative with $\frac{\partial y}{\partial x}$, applying the chain rule. $\frac{dy}{dx}=\frac{\partial y}{\partial x}+\frac{\partial y}{\partial z}\frac{dz}{dx}$. This makes sense. y is a function of x, but also of z (were it not the case, $\frac{\partial y}{\partial z}$ in the triple product wouldn't even make sense). Then, we write $$dz=\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}\frac{dy}{dx}dx$$ Moving now along a path where $dz=0$ $$\frac{\partial z}{\partial x}=-\frac{\partial z}{\partial y}\frac{dy}{dx} \tag{3} \label{3}$$ Now, to address my (1). Of course the right hand side is not $\frac{\partial z}{\partial x}$.$\frac{\partial z}{\partial x}$ would be $\frac{\partial z}{\partial y}\frac{\partial y}{\partial x}$, where the difference has been discussed earlier.
To adress (2), the relationship is in general not true, but, in this case differentiating the function $f$ with respect to $y$ and $x$ and comparing the two will show that $\frac{\partial z}{\partial x}\frac{\partial x}{\partial z}=1$ and so we write the "corrected" formula, $$-1=\frac{\partial x}{\partial z}\frac{\partial z}{\partial y}\frac{dy}{dx}$$ I hate thermodynamics

  • $\begingroup$ Historical footnote: The mathematisation of thermodynamics with all its partial derivatives is due in great part to Poincaré (a mathematician of the years 1880-1910 you may know) $\endgroup$
    – Jean Marie
    Jan 10, 2022 at 7:57
  • $\begingroup$ Besides, a similar question here with an interesting answer. $\endgroup$
    – Jean Marie
    Jan 10, 2022 at 8:01
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    $\begingroup$ I really don't like the notation of a partial derivative where you allow more than one variable to move. It's automatically misleading because there seems to be no sense in which you can do it precisely without relating the variables in question, in which case it is better represented by a one dimensional derivative, where you compose the function of several variables with the parametrization of every variable(except the one to which respect you're differentiating) with the variable you're taking a derivative out of. Honestly I might write a separate post to discuss that $\endgroup$ Jan 10, 2022 at 13:08

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