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Let $F(x,y,z)=0$. So $x,y,z$ are defined implicitly in function of the other variable, i.e. $x=x(y,z)$, $y=y(x,z)$ and $z=z(x,y)$. Now $$dx=\frac{\partial x}{\partial y}dy+\frac{\partial x}{\partial z}dz=\frac{\partial x}{\partial y}\left(\frac{\partial y}{\partial x}dx+\frac{\partial y}{\partial z}dz\right)+\frac{\partial x}{\partial z}dz$$ and thus $$\left(\frac{\partial x}{\partial y}\frac{\partial y}{\partial x}-1\right)dx+\left(\frac{\partial x}{\partial z}+\frac{\partial x}{\partial y}\frac{\partial y}{\partial z}\right)dz=0.$$ Since $dx$ and $dy$ are linearly independent, we finally get

\begin{align*} \frac{\partial x}{\partial y}\frac{\partial y}{\partial x}&=1 \tag{1}\\ \frac{\partial x}{\partial y}\frac{\partial y}{\partial z}&=-\frac{\partial x}{\partial z}\tag{2} \end{align*}

Equation $(1)$ is natural, but equation $(2)$ should be $\frac{\partial x}{\partial y}\frac{\partial y}{\partial z}=\frac{\partial x}{\partial z}\frac{\partial y}{\partial y}=\frac{\partial x}{\partial z},$ no ? So what's the matter here ? I know it's correct, but what does it mean exactly such a contradiction ? Because at the end I get $$\frac{\partial x}{\partial y}\frac{\partial y}{\partial z}\frac{\partial z}{\partial x}=-\frac{\partial x}{\partial z}\frac{\partial z}{\partial x}=-1$$ instead of $1$ (what should be expected). What is the mystery behind ? :)

Math Lover
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user380364
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    Remember, $\frac{\partial x}{\partial y}$ is not a fraction! There are some instances where it conveniently behaves like a fraction, but it is not a fraction and there are instances like this one where assuming it should have behaved like a fraction leads to contradictions. – JMoravitz Aug 10 '18 at 21:48
  • @JMoravitz: Could you tell a condition for that $\frac{\partial x}{\partial y}$ behaves as a fraction ? – user380364 Aug 11 '18 at 09:04
  • Related: https://math.stackexchange.com/questions/942457/understanding-frac-partial-x-partial-y-frac-partial-y-partial-z-frac – Hans Lundmark Aug 11 '18 at 09:40

2 Answers2

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Here is a more symmetric explanation of this computation that hopefully makes the result less surprising. Since $F$ is constant on our surface, we have $$0=dF=F_xdx+F_ydy+F_zdz.$$ Now, how does this relate to partial derivative like $\frac{\partial x}{\partial y}$? That partial derivative is just the coefficient of $dy$ when we write $dx$ as a linear combination of $dy$ and $dz$. Solving the equation above, we have $$dx=-\frac{F_y}{F_x}dy-\frac{F_z}{F_x}dz.$$ So, $\frac{\partial x}{\partial y}=-\frac{F_y}{F_x}$. Intuitively, this makes sense: if $z$ is held constant and we vary $y$ in one direction, we need to vary $x$ in the direction that makes $F$ move in the opposite direction, to keep $F$ equal to $0$. So since $F_x$ and $F_y$ represent the directions that $F$ varies when we change $x$ and $y$, the minus sign comes from needing the changes in $F$ from $x$ and $y$ to cancel out.

But now we see immediately where your surprising minus signs are coming from. If we compute $\frac{\partial x}{\partial y}\frac{\partial y}{\partial z}$ we would get $$\left(-\frac{F_y}{F_x}\right)\left(-\frac{F_z}{F_y}\right)=\frac{F_z}{F_x}.$$ Since we lost our minus sign, this is the negative of $\frac{\partial x}{\partial z}=-\frac{F_z}{F_x}$. Using the same intuition as before, we are multiplying the change in $x$ needed to counteract a change in $y$ (and keep $F$ constant) by the change in $y$ needed to counteract a change in $z$. The two "counteracts" cancel each other out, and we end up with the change in $x$ needed to duplicate a change in $z$, rather than to counteract the change in $z$.

Ultimately, the lesson here is that derivatives are not fractions, especially not partial derivatives. Ordinary derivatives often behave like fractions (because they are limits of fractions) via the chain rule. Partial derivatives do not (and the chain rule for them does not look like just multiplying fractions!), because as explained in J.G.'s answer they are extremely sensitive to what other variables are being held constant. In particular, in your partial derivatives $\frac{\partial x}{\partial y}$ and $\frac{\partial y}{\partial z}$, $z$ is being held constant for the first one while $x$ is being held constant for the second one. So these derivatives are being computed under different "background assumptions", and it's not particularly reasonable to expect them to behave like fractions the way single-variable derivatives do.

Eric Wofsey
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  • If we let $z=y^2+x$ how we can treat $y$ as a function of $x$. It is like saying that $y^2=x−z$ is a function. – user599310 May 20 '20 at 16:31
  • @user599310: We can treat $y$ as a function of $x$ locally near any point $(x_0,y_0,z_0)$ satisifying $z_0=y_0^2+x_0$ such that $x_0-z_0$ is positive. Namely, it will either be the function $x\mapsto \sqrt{x-z_0}$ or $x\mapsto-\sqrt{x-z_0}$, depending on the sign of $y_0$. When $y=0$ you cannot locally treat $y$ as a function of $x$ (and indeed, $\frac{\partial y}{\partial x}$ is not defined, but the formula $-\frac{F_x}{F_y}$ remains "valid" in that $F_y$ is $0$ at such points). – Eric Wofsey May 20 '20 at 19:44
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Partial derivatives are only defined once you specify what's held constant. Partial derivatives' product only allows chain-rule style cancellations if they're defined with the same thing(s) assumed constant. For example, if $x,\ y,\ z$ were non-constant differentiable functions of $w$, you'd have $$\left(\frac{\partial x}{\partial y}\right)_w \left(\frac{\partial y}{\partial z}\right)_w \left(\frac{\partial z}{\partial x}\right)_w = +1,$$ where the subscript indicates the constant-$w$ condition.

The result you're trying to understand is radically different; it's $$\left(\frac{\partial x}{\partial y}\right)_z \left(\frac{\partial y}{\partial z}\right)_x \left(\frac{\partial z}{\partial x}\right)_y=-1,$$ with a condition $f(x,\ y,\ z)=0$ existing and no fourth variable involved.

As a simple example of why the choice of what to hold constant matters, compare $2$-dimensional Cartesian and polar coordinates. Holding $y$ constant, $x^2=r^2-y^2$ implies $2x\left(\frac{\partial x}{\partial r}\right)_y=2r$ so $\left(\frac{\partial x}{\partial r}\right)_y=\frac{r}{x}$; holding $\theta$ constant, $x=r\cos\theta$ implies $\left(\frac{\partial x}{\partial r}\right)_\theta=\cos\theta=\frac{x}{r}$.

J.G.
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