Is $\frac{\partial x}{\partial y}=\frac{\partial x}{\partial z}\frac{\partial z}{\partial y}$?

Question

Suppose that $x=x(y,z)$, $y=y(z,x)$ and $z=z(x,y)$.

Do we have that $$\frac{\partial x}{\partial y}=\frac{\partial x}{\partial z}\frac{\partial z}{\partial y}$$ or it's false ? Moreover, I don't understand why $$\frac{\partial x(y,z)}{\partial y}\frac{\partial y(z,x)}{\partial z}\frac{\partial z(x,y)}{\partial x}=-1.$$

Answer 1

Take $z=xy$, then $x=\dfrac zy$ and $y=\dfrac zx$.

Then

$$\frac{\partial x}{\partial y}=-\frac z{y^2}\ne\frac{\partial x}{\partial z}\frac{\partial z}{\partial y}=\frac1yx.$$

Answer 2

Presumably you are differentiating implicit functions, subject to some equation $F(x,y,z)=0$ that you haven't told us about.

If you treat $x$ as a function of $y$ and $z$, and hold $z$ constant, you get $${\partial f\over\partial x}{\partial x\over\partial y}=-{\partial f\over\partial y}$$

Do this a few times, varying the roles of $x$, $y$ and $z$. Then the product of all the left-hand-sides is equal to the product of all the right-hand sides, from which you can derive the final relation you asked about.

But: This is a really sloppy way to do it, because $x$ stands both for a variable and an implicit function of $y$ and $z$, etc. Much better to let one symbol stand for one thing, as per Frieder's answer.

Answer 3

The second statement is true by the implicit function theorem.

If the second statement is true then the first statement is false. Proof:

Suppose $$\frac{\partial x}{\partial y}=\frac{\partial x}{\partial z}\frac{\partial z}{\partial y}$$.

Then $$\frac{\partial x}{\partial y}\frac{\partial z}{\partial x}\frac{\partial y}{\partial z} = 1$$, contradicting the second statement.

Answer 4

The following I find reassuring as it explicit in what exactly is held constant. If we have a function $f(x_1,...,x_n)$ with the $x_i$ restricted so that $f=0$, then

$$\frac{\partial f(x_1,...,x_n)}{\partial x_i}\mid_{x_1,...,x_{i-1}, x_{i+2},...,x_n}$$ $$=\frac{\partial f}{\partial x_i}\mid_{x_1,...,x_{i-1}, x_{i+1}, x_{i+2},...,x_n}+\frac{\partial f}{\partial x_{i+1}}\mid_{x_1,...,x_{i-1}, x_{i}, x_{i+2},...,x_n}\cdot \frac{\partial x_{i+1}}{\partial x_i}\mid_{x_1,...,x_{i-1}, x_{i+2},...,x_n} .$$

But because $f=0$, this is $0$, and so

$$\prod_{i=1}^n \frac{\partial x_{i+1}}{\partial x_i}\mid_{x_1,...,x_{i-1}, x_{i+2},...,x_n}= \prod_{i=1}^n-\frac{\partial f}{\partial x_i}\mid_{x_1,...,x_{i-1}, x_{i+1}, x_{i+2},...,x_n} \left(\frac{\partial f}{\partial x_{i+1}}\mid_{x_1,...,x_{i-1}, x_{i}, x_{i+2},...,x_n} \right)^{-1}=(-1)^n.$$