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I was studying this theorem and am struggling to understand the proof:

The proof I studied is as follows:

It is given that $f=f(x,y) $.

let $f=z \quad \rightarrow \quad dx=(\frac{\partial x}{\partial y})_z \ dy+(\frac{\partial x}{\partial z})_y \ dz \quad [1] \quad $ and $\quad dy=(\frac{\partial y}{\partial x})_y \ dx+(\frac{\partial y}{\partial z})_x \ dz \quad [2]$

Substituting [2] into [1] we get:

$dx=(\frac{\partial x}{\partial y})_z(\frac{\partial y}{\partial x})_z \ dx+[(\frac{\partial x}{\partial y})_z(\frac{\partial y}{\partial z})_x+(\frac{\partial x}{\partial z})_y] \ dz$

Now, if we hold x constant and so $dx=0$, we obtain the cyclic relation:

$(\frac{\partial x}{\partial y})_z(\frac{\partial y}{\partial z})_x(\frac{\partial x}{\partial y})_z=-1$

What I don't understand here is that, how can our expression include $(\frac{\partial x}{\partial y})_z$ when we've stated, in our proof, that $x$ must be taken to be a constant? Wouldn't this factor then be equal to $0$ since $x$ isn't changing?

Amin Parvaresh
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1 Answers1

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No, the correct setting here is some implicit relation defining a surface in $xyz$-space. You should start with $f(x,y,z)=0$ and, assuming the partial derivatives are all non-zero, you can say that on this surface locally each variable is given as a differentiable function of the remaining two.

Before the last step, you also need to prove that $\left(\frac{\partial x}{\partial z}\right)_y \left(\frac{\partial z}{\partial x}\right)_y = 1$. This is false if you omit the $(\ )_y$.

You don't need to set $dx=0$ at the end. You just have to say that if $A\,dx + B\,dz = 0$, then, since $x$ and $z$ are independent variables, we must have $A=B=0$. Note that because of my comment in the previous paragraph, when you rearrange terms here you do in fact have $A=0$.

I'm not very fond of this proof. I would rather use implicit differentiation (i.e., the chain rule) with $f(x,y,z) = 0$ to deduce that, for example, $$\left(\frac{\partial z}{\partial x}\right)_y = -\frac{\partial f/\partial x}{\partial f/\partial z}.$$ Putting the three formulas together gives you your cyclic rule. This is, in fact, a useful formula for implicit differentiation.

Ted Shifrin
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