# understanding $\frac{\partial x}{\partial y}\frac{\partial y}{\partial z}\frac{\partial z}{\partial x}=-1$

In particular, when $x=y=z$ isn't the above expression equal to 1 and not -1?

• try reading the above as "a change in x due to a change in y divided by a change in y times a change in y due to a change in z divided by a change in z times a change in z due to a change in x divided by a change in x". Notice that, for example, that "a change in y due to a change in z" and "a change in y" do not represent the same quantity. – John Joy Sep 23 '14 at 14:06
• I'll have to think about it more, thanks. – user174513 Sep 24 '14 at 6:19

I'm afraid that your comment about letting $x=y=z$ makes no sense at all here.
The formula ("Barkhausen's tube formula") concerns the situation when there are three quantities $x$, $y$ and $z$ constrained by a relation $F(x,y,z)=0$, so that one can view either one of them as a function of the two others: $x=x(y,z)$ or $y=y(x,z)$ or $z=z(x,y)$.
(At least this holds locally near some point $(x,y,z)=(a,b,c)$ such that $F(a,b,c)=0$, provided that the partial derivatives $F'_x$, $F'_y$ and $F'_z$ all are nonzero at this point, and $F$ is of class $C^1$; this is the implicit function theorem.)
To avoid confusion, write $x=f(y,z)$, $y=g(x,z)$ and $z=h(x,y)$ instead. Then instead of $\partial x/\partial y$ we write $f'_y$, and in particular at the point $(x,y,z)=(a,b,c)$ we write $f'_y(b,c)$, etc. Implicit differentiation shows that $$\left( \frac{\partial x}{\partial y} = \right)\quad f'_y(b,c)=- \frac{F'_y(a,b,c)}{F_x'(a,b,c)} ,$$ $$\left( \frac{\partial y}{\partial z} = \right)\quad g'_z(a,c)=- \frac{F'_z(a,b,c)}{F_y'(a,b,c)} ,$$ $$\left( \frac{\partial z}{\partial x} = \right)\quad h'_x(a,b)=- \frac{F'_x(a,b,c)}{F_z'(a,b,c)} .$$ The product of these expressions is $-1$, and that's your formula.