# Given $F(x,y,z)=0$, $\frac{\partial x}{\partial y} \frac{\partial y}{\partial z}\frac{\partial z}{\partial x}$ should be -1, but I got 1

What went wrong in my calculation?

Given $F(x,y,z)=0$ , $F_x\neq 0 , F_y\neq 0, F_z\neq 0$, and $z=f(x,y), y=g(x,z), x=h(y,z)$

$F_x\quad\,+F_y\frac{\partial y}{\partial x} + F_z \frac{\partial z}{\partial x} = 0\\ F_x\frac{\partial x}{\partial y}+F_y\quad\,+F_z\frac{\partial z}{\partial y}=0$

Then we have:

$F_y(\frac{\partial x}{\partial y}\frac{\partial y}{\partial x}-1)+F_z(\frac{\partial z}{\partial x}\frac{\partial x}{\partial y}-\frac{\partial z}{\partial y}) =0$

Let $P(x,y)=F(x,y,f(x,y))\\ P_x+P_y\frac{\partial y}{\partial x}=0\\ \frac{\partial y}{\partial x} = -\frac{P_x}{P_y}, similarly, \frac{\partial x}{\partial y} = -\frac{P_y}{P_x}, \frac{\partial x}{\partial y}\frac{\partial y}{\partial x} = 1$

$F_y(1-1)+F_z(\frac{\partial z}{\partial x} \frac{\partial x}{\partial y} - \frac{\partial z}{\partial y}) = 0\\ \frac{\partial z}{\partial x}\frac{\partial x}{\partial y} - \frac{\partial z}{\partial y}=0\\ \frac{\partial z}{\partial x}\frac{\partial x}{\partial y}\frac{\partial y}{\partial z} -\frac{\partial z}{\partial y}\frac{\partial y}{\partial z} = 0\\ \frac{\partial z}{\partial x}\frac{\partial x}{\partial y}\frac{\partial y}{\partial z} = 1$

The usual partial differentiation notation is ambiguous, and you are making conflicting assumptions about what the notation means.

For example, in this context, $\frac{\partial y}{\partial x}$ could mean both:

1. the derivative of $y$ with respect to $x$ while holding $z$ constant
2. the derivative of $y$ with respect to $x$ while holding $y$ constant

(it could mean lots of other things as well, but these two interpretations are the only ones you're supposed to consider for this problem)

The first equation you computed takes conflicting meanings: the equation

$$F_x + F_y \frac{\partial y}{\partial x} + F_z \frac{\partial z}{\partial x} = 0$$

could mean one of two things:

• Both derivatives are taken while holding $y$ constant
• Both derivatives are taken while holding $z$ constant

However, you've made inconsistent interpretations: you interpreted $\frac{\partial y}{\partial x}$ as holding $z$ constant, and $\frac{\partial z}{\partial x}$ as holding $y$ constant.

Under the two correct interpretations, the equation simplifies to either

$$F_x + F_y \cdot 0 + F_z \cdot \left. \frac{\partial z}{\partial x}\right|_{y \text{ const}} = 0$$ $$F_x + F_y \cdot \left. \frac{\partial y}{\partial x}\right|_{z \text{ const}} + F_z \cdot 0 = 0$$

And you get

$$\left. \frac{\partial z}{\partial x}\right|_{y \text{ const}} = -\frac{F_x}{F_z}$$

Ultimately, you get

$$\left. \frac{\partial z}{\partial x}\right|_{y \text{ const}} \left. \frac{\partial x}{\partial y}\right|_{z \text{ const}} \left. \frac{\partial y}{\partial z}\right|_{x \text{ const}} = \left(-\frac{F_x}{F_z} \right) \left(-\frac{F_y}{F_x} \right) \left(-\frac{F_z}{F_y} \right) = -1$$