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

If $$f(x, y, z) = 0$$ and $$\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} , \frac{\partial f}{\partial z} \neq 0$$, show that

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

I understand that posting questions without showing your work is frowned upon. I just couldn't come up with anything.

• If you understand you shouldn't do something, then don't do it... Furthermore, maybe you are using an odd notation, or the question makes no sense. You put some assumptions on a function $f$, and then in the conclusion, the formula doesn't even depend on $f$. So I do not understand the question. – A. Pongrácz Apr 27 '19 at 22:27
• Also, if $x,y,z$ are supposed to be variables, then writing $f(x,y,z)=0$ means that $f$ is the constant zero function. In that case, the partial derivatives of the function are all zero. So strictly speaking, I have just solved your problem: the assumption is empty, so whatever conclusion is claimed, the statement is true. On a more serious note: check the problem (you probably didn't copy it correctly), and make some effort in solving it. – A. Pongrácz Apr 27 '19 at 22:30
• Me neither. But it's from the exercise list given by my multivariable calculus teacher and he's a Physics PhD. He is known for giving students these crazy hard exercises but I assume that this has a solution. – Sigma Apr 27 '19 at 22:31

## 2 Answers

You question makes sense when $$x=x(y, z)$$, $$y=y(x, z)$$, $$z=z(x, y)$$ are functions which are implicitly defined by $$F(x, y, z)=0$$. We can indeed assure that's the case since the implicit function theorem tells us that if the partial derivatives are non-zero, as given, then $$F(x, y, z)=0$$ defines each variable as an implicit function of the others. Then, using the classic result from the same theorem: $$\frac{\partial x}{\partial y}=-\frac{F_y}{F_x}, \qquad \frac{\partial y}{\partial z}=-\frac{F_z}{F_y}, \qquad \frac{\partial z}{\partial x}=-\frac{F_x}{F_z}$$ We get: $$\frac{\partial x}{\partial y}\frac{\partial y}{\partial z}\frac{\partial z}{\partial x}= (-1)^3\frac{F_y}{F_x}\cdot\frac{F_z}{F_y}\cdot\frac{F_x}{F_z}=-1$$ As desired.

• Would I be correct in saying that the Implicit Function Theorem is a rather advanced topic and that is usually covered in a real analysis course, and not in a common multivariable calc course? For example, Thomas Calculus, a standard textbook, only mentions the implicit function theorem as a "topic in advanced calculus", without actually covering it. – Sigma Apr 27 '19 at 22:50
• @VictorS. I believe so. However, the question itself is about implicitly-defined functions, so talking about it without the context of the implicit function theorem makes no sense, really. – Yuval Gat Apr 27 '19 at 22:58
• The problem set from which this question is from a problem set about implicit functions. However, this is a calc 3 course and the implicit function theorem isn't given at this level.. I suppose. – Sigma Apr 27 '19 at 23:07
• @VictorS. You don't need the implicit function theorem, just the chain rule for partial derivatives as shown in this answer – Mark S. Apr 27 '19 at 23:37

$$\begin{array}{l}given:\left\{ \begin{array}{l}f\left( {x,y,z} \right) = 0\\z = z\left( {x,y} \right)\end{array} \right.\\\left\{ \begin{array}{l}dz = \frac{{\partial z}}{{\partial x}}dx + \frac{{\partial z}}{{\partial y}}dy\\\left\{ {dx = \frac{{\partial x}}{{\partial y}}dy + \frac{{\partial x}}{{\partial z}}dz} \right.\\ = \frac{{\partial z}}{{\partial x}}\left( {\frac{{\partial x}}{{\partial y}}dy + \frac{{\partial x}}{{\partial z}}dz} \right) + \frac{{\partial z}}{{\partial y}}dy\\ = \left( {\frac{{\partial z}}{{\partial x}}\frac{{\partial x}}{{\partial z}}} \right)dz + \left( {\frac{{\partial z}}{{\partial x}}\frac{{\partial x}}{{\partial y}} + \frac{{\partial z}}{{\partial y}}} \right)dy\\\left\{ {\left( {\frac{{\partial z}}{{\partial x}}\frac{{\partial x}}{{\partial z}}} \right) = 1} \right.\\ = dz + \left( {\frac{{\partial z}}{{\partial x}}\frac{{\partial x}}{{\partial y}} + \frac{{\partial z}}{{\partial y}}} \right)dy\end{array} \right.\\ \Downarrow \\\left( {\frac{{\partial z}}{{\partial x}}\frac{{\partial x}}{{\partial y}} + \frac{{\partial z}}{{\partial y}}} \right)dy = 0\\\frac{{\partial z}}{{\partial x}}\frac{{\partial x}}{{\partial y}} + \frac{{\partial z}}{{\partial y}} = 0\\\frac{{\partial z}}{{\partial x}}\frac{{\partial x}}{{\partial y}} = - \frac{{\partial z}}{{\partial y}}\\\frac{{\partial z}}{{\partial x}}\frac{{\partial x}}{{\partial y}}\frac{{\partial y}}{{\partial z}} = - 1\end{array}$$

This is a note from a physics teacher. (Not my work, I rearrange it.)

I dont guarantee this is correct, but I think it is.