# Show $\bigg(\frac{\partial x}{\partial y}\bigg) \bigg(\frac{\partial y}{\partial z}\bigg)\bigg(\frac{\partial z}{\partial x}\bigg) = - 1,$ [duplicate]

Question statement: Let $F(x,y,z)$ is a continuously differentiable function with nonvanishing partials at $(0,0,0).$ Define $x = x(y,z), \; y = y(x,z), \; z = z(x,y)$ as the solutions of the equation $F(x,y,z) = F(0,0,0)$ in the neighborhood of $(0,0)$ in the corresponding variables. Prove that $$\bigg(\frac{\partial x}{\partial y}\bigg) \bigg(\frac{\partial y}{\partial z}\bigg)\bigg(\frac{\partial z}{\partial x}\bigg) = - 1,$$ where the three partial derivatives are taken at point $(0,0)$ in the corresponding pair of variables.

I know I have seen a couple weak proofs of this, one in thermodynamics where my professor referred to this as "The -1 Rule", and another in the back of a long lost textbook of mine. However, I am interested in a detailed, unshakable proof of this explicitly using the implicit function theorem. I've tried several times now and I have been unable to produce one. Any helpful comments/answers would be appreciated!

• Hint: implicit function theorem. See "regularity" paragraph in the section "statement of the theorem" in the Wikipedia article – user258700 Jun 8 '16 at 16:31
• This is cyclic rule, refer (en.wikipedia.org/wiki/Triple_product_rule) – Prasanna Jun 8 '16 at 16:36
• According to the comment of Hussein, you have for example $\left(\frac{\partial x}{\partial y}\right)=-\frac{F_y}{F_x}$, where $F_x:=\partial F/\partial x$ etc. Make the product of the three franction, and you obtain the result (including the $(-1)$!) – guestDiego Jun 8 '16 at 16:43
• @AhmedHussein I see how that would be of great help, then the question would become, how to prove the formula given in the section on "regularity", since that basically amounts to proving the original question. Something I'll definitely look into, thanks for the comment! – Merkh Jun 8 '16 at 16:45
• @Prasanna Thats where I've see this before online! Triple product rule... I was a little skeptical of the proofs given there, and for my purposes (studying for a qualifier), I would like to know all the details (I feel like there is more to it than what is shown on the wiki page). However, this provides a good place to start, thanks! – Merkh Jun 8 '16 at 16:49

I think that first, you have to justify why these three partial derivatives exist, which can be done by using the implicit function theorem, provided none of the three derivatives $\frac{\partial F}{\partial x}$, $\frac{\partial F}{\partial y}$, $\frac{\partial F}{\partial z}$ is equal to zero at $(0,0,0)$. This theorem asserts that there exists a continously differentiable function $(y,z) \mapsto \phi_1(y,z)$, such that $$F(x,y,z)=F(0,0,0) \Longleftrightarrow x=\phi_1(y,z),$$ on a suitable neighborhood of the origin. Differentiating $\displaystyle F(\phi_1(y,z),y,z)=F(0,0,0)$ with respect to $y$ yields to $\displaystyle \dfrac{\partial \phi_1}{\partial y}=- \dfrac{\dfrac{\partial F}{\partial y}}{\dfrac{\partial F}{\partial x}}.$ Here, the derivative $\dfrac{\partial \phi_1}{\partial y}$ is the rigourous definition of your $\frac{dx}{dy}$. In the same manner, you can obtain two functions $\phi_2$ and $\phi_3$ such that $\displaystyle F(x,\phi_2(x,z),z)=F(0,0,0)$ and $\displaystyle F(x,y,\phi_3(y,z))=F(0,0,0)$, from which you get $$\dfrac{\partial \phi_2}{\partial z}=- \dfrac{\dfrac{\partial F}{\partial z}}{\dfrac{\partial F}{\partial y}} \textrm{ and } \dfrac{\partial \phi_3}{\partial x}=- \dfrac{\dfrac{\partial F}{\partial x}}{\dfrac{\partial F}{\partial z}}.$$ And you're tring to calculate the value of $\displaystyle \dfrac{\partial \phi_1}{\partial y} \times \dfrac{\partial \phi_2}{\partial z} \times \dfrac{\partial \phi_3}{\partial x}.$