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I would like to prove the following identity, where $x,y,z$ are linked by a functional relation $f(x,y,z)=0$ and where the parentheses denote differentiation while keeping the indicated variable constant:

\begin{equation} \left(\frac{\partial x}{\partial y}\right)_z \left(\frac{\partial y}{\partial z}\right)_x \left(\frac{\partial z}{\partial x}\right)_y=-1 \end{equation}

Can you help?

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For searching purposes: this chain relation is often attributed to Euler; thus the terms "Euler relation" or "Euler chain relation" are sometimes used. –  J. M. Apr 15 '12 at 15:58

1 Answer 1

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Let $A_t=\dfrac{\partial f}{\partial t}$ for $t$ in $\{x,y,z\}$, then $$ \left(\frac{\partial x}{\partial y}\right)_z=-\frac{A_y}{A_x},\qquad\left(\frac{\partial y}{\partial z}\right)_x=-\frac{A_z}{A_y},\qquad\left(\frac{\partial z}{\partial x}\right)_y=-\frac{A_x}{A_z}. $$ The result follows.

To prove the first identity for example, assume that $x=x^*+a$, $y=y^*+b$ and $z=z^*$ solve $f(x,y,z)=0$ with $a\to0$ and $b\to0$. Then $\dfrac{a}b\to\left(\dfrac{\partial x}{\partial y}\right)_z$ and $$ 0=f(x^*+a,y^*+b,z^*)-f(x^*,y^*,z^*)=aA_x+bA_y+o((a,b)), $$ hence $$ \dfrac{a}b\to-\dfrac{A_y}{A_x}. $$

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isn't $\dfrac{a}b$ going to $+\dfrac{A_y}{A_x}$ instead of $-\dfrac{A_y}{A_x}$? –  sjdh Dec 10 '12 at 9:33
    
@sjdh No. Solve the linear case. –  Did Dec 10 '12 at 9:35
    
sorry, I copied the wrong formulae from you post. I meant Isn't $\dfrac{a}b\to$ going to $+\left(\dfrac{\partial x}{\partial y}\right)_z$ instead of $-\left(\dfrac{\partial x}{\partial y}\right)_z$ –  sjdh Dec 10 '12 at 9:49
    
@sjdh You are right. Well done! (Thanks, post modified.) –  Did Dec 10 '12 at 14:06

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