# Proving the reciprocal relation between partial derivatives

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1. Mar 9, 2015

### "Don't panic!"

If three variables $x,y$ and $z$ are related via some condition that can be expressed as $$F(x,y,z)=constant$$ then the partial derivatives of the functions are reciprocal, e.g. $$\frac{\partial x}{\partial y}=\frac{1}{\frac{\partial y}{\partial x}}$$ Is the correct way to prove this the following.

As $x,y$ and $z$ are related by $F(x,y,z)=constant$, at most only two of the variables can be independent (as the third can be expressed in terms of the other two). Consider the differentials $$dx=\frac{\partial x}{\partial y}dy+\frac{\partial x}{\partial z}dz$$ $$dy=\frac{\partial y}{\partial x}dx+\frac{\partial y}{\partial z}dz$$
Substituting the second expression into the first gives $$dx=\frac{\partial x}{\partial y}\left(\frac{\partial y}{\partial x}dx+\frac{\partial y}{\partial z}dz\right)+\frac{\partial x}{\partial z}dz \\ \Rightarrow\;\;\left(1-\frac{\partial x}{\partial y}\frac{\partial y}{\partial x}\right)dx=\left(\frac{\partial x}{\partial y}\frac{\partial y}{\partial z}+\frac{\partial x}{\partial z}\right)dz$$

Now, we choose $x$ and $z$ to be the independent variables, and as such, for this equality to be true $\forall\; x,z$ it must be that the terms in the brackets vanish identically. We see then, from the left-hand side of the equality, that $$\left(1-\frac{\partial x}{\partial y}\frac{\partial y}{\partial x}\right)=0\;\;\Rightarrow\;\;\frac{\partial x}{\partial y}=\frac{1}{\frac{\partial y}{\partial x}}$$ and from the right-hand side $$\left(\frac{\partial x}{\partial y}\frac{\partial y}{\partial z}+\frac{\partial x}{\partial z}\right)=0\;\;\Rightarrow\;\;\frac{\partial x}{\partial y}\frac{\partial y}{\partial z}=-\frac{\partial x}{\partial z}$$ and hence together this gives the relation $$\frac{\partial x}{\partial y}\frac{\partial y}{\partial z}\frac{\partial z}{\partial x}=-1.$$

Last edited: Mar 9, 2015
2. Mar 14, 2015

### Greg Bernhardt

Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

3. Mar 17, 2015

### Hawkeye18

You should just use implicit function theorem here. You calculations are too complicated, and I do not see an easy way to justify them. In particular the phrase "at most only two of the variables can be independent" is too vague.

To be rigorous one should work in terms of functions and not variables.

The implicit function theorem gives you that if at some point $F_x(x,y,z)\ne 0$ then in a neighborhood of this point $x$ can be expressed in terms of $y$ and $z$. Writing $$F(x(y,z),y,z)=C$$ and taking partial with respect to $y$ you get using chain rule that
$$F_x \frac{\partial x}{\partial y} + F_y =0,$$ so $$\frac{\partial x}{\partial y} = - F_y/F_x$$ Similarly $$\frac{\partial y}{\partial x} = - F_x/F_y,$$ and you get reciprocity. Your last identity also can be easily obtained this way (it holds when all 3 partials $F_x$, $F_y$ and $F_z$ are non-zero).