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When there are two variables x, and p related via any differentiable function, we have the following identity:

[itex]\frac{dx}{dp}\frac{dp}{dx}=1[/itex]

When it comes to partial derivatives we canNOTsay [itex]\frac{\partial x}{ \partial t}\frac{\partial t}{\partial x}=1[/itex] . However I have seen the following identity which seems to be the generalization of the above identity for variables (x,y,z) and (p,q,r):

\begin{equation}

\left( \begin{array}{ccc}

\frac{\partial x}{ \partial p} & \frac{\partial y}{ \partial p} & \frac{\partial z}{ \partial p}\\

\frac{\partial x}{ \partial q} & \frac{\partial y}{ \partial q} & \frac{\partial z}{ \partial q}\\

\frac{\partial x}{ \partial r} & \frac{\partial y}{ \partial r} & \frac{\partial z}{ \partial r}\end{array} \right)

\left( \begin{array}{ccc}

\frac{\partial p}{ \partial x} & \frac{\partial q}{ \partial x} & \frac{\partial r}{ \partial x}\\

\frac{\partial p}{ \partial y} & \frac{\partial q}{ \partial y} & \frac{\partial r}{ \partial y}\\

\frac{\partial p}{ \partial z} & \frac{\partial q}{ \partial z} & \frac{\partial r}{ \partial z}\end{array} \right)=

\left( \begin{array}{ccc}

1 & 0 & 0\\

0 & 1 & 0\\

0 & 0 & 1\end{array} \right)

\end{equation}

Is the above identity true in general? If so, does anyone know of an elegant proof for that? I have two proofs but they I am looking for better ones. One mentions a function f(p,q,r) for the proof which I think it is unnecessary and also confusing, Another proof ( my own) is that we write

[itex]\frac{\partial p}{\partial p}=1 \rightarrow \frac{\partial p}{\partial x}\frac{\partial x}{\partial p}+\frac{\partial p}{\partial y}\frac{\partial y}{\partial p}+\frac{\partial p}{\partial z}\frac{\partial z}{\partial p}=1[/itex]

[itex]\frac{\partial p}{\partial q}=0 \rightarrow \frac{\partial p}{\partial x}\frac{\partial x}{\partial q}+\frac{\partial p}{\partial y}\frac{\partial y}{\partial q}+\frac{\partial p}{\partial z}\frac{\partial z}{\partial q}=0[/itex]

and so on. From nine equations we can get the identity but the proof is insulting!

I would be grateful if you share a better proof.

P.S. The left matrix is called the jacubian. Does the other matrix, which in fact is the inverse if jacubian, have a particular name?

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# Inverse of jacubian

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