# Differentiation of a vector (in index notation) with respect to its square

I have a formula which contains the derivative $\partial_{z^2}$(with respect to the square of a vector $z^\mu$). However, as the functions inside the formula might not directly depend on $z^2$ but on $z^\mu$, I am trying to simplify this using the chain rule (assuming Einstein summation):

$$\partial_{z^2} = \dfrac{\partial z^\mu}{\partial z^2} \partial_{z^\mu}.$$

Then I use the inverse function theorem:

$$\dfrac{\partial z^\mu}{\partial z^2} = \left(2 z_\mu \right)^{-1}$$

and use the fact that $\dfrac{z_\mu}{z^2}$ is an inverse of $z^\mu$ wrt index contraction, to get:

$$\partial_{z^2} = \dfrac{z^\mu}{2 z^2} \partial_{z^\mu}.$$

Now I wonder if this is correct?

1) I am not sure if I am allowed to use the inverse function theorem for functions $\mathbb{R}^n \rightarrow \mathbb{R}$ (in this case $z^\mu\rightarrow z^2$)

2) the inverse of $z^\mu$ I use here is not unique (there exist an infinite number of vectors $v$ that satisfy $v\cdot z=1$).

But if this is not right, how then can I calculate $\dfrac{\partial z^\mu}{\partial z^2}$?

Thank you!

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This makes no sense because you haven't specified what you're holding constant in taking that partial derivative. My guess would be that the idea is to hold the direction constant while changing the magnitude? – joriki Mar 20 '13 at 13:39
Maybe I am entirely wrong here. I try to prove the following formula: $\partial_\mu \partial_\nu = 2 g_{\mu\nu} \partial_{z^2} + 4 z_\mu z_\nu \partial^2_{z^2}$ where $z^\mu$ are my coordinates. I am totally confused by the $\partial_{z^2}$ , so in the question above, I describe my steps until now. But I am not certain I am allowed to do so. Thanks – freddieknets Mar 20 '13 at 13:58
I have nothing to add to my previous comment; it applies equally to your comment as it does to the question. It makes no sense to ask whether you're allowed to do something when you haven't defined what it is you're doing. – joriki Mar 20 '13 at 14:12