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Let $t \in$ $\mathbb{R}$ and $z = f(t)$ $\in \mathbb{R}^{n}$.

How to calculate $\frac{\partial}{\partial z} \frac{\partial z}{\partial t}$?

Let's start by a 1D example and total derivative first:

$z = t^{2}$

$\frac{d}{dt} t^{2} = 2t$

$\frac{d}{dz} 2t = \frac{d}{dz}\pm 2\sqrt{z} = \pm \frac{1}{\sqrt{z}} = \frac{1}{t}$

Now I am not sure how this translates to $n \neq 1$ and partial derivatives. When taking the partial derivative, we usually treat all but one variable as constants. Since $\frac{\partial}{\partial t}f(t)$ does not contain z, one could argue that the partial derivative $\frac{\partial}{\partial{z}}\frac{\partial}{\partial{t}} f(t)$ is zero.

But what about the dependency? How to use the chain rule here? The weird thing is, that we are not taking the derivative wrt. the dependent variable t.

Edit: Modifided notation to make it more clear

My attempt: Define $u(z,t) = \frac{\partial z}{\partial t}$

$\frac{\partial }{\partial z} u = u_{z} + u_{t} \frac{\partial{t}}{\partial z} = 0 + \frac{\partial^{2}z}{\partial t^{2}} \big(\frac{\partial{z}}{\partial t}\big)^{-1}$

Is the chain rule used correctly here, or even applicable?

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  • $\begingroup$ Since $t$ does contain $z$, you can't treat $t$ as a constant when taking the partial with respect to $ z$. The partial derivative is only interested in independent variables, as it tells you the change in one direction. $\endgroup$ – Kaynex Oct 6 '17 at 15:15

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