TITLE: Derivative of a contravariant tensor
QUESTION [1 upvotes]: Let $T$ be a contravariant tensor so it transforms under change of coordinates like
$$
T^{i'} = T^i\ \frac{\partial x^{i'} }{\partial x^i}
$$
In this it seems $T^{i'}$ is a function of the "primed" coordinates, i.e. $T^{i'}(x')$, so it should be possible to calculate the partial derivative $\partial/ \partial x^{j'}$ by using the product rule:
$$
\frac{\partial T^{i'}}{\partial x^{j'}} = \left( \frac{\partial T^i}{\partial x^{j}} \frac{\partial x^{j} }{\partial x^{j'}} \right) \frac{\partial x^{i'} }{\partial x^i}
+
T^i \left(\frac{\partial}{\partial x^{j'}}\frac{\partial x^{i'}}{\partial x^i}\right)
$$
However, I don't really know what to do with the expression in the second pair of brackets, since it has "primed" and "unprimed" partials?

REPLY [0 votes]: No, actually $T^{i'}$ is a function in not primed coordinates. It's components show which value the primed coordinate will have depending of not primed one. Example: $$x=r \cos(\phi), y = r \sin(\phi)$$
Here primed coordinates are x and y. If you will take contravariant tensor of form $T=(f^1(\phi, r), f^2(\phi, r))^T$, you will get $T'=(\cos(\phi)f^1(\phi, r) -r \sin(\phi) f^2(\phi, r), f^1(\phi, r) \sin(\phi) + f^2(\phi, r)r \cos(\phi))$.
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