Might anyone offer a derrivation? My attempt bellow
($ x_{i'}$ is counting through the transformed coordinates)
$\displaystyle\frac{\partial }{\partial x_{i'}}= \displaystyle\frac{\partial x_i}{\partial x_{i'}} \displaystyle\frac{\partial }{\partial x_i}$
$ dx^{n'} = \displaystyle\frac{\partial x^{n'} }{\partial x_n}dx^n$
Knowing this, and going about it like with transformations with vectors, and the dual
$ A^ie_i=A^{i'}e_{i'} = A^{i'}M^i_{i'}e_i $ $ => A^{i'}=(M^{-1})^{i'}_iA^i $
we get
$ A^i\displaystyle\frac{\partial }{\partial x_i} = A^{i'}\displaystyle\frac{\partial }{\partial x_{i'}} = A^{i'}\displaystyle\frac{\partial x_i}{\partial x_{i'}} \displaystyle\frac{\partial }{\partial x_i}$
is the following OK? And if so, why
=> $A^i= A^{i'}\displaystyle\frac{\partial x_i}{\partial x_{i'}}$ => $A^{i'}= A^{i}\displaystyle\frac{\partial x_{i'}}{\partial x_{i}}$
I know I can treat it( $\displaystyle\frac{\partial x_i}{\partial x_{i'}}$ ) as a matrix, but the inverse of one usually isn't just the inverse of each of its components
The other way I went about it, is demanding that a tensor product is equivalent under the transformation, and computing the needed transformations of coeficients (already knowing those for the basis), so that everything cancels out. Leaving the product equivalent under the transformation.
I am an amateur mathematician, so I apologise for any lapses of terminology.
I'll write the two coordinate systems as $x_j$ and $y_k$, to better distinguish them. Then we know that $a_j \frac{\partial}{\partial x_j}=b_k \frac{\partial}{\partial y_k}$. So, by applying both sides to the coordinate function $x_i$, we know that $a_j \frac{\partial x_i}{\partial x_j}=b_k \frac{\partial x_i}{\partial y_k}$. By definition, $\frac{\partial x_i}{\partial x_j}=\delta^i_j$, so on the left-hand side, we only get $a_i$, and so $a_i=b_k \frac{\partial x_i}{\partial y_k}$. Similarly, we could get that $b_i=a_j\frac{\partial y_i}{\partial x_j}$. You are correct that the matrices $A_1=(\frac{\partial x_i}{\partial y_k})$ and $A_2=\frac{\partial y_i}{\partial x_j}$ are inverses of each other, but keep in mind that it doesn't need to be the case (and in general is definitely not the case) that $\frac{\partial x_i}{\partial y_j}=(\frac{\partial y_j}{\partial x_i})^{-1}$. That is, you shouldn't think of it as inverting each component.