If we have a (non-singular) coordinate transformation from the coordinates $\{x^i\}$ to the coordinates $\{x^{\prime i}\}$, what is the relation between the partial derivatives $\frac{\partial x^{\prime j}}{\partial x^i}$ and $\frac{\partial x^i}{\partial x^{\prime j}}$?
Or where can I get more information about this?
Thanks for reading!
I assume that as a coordinate transformation you understand a $ C^1$-Diffeomorphism
$$ \phi: U \rightarrow V, \{x^i\} \mapsto \{x^{'i}\}, $$
where $U,V$ are open subsets of $\mathbb{R}^n$. Then $\phi$ has per definition a differentiable inverse and we get
\begin{align*} \phi \circ \phi^{-1} & = id_V\\ \phi^{-1} \circ \phi & = id_U \end{align*}
Differentiating both sides and using the chain rule we yield \begin{align*} J\phi \circ J(\phi^{-1}) & = id_{\mathbb{R}^n} \\ J(\phi^{-1}) \circ J\phi & = id_{\mathbb{R}^n}, \end{align*} where $J\phi$ denotes the Jacobi-Matrix, i.e. $J \phi = \left ( \frac{\partial x^{'i}}{\partial x^{j}} \right )_{i,j} $ and $J (\phi^{-1}) = \left ( \frac{\partial{x^i}}{\partial x^{'j}} \right )_{i,j}$.
If you want to omit the Matrices you get the relation mentioned in the comments above, which is
$$ \sum_{k = 1}^n \frac{\partial x^{'i}}{\partial x^{k}} \cdot \frac{\partial x^{k}}{\partial x^{'j}} = \delta^i_j $$
EDIT:
$$ \frac{\partial r}{\partial x} = \frac{\partial \sqrt{x^2 + y^2}}{\partial x} = \frac{x}{\sqrt{x^2+y^2}} = \frac{x}{r} = \frac{r \cos{\phi}}{r} = \cos(\phi), $$
which shows, that your computation was correct. Sorry, for that.
Nevertheless in general you do not have the equation $\frac{\partial x}{\partial x'} = \frac{\partial x'}{\partial x}$. One counterexample is the following:
\begin{align*} x' & = 2\cdot x \\ y' & = 3\cdot y \end{align*}
This is a $C^1$-Diffeomorphism with partial derivatives:
\begin{align*} \frac{\partial x'}{\partial x} = 2 \neq \frac{1}{2} = \frac{\partial x}{\partial x'} \\ \frac{\partial y'}{\partial y} = 3 \neq \frac{1}{3} = \frac{\partial y}{\partial y'} \end{align*}
I hope I could help you.