Problem:
Considering $x$ and $y$ as independent variables, find $\dfrac{\partial r}{\partial x}, \dfrac{\partial r}{\partial y}, \dfrac{\partial \theta}{\partial x}, \dfrac{\partial \theta}{\partial y}$ when $x = e^{2r} \cos \theta, y = e^{3r} \sin \theta$.
Solution:
First differentiate the given relations with respect to $x$:
$1 = 2e^{2r} \cos \theta \dfrac{\partial r}{\partial x} - e^{2r} \sin \theta \dfrac{\partial \theta}{\partial x}$ and $0 = 3e^{3r}\sin \theta \dfrac{\partial r}{\partial x} + e^{3r} \cos \theta \dfrac{\partial \theta}{\partial x}$.
Then solve simultaneously to obtain $\dfrac{\partial r}{\partial x} = \dfrac{\cos \theta}{e^{2r}(2+\sin^{2} \theta)}$ and $\dfrac{\partial \theta}{\partial x} = - \dfrac{3 \sin \theta}{e^{2r}(2+sin^2 \theta)}$
Question:
(1) So first of all, why does differentiating with respect to $x$ result in $\dfrac{\partial r}{\partial x}$ and $\dfrac{\partial \theta}{\partial x}$ (by the way what are these called?)? Is this because the problem says "$x$ and $y$ as independent variables" ? My initial reaction was to do $r$ and $\theta$ separately while regarding all other variables as constants... This is implicit (partial?) differentiation, right?
How should I understand what is being done here? A lot of times it seems that things like this turn out to really just be a mapping. Can I think of this that way as well? I couldn't even say what the domain and codomain would be...
Whenever I perform this operation, do I just take partial derivatives of both sides treating all independent variables (other than the variable with respect to which I am differentiating) as constants and all non-independent (dependent?) variables as variables that need to be differentiated and will have that partial derivative symbol? Once I accept that I can see how they got the first implicit partial differentiation (if that's what it is called).
(2) What do they mean by "solve simultaneously" ? I tried to solve for each (again I don't know what they're called yet) "partial differential" resulting in: $$\dfrac{\partial \theta}{\partial x} = -\dfrac{\sin \theta \frac{\partial r}{\partial x}}{\cos \theta}$$ and $$\dfrac{\partial r}{\partial x} = \dfrac{1+e^{2r}\sin \theta \frac{\partial \theta}{\partial x}}{2e^{2r}\cos \theta}$$
But I couldn't get the same answer... I tried to substitute this further, but looking at that answer, I was sure I was missing something... Could somebody please show me what to do?
Thank you in advance for any help!