I am trying to clear up a consistent source of confusion for myself when finding the partial derivatives of an implicitly defined function. I think I have gotten the hang of the process but I often get confused because I don't think I actually know what I'm actually doing and why.
As an example: Let u be defined as a function of $x,y$ by means of the equation $$ u=F(x+u,yu) $$ Find $\frac{\partial u}{\partial x}$ in terms of the partials of $F$.
The method seems to be "ok, I can define some function $g(x,y)=u$ and then I can rewrite the arguments of $F$ in a way that doesn't lead to an equation where all of the $\frac{\partial u}{\partial x}$'s cancel (this step in particular is confusing, why do they cancel sometimes?) as $F(u_1(x,y),u_2(x,y))$ and then we get $$ \frac{\partial u}{\partial x}=D_1F\frac{\partial u_1}{\partial x}+D_2F\frac{\partial u_2}{\partial x}\\ =D_1F(1+\frac{\partial g}{\partial x})+D_2Fy\frac{\partial g}{\partial x}\\ =D_1F(1+\frac{\partial u}{\partial x})+D_2Fy\frac{\partial u}{\partial x} $$ Which we can now solve for $\frac{\partial u}{\partial x}$ as normal.
Any explanation for why the steps are necessary or any theoretical underpinnings I am missing would be helpful. If this is really just best understood as a process, that's ok too I just want to know. I have really had trouble finding good explanations.
Just let $c(x,y) = \langle c^1=x+u,c^2=yu\rangle$ then using Chain Rule you have:
$$\frac{\partial u}{\partial x}=\frac{\partial}{\partial x} (F \circ c) = \nabla F\Bigr|_{c(x,y)} \cdot \left\langle \frac{\partial c^1}{\partial x} \Bigr|_{(x,y)}, \frac{\partial c^2}{\partial x}\Bigr|_{(x,y)} \right\rangle = \frac{\partial F}{\partial x^1}\Bigr|_{c(x,y)}\frac{\partial c^1}{\partial x^1} \Bigr|_{(x,y)} +\frac{\partial F}{\partial x^2}\Bigr|_{c(x,y)}\frac{\partial c^2}{\partial x^1} \Bigr|_{(x,y)}$$
Edit: Let $f(x,y)$ be differentiable and $c(x,y) = \langle c^1,c^2 \rangle$ be differentiable as well. Suppose also that $f \circ c$ is defined (then it is differentiable). Let $ f\circ c$ be differentiable at $(a,b)$ then:
$$\frac{\partial (f \circ c)}{\partial x}\Bigr|_{(a,b)} = \frac{d}{dt}\Bigr|_{t=a} (f \circ c)(t,b) = \nabla f\Bigr|_{c(a,b)} \cdot \left\langle\frac{\partial c^1}{\partial x}\Bigr|_{(a,b)},\frac{\partial c^2}{\partial x}\Bigr|_{(a,b)}\right\rangle$$