Need help with identifying dependent variable in change of variables, non-independent variables problem(s) I understand the idea behind non-independent variables and have been able to solve problems of this form: we have a function $f$, and a constraint equation that relates the variables of $f$ to each other.
e.g. $w=x^2+y^2+z^2$, where $z=x^2+y^2$.
To solve for $(∂w/∂z)_y$ we do chain rule:
$(∂w/∂z)_y$ = $2x(∂x/∂z)_y + 2z$. We then solve for $(∂x/∂z)_y$ and get the answer we want: $1+2z$.
But, I am attempting to do this problem and have been quite stuck.
$$w=u^3-uv^2, u = xy, v = u + x$$ Find $(∂w/∂u)_x$
I'm stuck here because there are a lot more variables than I'm used to. Also, which is the dependent variable?
I think this is quite similar to an example brought up in class that I didn't understand.
$$f(x, y) = x + y, ∂f/∂x = 1$$
Change of variables:  $$x = u, y = u + v$$
Then, $$f = 2u + v, ∂f/∂u = 2$$
$$x = u \text{, but } ∂f/∂x = ∂f/∂u$$
I think the problem I'm having with both examples is that I don't know what the dependent variable is. Is it x,y,u, or v? If someone can explain both examples, I'd appreciate it. Thank you.
 A: $$\begin{align}w&=u^3-uv^2,u=xy,v=u+x\\1)\quad\frac{\partial w}{\partial u}&=3u^2-v^2-2uv\cdot\frac{\partial v}{\partial u}\\2)\quad\frac{\partial v}{\partial u}&=1+\frac{\partial x}{\partial u}\\3)\quad\frac{\partial x}{\partial u}&=\frac{1}{y},\text{ since }x=\frac{u}{y}\end{align}\\\therefore\frac{\partial w}{\partial u}=3u^2-v^2-2uv\cdot(1+\frac{1}{y})$$
$1)$ Is true by the chain rule and the product rule of differentiation: $\partial/\partial u[uv^2]=v^2\cdot\partial/\partial u[u]+u\cdot\partial/\partial u[v^2]\text{ (product rule) }=v^2+2uv\cdot\partial/\partial u[v]\text{ (chain rule). }$
When the variables are independent of each other, their partial derivatives with respect to one another are zero. For example, let $z=6x^2+y^2$ and $x,y$ have no relation to one another. Then, doing things the long way round as I have done in $1),2)$ and $3)$, $\partial/\partial x[z]=12x+2y\cdot\partial/\partial x[y]=12x+2y\cdot0=12x,$ which is the answer you'd expect from independent variables $x,y$, crucially because $\partial/\partial x[y]=0$ since they are not related to one another. It is only when there is a relationship between them that these are non-zero: we cannot let $x$ vary and hold $y$ constant for the purposes of differentiation if $y$ changes as $x$ varies!
In your in-class example, initially $x,y$ had zero relationship between them, and so the differentiation was trivial. However the reason $\partial f/\partial x$ appeared to be both $2$ and $1$ at the same time after you'd introduced $u$ is because you gave $y$ a structure, a relationship that was in common with $x$, because $x=u,y=u+v$ means $y=x+v$, so things are more complicated. Things would be a whole lot better if we used explicit function notation... e.g. $x(u)=u,y(u)=u+v$ in the second half of the example, showing that $x,y$ both depend on the same variable, $u$, but initially $x,y$ were not functions of anything and depended on nothing.
If you can write everything as a simple function of one variable, do so!
More intuition for the last example: changing variables to the new $u,v$ would be like a change of coordinates on the graph: all the same points are reached, but with new axes and a new relationship between the function and those axes... so $\partial f(u,v)/\partial u\neq\partial f(x,y)/\partial x$... because what $f$ is and how it behaves has changed - its behaviour depends on what variables it is taking in and what it does to them.
