Integral of chain rule of total derivative I am trying to understand how you would 'reverse' the chain rule for a derivative.
If I have a function $$f(x(r,\theta),y(r,\theta))$$ then differentiating w.r.t $r$ gives: $$\frac{\partial f}{\partial r}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial r}$$ 
Is there a formal way of reversing this rather than 'by inspection'? I thought integrating w.r.t $r$ would be the way but I expect I must be making an error in my integration:
$$\partial f = \frac{\partial f}{\partial x}\partial x + \frac{\partial f}{\partial y}\partial y$$
Now would the $\partial x$s and the $\partial y$s 'cancel'? Wouldn't that leave me with $\partial f= 2\partial f$?
 A: I see a $\frac{\partial f}{\partial r}$, $\frac{\partial f}{\partial x}$, and a $\frac{\partial f}{\partial y}$. Many books write out the chain rule like this. Technically, it's incorrect. When you take $f: (u,v) \mapsto \mathbb{R}$, the outside function, and form the composition $f \circ g$ where $g: (r,\theta) \mapsto (u,v)$, this composition is a new function and should be given a new name like $z = f\circ g: (r,\theta) \mapsto \mathbb{R}$. Notice how I'm using $(u,v)$ and not $(x,y)$. You already used $x$ and $y$ as function names. I shouldn't write $x = x(r,\theta)$ and $y = y(r,\theta)$ as is done frequently (I actually do this a lot). Instead, I labeled the outputs $u = x(r, \theta)$ and $v = y(r,\theta)$. 
The chain rule should always be stated: "derivative of the composition = ( derivative of the outside function )( derivative of the inside function )." As stated, you have the derivative of the outside function on the left? Making me confused on how you are defining $f$. It makes it seem like $f$ is a 3 input function. With this notation, the chain rule should be written
$$\frac{\partial z}{\partial r} = \frac{\partial f}{\partial u}\frac{\partial x}{\partial r} + \frac{\partial f}{\partial v}\frac{\partial y}{\partial r}  $$
It looks weird, mainly because I'm objecting to writing $x = x(r,\theta)$. Because, technically since you used $x$ as a function name, $\frac{\partial }{\partial x}$ doesn't make sense. This would be like, in single variable calculus, writing $\frac{d}{df}$ when $f$ is the function name. However given a single variable function $f$, producing outputs of $f(x)$, $\frac{d}{d(\;f(x)\;)}$ does make sense (this would be a derivative in the range of the function - or the domain of the inverse function - i.e. $\frac{d}{dy}f^{-1}$ where $y = f(x)$)
To undo the chain rule:
$$\int \frac{\partial z}{\partial r}\;dr = z(r,\theta) + F(\theta)  $$
If you wanted to see this, in the way that you are doing, you'd have to do a u-substitution. If I give you
$$\int \frac{d}{d(g(x))}f(g(x)) \frac{dg}{dx}dx $$
You let $u = g(x)$ and proceed to find $ \int f'(u) du = f(u) = f(g(x))$. You'd do the same thing here. Actually nevermind, you will need multivariable calculus and just go in reverse (the following is a repeat of what I've already written down above)
$$ \int \nabla f(x(r,\theta), y(r,\theta)) \cdot \Big<\frac{\partial x}{\partial r}, \frac{\partial y}{\partial r}\Big>dr \\
= \int \frac{\partial }{\partial r}f(x(r,\theta), y(r,\theta)) dr \\
= f(x(r,\theta), y(r,\theta)) + F(\theta)
$$
