How to prove that there is no differentiable function with given partial derivatives

Let $U=\{(x,y)\in\mathbb{R}^2:(x,y)\neq(0,0)\}$. Show that there is no differentiable function $f:U\rightarrow\mathbb{R}$ satisfying $$\frac{\partial f}{\partial x}=\frac{y}{x^2+y^2}, \frac{\partial f}{\partial y}=-\frac{x}{x^2+y^2}.$$

My initial thought was to show that these partials are not continuous at some point $(x,y)\neq(0,0)$, but since it is possible to have a differentiable function that is not $C^1$, this is not enough. Any suggestions are greatly appreciated.

• Was this a Calculus II exercise? It looks like it's a hard one: it is easy to get the right idea if you have seen something more advanced, but to someone who is just learning partial derivatives and has never heard about exact/closed forms it is difficult to come up with it. Aug 26, 2018 at 6:47

In my opinion, I think this exercise is the most important exercise in calculus since it is a starting point of the de Rham cohomology, which shows an interaction between topology and functions on it. Anyway, this question is equivalent to the following: can we find $f:U \to \mathbb{R}$ s.t. $$\nabla f = \left(\frac{y}{x^{2}+y^{2}}, -\frac{x}{x^{2}+y^{2}}\right)?$$ If such $f$ exists, the vector field on the right-hand side will be called as a conservative field, and by the fundamental theorem of a line integral, if we integrate the vector field over any given (simple) closed curve, the result should be zero. However, if you try to integrate it over a unit circle centered at origin, the result became $2\pi$.
• I'm getting $-2\pi$ when I integrate over the unit circle centered at the origin. It's relatively straightforward to convert to polar and simplify, but I get $-\int_0^{2\pi}dt$. Is there some reason this must be positive? Aug 25, 2018 at 17:58
• @Atsina Yes you are right, it gives $-2\pi$. The important point is that 1-form $$\frac{y}{x^{2}+y^{2}}dx - \frac{x}{x^{2}+y^{2}}dy$$ is a closed 1-form which is not exact. Also, this is a basis of the de Rham cohomology group $H^{1}(U;\mathbb{R})\simeq \mathbb{R}$. Aug 25, 2018 at 18:17
Assume that such a $f$ exists and let $\gamma\colon[0,1]\to U$ defined by $\gamma(t)=(\cos(2\pi t),\sin(2\pi t))$, then: $$\int_\gamma\mathrm{d}f=f(\gamma(1))-f(\gamma(0))=0,$$ however using the given formula for $\mathrm{d}f$, one finds that: $$\int_\gamma\mathrm{d}f=\int_0^1 \mathrm{d}f_{\gamma(t)}\cdot\dot{\gamma}(t)\,\mathrm{d}t=2\pi\int_0^1\frac{-\sin(2\pi t)^2-\cos^2(2\pi t)}{\cos^2(2\pi t)+\sin^2(2\pi t)}=-2\pi,$$ whence a contradiction.
The intuition behind this reasoning is that the angular form $\frac{y\mathrm{d}x-x\mathrm{d}y}{x^2+y^2}$ is not exact on $U$.