When using polar coordinates the angle is often defined as $\theta(x,y) = \arctan\left(\frac{y}{x}\right) \quad(1)$
However, this only holds in the first quadrant. In general, it holds $${\displaystyle \theta(x,y) ={\begin{cases}\arctan {\frac {y}{x}}&\mathrm {f{\ddot {u}}r} \ x>0,\ y\geq 0\\\arctan {\frac {y}{x}}+2\pi &\mathrm {f{\ddot {u}}r} \ x>0,\ y<0\\\arctan {\frac {y}{x}}+\pi &\mathrm {f{\ddot {u}}r} \ x<0\\\pi /2&\mathrm {f{\ddot {u}}r} \ x=0,\ y>0\\3\pi /2&\mathrm {f{\ddot {u}}r} \ x=0,\ y<0\\\end{cases}}}\qquad (2)$$
Now, in my lecture notes we define $\theta$ via $(1)$ for all $(x,y)\in\mathbb{R}^2$ with $(x,y)\neq(0,0)$ but only work with the gradient $\nabla\theta$ and not $\theta$. If $(2)$ would be differentiable I could justify the usage of definition $(1)$ for all quadrants. However, $(2)$ is not differentiably (since it's not continuous). How then can we justify the use of $(1)$ to derive $\nabla\theta$ for all $(x,y)\in\mathbb{R}^2$, $(x,y)\neq(0,0)$?