# Why this vector field $f$ belongs to $C^1({\bf R}^2\times {\bf R})$?

The following system is an example in a book of dynamical system(in the section about Hopf Bifurcation).

\begin{align} \dot{x}=\mu x- y-x\sqrt{x^2+y^2} \\ \dot{y}=x + \mu y-y\sqrt{x^2+y^2} \end{align}

But I don't understand why it can be called a $C^1$-system, i.e., the vector field $f:R^2\times R\to R^2$ defined by this systemm belongs to $C^1(R^2\times R)$, i.e., all of the first partial derivatives with respect to $x,y$ and $\mu$ are continuous for all $x,y$ and $\mu$.

I set it as an exercise, and found that $$\frac{\partial f}{\partial x}= \left( \begin{array}{c} \mu-(\sqrt{x^2+y^2}+\frac{x^2}{\sqrt{x^2+y^2}}) \\ 1-\frac{xy}{\sqrt{x^2+y^2}} \\ \end{array} \right)$$ This is not even defined at $(0,0,\mu)$. Then why $f\in C^1(R^2\times R)$?

Edit: The original question finally boils down to another question:

What exactly is the definition of $C^1$ functions, or more generally $C^k$ functions? Are the $k$-th derivatives allowed to have removable-discontinuity points?

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You have a function $f:(x,y) \mapsto f(x,y)=(u(x,y),v(x,y))$ defined by expressions containing square roots (or maybe denominators). At all points $(x_0,y_0)$ where the radicand (the denominator) is nonzero you may calculate the partial derivative ${\partial f\over \partial x}$ using the rules of calculus. At points $(x_0,y_0)$ where the radicand is zero the function $f$ may fail to be differentiable because the function $t\mapsto \sqrt{t}$ is not differentiable at zero. It follows that we have to check such points separately (hoping that a "special effect" comes to our help). In the example at hand the only special point is the origin. Here we use the actual definition of partial derivative and obtain, e.g., $${\partial u \over\partial x}(0,0)=\lim_{x\to 0}{u(x,0)-u(0,0)\over x} = \lim_{x\to 0}{\mu x -x|x| - 0 \over x} =\mu\ .$$ This shows that ${\partial u\over\partial x}$ is also defined at $(0,0)$, so it is defined everywhere. We still have to check that it is continuous at $(0,0)$. Now for $(x,y)\ne(0,0)$ you have obtained $${\partial u\over\partial x}=\mu-\sqrt{x^2+y^2}\Bigl(1+{x^2\over x^2+y^2}\Bigl )\ .$$ Here the last factor has a value between $1$ and $2$. As $\lim_{(x,y)\to(0,0)}\sqrt{x^2+y^2}=0$ it follows that $$\lim_{(x,y)\to(0,0)}\ {\partial u\over\partial x}=\mu={\partial u\over\partial x}(0,0)\ .$$ This proves that $u(\cdot,\cdot)$ is indeed $C^1$ on all of ${\mathbb R}^2$.