I have a puzzling situation involving derivatives. I want to derivate:
$$
\frac{d}{dx}| \mathbf F(x)|
$$
This was actually something involving physics. Lets be 2-dimensional for simplicity. Let a particle be at position $\mathbf r = (x, y)$. The distance $s$ of the particle from point $(0, 0)$ is simply $s = |\mathbf r|$. I want to calculate how that distance changes over time.
$$
\frac{ds}{dt} =
\frac{d}{dt}|\mathbf r| =
\frac{d}{dt}\sqrt{x(t)^2 + y(t)^2} =
\frac{1}{\sqrt{x(t)^2 + y(t)^2}}\left(x\frac{dx}{dt} + y\frac{dy}{dt}\right) = \frac{1}{|\mathbf r|}\left(x\frac{dx}{dt} + y\frac{dy}{dt}\right)
$$
As you can see, $ds/dt$ is not defined when $|\mathbf r| = 0$. I can't see why. On physics point of view, the particle should always travel continuously in the plane (assuming the path it makes is continuous and fully differentiable). Why is the distance variation undefined? Assume for instance, I have a table, and $(x, y)$ is the position of my fingers. I can't see why it wouldn't exist.
Hypothesis: Notice that, by description I told, the curve $(x, y)$ is continuous on all points, and smooth/differentiable on all points. Thus, $x(t), y(t), x'(t), y'(t)$ is well defined, for all points. If you want, consider them to be class $C^\infty$.
My question: Does this derivative exist or not when $|\mathbf r| = 0$? What is the value/evaluation of such derivative in an arbitrary given period $t_0$ when $|\mathbf r| = 0$?
Considering $x(t) = t^2$ and $y(t) = t^2$, we get $s$ proportional to $t^2$, and thus its derivative exists at $t=0$ with the derivative having a well defined value of zero.
Best Answer
Note that for $\vec r \ne 0$, we can write
$$\begin{align} \frac{ds(t)}{dt}&=\frac{\vec r(t)\cdot \frac{d\vec r(t)}{dt}}{|\vec r(t)|}\\\\ &=\bbox[5px,border:2px solid #C0A000]{\hat r(t)\cdot \frac{d\vec r(t)}{dt}} \tag 1 \end{align}$$
where in $(1)$, $\hat r(t)=\frac{\vec r(t)}{|\vec r(t)|}$ is the position unit vector. However, the unit vector $\hat r$ is undefined at the origin.
This fact does not automatically imply that the derivative $s'(t)$ fails to exist at $\vec r=0$. In the ensuing analysis, we will explore whether $s'(t)$ exists at $\vec r=0$.
Assume that at $t_0$, $\vec r(t_0)=0$. We assume that $\vec r''(t)$ exists. Then, the derivative of $s(t)$ at $t_0$, if it exists, is given by
$$\begin{align} s'(t_0)&=\lim_{h\to 0}\left(\frac{\left|\vec r(t_0+h)\right|-\left|\vec r(t_0)\right|}{h}\right)\\\\ &=\lim_{h\to 0}\frac{\left|\vec r(t_0+h)\right|}{h} \\\\ &=\lim_{h\to 0}\frac{\left|\vec r'(t_0)h+O(h^2)\right|}{h} \\\\ &=\bbox[5px,border:2px solid #C0A000]{\lim_{h\to 0}\left(\frac{|h|}{h}\,\left|\vec r'(t_0)+O(h)\right|\right)} \tag2 \end{align}$$
If $\vec r'(t_0)=0$, then from $(2)$ we see that $s'(t_0)=0$ also. However, if $\vec r'(t_0) \ne 0$, then the limit fails to exist since the limits from the right-hand side and left-hand side are unequal.