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I appreciate your help. In this problem, a function is dotted with its own derivative. I'm not sure if this is a case of directional derivative or just scaling. The problem is as follows:

Let $\vec{x}: \mathbb{R} \to \mathbb{R}^3$ be a differentiable function and $r: \mathbb{R} \to \mathbb{R}$ be the function $r(t) = \lVert \vec{x}(t) \rVert$ denote the $l_2$ length. Let $t_0$ be a real number and $r(t_0) \neq 0$, then $r$ is differentiable at $t_0$ and $$ r'(t_0) = \frac{\vec{x}'(t_0) \cdot \vec{x}(t_0)}{r(t_0)} $$

I appreciate your help.

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  • $\begingroup$ You could call this the directional derivative of the function $f(\vec{y})=\|\vec{y}\|$ along the tangent of the curve $t\mapsto \vec{x}(t)$. $\endgroup$
    – Kurt G.
    May 7, 2022 at 20:07
  • $\begingroup$ @KurtG. Thanks! It makes sense. I think the denominator is just for normalization. I'm now thinking of as a normal derivative after reading Wikipedia. $\endgroup$
    – abenol
    May 7, 2022 at 20:42
  • $\begingroup$ In this particular case the denominator is a consequence of the function $f$ you are deriving. I find it a coincidence that we have the term $\vec{x}(t_0)/r(t_0)$ here which has length one. What kind of normalization do you get for other functions $f$ ? I don't see a deep result here. $\endgroup$
    – Kurt G.
    May 8, 2022 at 14:35

2 Answers 2

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Perhaps think of it as the composition of two functions, with $n(x) = \|x\|$ we have $r = n \circ x$. Then $Dr(t)h= Dn(x(t)) D x(t)$. Since $Dn(x) = {1 \over \|x\|} x^Th$, we have $Dr(t) = {1 \over r(t)} x(t)^T x'(t)$.

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From the definition of norm,
$$ r(t)^2 = \lVert \vec{x}(t) \rVert^2 = \vec{x}(t) \cdot \vec{x}(t), $$ so by differentiating with respect to $t$ (using product rule for dot products), $$ 2 \, r(t) \, r'(t) = \vec{x}'(t) \cdot \vec{x}(t) + \vec{x}(t) \cdot \vec{x}'(t) = 2 \, \vec{x}'(t) \cdot \vec{x}(t). $$ Evaluate at $t = t_0$, and since $r(t_0) \neq 0$, we can divide by $2 \, r(t_0)$ to obtain $$ r'(t_0) = \frac{\vec{x}'(t_0) \cdot \vec{x}(t_0)}{r(t_0)}, $$ as desired.

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  • $\begingroup$ This assumes one of the things to be proved — that $r$ is in fact differentiable. $\endgroup$ May 10, 2022 at 1:26
  • $\begingroup$ Well $r$ is the composition of differentiable functions, so it must be differentiable: $r = n \circ (x \times x) \circ \Delta$, where $\Delta: \Bbb{R} \to \Bbb{R}^2,\, t \mapsto (t, t)$, and $n: \Bbb{R}^3 \to \Bbb{R},\, v \mapsto \sqrt {v \cdot v}$. $\endgroup$ May 10, 2022 at 1:31
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    $\begingroup$ Well, yes, but differentiability of $n$ requires justification. Your proof (by implicit differentiation) is fine for a physics-y course, not for a multivariable analysis course. At the very least, you should include your statements in the proof (and admit that you’ve left out details). $\endgroup$ May 10, 2022 at 1:34

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