# Directional derivative dotted with function

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 $$r'(t_0) = \frac{\vec{x}'(t_0) \cdot \vec{x}(t_0)}{r(t_0)}$$

• You could call this the directional derivative of the function $f(\vec{y})=\|\vec{y}\|$ along the tangent of the curve $t\mapsto \vec{x}(t)$. May 7, 2022 at 20:07
• @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. May 7, 2022 at 20:42
• 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. May 8, 2022 at 14:35

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 $$Dr(t) = {1 \over r(t)} x(t)^T x'(t)$$.
$$r(t)^2 = \lVert \vec{x}(t) \rVert^2 = \vec{x}(t) \cdot \vec{x}(t),$$ 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.
• This assumes one of the things to be proved — that $r$ is in fact differentiable. May 10, 2022 at 1:26
• 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 $n: \Bbb{R}^3 \to \Bbb{R},\, v \mapsto \sqrt {v \cdot v}$. May 10, 2022 at 1:31
• 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). May 10, 2022 at 1:34