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I was reading my class notes about direction derivatives, and my teacher stated the following:

Let $f: \mathbb{R}^n \to \Bbb R$ be a function, $p,v \in \mathbb R ^n$ and let $f'(p,v)$ denote the directional derivative of $f$ at $p$ relatively to $v$, this is:

$$f'(p,v) = \lim_{h \to 0} \frac{f(p + hv) - f(p)}{ h}$$

Then he added the following note:

  • If $v \neq 0$, the direction derivative only depends on $p$ and on the direction of $v$.

This would mean that for any $k > 0$, $f'(p,v) = f'(p,kv)$ since multiplying a vector by a scalar doesn't change the direction of the vector. But:

$$f'(p,kv) = \lim_{h \to 0} \frac{f(p + hk \cdot v) - f(p)}{ h}$$

If we change do the following change of variables: $\lambda = hk$ we get:

$$f'(p,kv) = k \lim_{\lambda \to 0}\frac{f(p + \lambda v) - f(p)}{ \lambda} = k \cdot f'(p,v) $$

So : $f'(p,kv) = k \cdot f'(p,v)$

Am I doing some mistake here, or does the value of the directional derivative change not only with the direction of $v$ but only with its magnitude?

Eduardo Magalhães
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  • I think it is possible that your teacher has assumed that $|v|=1$ which corresponds to $k=1$. I think 1 and 2 explains it well why this choice is made. –  Jul 22 '21 at 23:56
  • I don't think that's the case. He never mentioned that $||v|| = 1$ and, in the very first example he showed us, he calculated the direction derivative with $v = (1,1)$ @AndréArmatowski. Is it that common to assume that $||v|| = 1$ when talking about directional derivatives? – Eduardo Magalhães Jul 23 '21 at 00:03
  • If you do not assume that $|v|=1$ then you will have directional derivatives (multiple) in a certain direction, since it would depend on the length on the vector chosen. Easy example, $f(x,y)=x$ relative vectors along the $x$-axis of different norm. So for benefit of talking about a singular value for a specific direction, most sources I have met assumes $|v|=1$. –  Jul 23 '21 at 00:47
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    With that said, it is entirely possible that it is not assumed that $|v|=1$. It was only a guess as to why your lecture notes are as they are. –  Jul 23 '21 at 01:03
  • Thank you @AndréArmatowski – Eduardo Magalhães Jul 23 '21 at 10:44

1 Answers1

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Suppose that $f:\mathbb{R}^{n}\to\mathbb{R}^{m}$ is differentiable at $p\in\mathcal{O}\subseteq\mathbb{R}^{n}$, where $p$ is a limit point and $\mathcal{O}$ is open.

This means there exists a linear mapping $L:\mathbb{R}^{n}\to\mathbb{R}^{m}$ such that \begin{align*} \lim_{h\to 0}\frac{\|f(p + h) - f(p) - L(h)\|_{\mathbb{R}^{m}}}{\|h\|_{\mathbb{R}^{n}}} = 0 \end{align*}

In particular, if we take $h = tv$ such that $p + tv\in\mathcal{O}$, it results that \begin{align*} \lim_{t\to 0}\frac{\|f(p + tv) - f(p) - L(tv)\|_{\mathbb{R}^{m}}}{\|tv\|_{\mathbb{R}^{n}}} & = \lim_{t\to 0}\frac{1}{\|v\|_{\mathbb{R}^{n}}}\left\|\frac{f(p + tv) - f(p)}{t} - L(v)\right\|_{\mathbb{R}^{m}} = 0\\\\ & \Rightarrow \lim_{t\to 0}\frac{f(p + tv) - f(p)}{t} = L(v) \end{align*}

What does this result tell us?

It means the directional derivative of $f$ along $v$ at the point $p$ is given by $f'(p)v$.

Based on such result, we can answer your question as follows: \begin{align*} D_{kv}f(p) = f'(p)(kv) = kf'(p)v = kD_{v}f(p) \end{align*}

Hopefully this helps!

user0102
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