# Do directional derivatives require direction vectors to have unit length? If so, why?

I have just started studying directional derivative from the book Mathematical Analysis by T.M. Apostol.

The directional derivative is the generalization of partial derivative. The partial derivative represents the rate of change of a function due to small change of one of the independent variables involved, whereas the directional derivative represents the rate of change of the function due to the small change of a point in it's domain along any arbitrary direction.

The concept of directional derivative is as follows :

Let $$\mathbf{f} : S \longrightarrow \mathbb R^{m}$$ be a vector valued function defined over $$S \subset \mathbb R^{n}$$. Suppose we are to find out the rate of change of $$\mathbf{f}$$ when we move from a point $$\mathbf{c}$$ of $$S$$ to a nearby point $$\mathbf{c} + \mathbf{u}$$ along a line segment. Since each point of the line can be taken as $$\mathbf{c} + h\mathbf{u}$$ for some $$h \in \mathbb R$$, we can take $$h$$ sufficiently small so that $$\mathbf{c} + h\mathbf{u}$$ is in $$S$$. Then the quantity

$$\lim_{h \rightarrow 0} \frac {\mathbf{f}(\mathbf{c} + h\mathbf{u}) - \mathbf{f}(\mathbf{c})} {h}$$

if it exists is called the directional derivative of $$\mathbf{f}$$ at $$\mathbf{c} \in S$$ in the direction of $$\mathbf{u}$$.

I am having some difficulty here. According to my teacher's lecture notes, $$||\mathbf{u}|| = 1$$. For this reason the directional derivative of a given function $$\mathbf{f}$$ at a point along some certain direction may differ.

Is there any significance of considering $$||\mathbf{u}|| = 1$$?

If the answer to my question is affirmative, then why?

• Some authors require normalization; others don't. The advantage to not normalizing is that it makes the directional derivative a linear operator. I have answered this question here: math.stackexchange.com/questions/809376/… Dec 27, 2016 at 6:20
• Yes, there is a significance of considering $||u|| = 1.$ Because, $\frac{u}{||u||}$ is the unit vector, $\frac{hu}{||u||}$ represents the vector displacement while moving a distance of $h*||u|| = h$. Dec 27, 2016 at 6:26
• I find difficulty in understanding the concept in the link provided by you @symplectomorphic.In fact I fail to compare my defination of directional derivative to your's. Please be more explicit reminding the matter that I am a novice to this concept. Dec 27, 2016 at 9:01

The directional derivative $\mathrm D_u f(x)$ can be seen as the derivative of $$\gamma(t) = f(x + tu)$$ at $0$. If we allow $\lVert u \rVert \neq 1$, we can define $\phi(t) = f(x + tu/\lVert u\rVert)$. Then $$\gamma'(t) = \frac{\mathrm d}{\mathrm dt}f(x + tu) = \frac{\mathrm d}{\mathrm dt}f\left(x + (t\lVert u \rVert) \frac{u}{\| u\|} \right) = \|u\| \phi'(t)$$ which results in different directional derivatives along the same direction. This might be undesirable.
• Do you mean @Henry W. $f'(x + tu) = D_u f(x + tu)$? Dec 27, 2016 at 9:18
• Sorry,I also feel trouble to understand the fact that $\gamma' (t) = ||u|| \phi' (t)$.Please be more explicit. Dec 27, 2016 at 9:42
It's a very nice idea to view $D_u f(x)$ as a function of $u$, with $x$ fixed. In convex analysis, it turns out that if $f$ is convex then the function $u \mapsto D_u f(x)$ is a convex function of $u$, and in fact is the support function of the set $\partial f(x)$ (under mild assumptions). This is a beautiful connection between two different notions of "derivative" that are useful in convex analysis. But we could not make this statement if $u$ were required to be a unit vector.
So I think the elegant / beautiful thing to do is to allow $u$ to be any vector, not necessarily a unit vector.