# Directional derivatives a doubt about the definition

In the calculation books it usually appears that the norm of the directional vector is $$||y||=1$$. Does anyone know a compelling justification of why they do this? Because in the books of analysis, it is said that $$y$$ does not have to be unitary.

• Your title mentions partial derivatives, directional derivatives, and "a double about the definition". But in the body of your question I see no mention of partial derivatives or directional derivatives. If you are not getting adequate answers, you might get better ones if you edit the question to be more explicit about which definition(s) are in doubt, what those definitions are, and how this relates to partial derivatives and directional derivatives. Examples with details are often helpful. – David K May 13 at 15:01
• In calculus courses, one wants the directional derivative to depend just on the direction. In the setting of multivariable analysis, one is more interested in a linear map, and hence the directional derivative $D_{\mathbf v}f(\mathbf a)$ really becomes the rate of change at $\mathbf a$ as one moves with instantaneous velocity vector $\mathbf v$. – Ted Shifrin May 13 at 16:48
• @DavidK I edit my question think now my question is clear – Ricardo Freire May 13 at 19:40
• Yes, it is a much more specific question now. We know what notation you know, which helps. By the way, some related questions (showing how and why some people use unit vectors and some don't) are math.stackexchange.com/questions/2073264, math.stackexchange.com/questions/809376, math.stackexchange.com/questions/1952498, math.stackexchange.com/questions/2291302, and math.stackexchange.com/questions/1486767 – David K May 14 at 1:15

If a vector $$u$$ describes a direction, then $$\alpha u$$ describes the same direction for all $$\alpha \in \mathbb{R}$$, $$\alpha > 0$$. Thus, as long as we are only talking about the direction, we can normalize the vector without loosing information. Some formulae in vector calculus are based on the fact that direction vectors are normalized, therefore it is usually assumed in this context.
But again, it doesn't make any difference as long as you simply want to talk about a direction.

In the definition in the question, you must already have a definition of a kind of derivative written $$f'(a;y)$$ where $$a$$ is a point in the domain of $$f$$ and $$y$$ is a vector. Otherwise the definition shown in the question would be meaningless.

Presumably, the definition of $$f'(a;y)$$ is something equivalent to

$$f'(a;y) = \lim_{h\to0} \frac{f(a + hy) - f(a)}{h}.$$

Some authors might like to call this a directional derivative. Your textbook apparently calls it something else, because now we see the already-defined derivative $$f'(a;y)$$ being used to define a directional derivative.

This is something that happens in mathematics: in different books you will sometimes see the same words used in different ways. This occurs because not everyone is equally interested in all the different ways you can analyze some mathematical problems. Some people prefer to concentrate on one approach, some on another. But we have a limited tolerance for learning new words and new symbols, so instead people use the same words and sometimes the same symbols in slightly different ways.

The important thing is that as long as you're doing one "kind" of mathematics--the mathematics from a particular textbook, or a particular research paper, or a particular module of a course--you work from clearly stated definitions and the definitions remain the same while you do that "kind" of mathematics.

Some people have a geometric intuition about the slope of a plane tangent to a multidimensional graph, and measuring the slope in different directions by cutting it with different planes as in this question. For that kind of slope, once you have the orientation of the cutting plane you want a unique answer, which you don't get if you can use $$f'(a;y)$$ for any $$y$$ parallel to in the cutting plane. (If $$y$$ is parallel then so is $$2y,$$ but $$f'(a;2y) = 2f'(a;y).$$) Since the usual measure of slope is number of units of change in the function per unit distance traveled, the obvious way to get a unique answer is to choose a unit vector.

But another approach is to more generally apply the derivative $$f'(a;y),$$ allowing $$y$$ to be any kind of vector. This kind of derivative has the nice property that $$f(a;pu + qv) = p f(a;u) + q f(a;v)$$ where $$u$$ and $$v$$ are any vectors and $$p$$ and $$q$$ are any real numbers. If you are doing a lot of mathematics using facts like this, you'd like to be able to use words sometimes to describe this kind of derivative. You might then use the words directional derivative for this kind of derivative, especially if you very rarely have any reason to want thee $$y$$ in $$f(a;y)$$ to be a unit vector. That kind of mathematics is exhibited in many of the answers to this question.

By choosing a definition of directional derivative in which the vector must be a unit vector, the author of your definition has indicated something about which parts of the mathematics of multidimensional functions they're mainly going to be doing. Another author who chose a different definition has indicated that they're mostly going to be doing other parts of the mathematics of those functions.