In learning multivariable calculus, I've often seen the gradient introduced before the directional derivative. To me this is backwards. Once we have partials, we treat then as rates of change in the direction of the axes. We should then naturally ask "what if the direction were not aligned with the axes?". From there, a natural question of "in which direction is the rate of change maximal?" Seems to arise. In the development of calculus, which came first?

I find that by introducing the gradient before directional derivatives it seems like some arbitrary vector that was pulled out of no where. It only seems to make sense once we understand directional derivatives.


The following is a partial answer based on my own reading in Kline's Mathematical Thought from Ancient to Modern Times and the Jeff Miller's website on the earliest uses of the symbols and words involved.

The development of vectors and their associated calculus has a rather fractious history in the nineteenth century. It seems best to leave the divergence theorem and its friends out of this, since that history is complicated enough as it is.

Vectors and vectorial thinking are a very late development in the century. To have an idea of how people thought before, a reasonable place to look is that throwback, Green's theorem, which is stated as $$ \int P \, dx + Q \, dy = \iint \left( \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} \right) \, dx \, dy. $$ Here $P$ and $Q$ are just functions of $x$ and $y$, with no relationship between them. This formula is quite annoying, since it's pretty hard to remember where the minus sign goes. But this is how people worked: you can look in the paper on the divergence theorem I linked above to see this in more detail.

How about derivatives, then? Lagrange is credited with the form of Taylor's theorem in two variables. Are there directional derivatives in this? Not really: it's a function of two variables, not a vector.

The first "vectory" thing that is produced is the quaternions, by Hamilton in the 1840s (that they were written down by Gauss and not published years earlier is both typical and not really relevant, since Gauss's unpublished work rarely leaks out into common mathematical parlance in his lifetime, at least). This is the beginning of a shift towards what becomes abstract algebra: this, Grassmann's work, and shortly afterward Cayley and others, push away from things that have to be grounded in reality to things that rely on the rules of symbol-pushing. Abstract vector spaces come late enough that they are not worth talking about here. Concrete vectors in the Gibbs–Heaviside model also come along much later, initially as an offshoot of quaternions that takes the physically useful parts and strips off the rest. There's an essay that describes the debate between the emerging vectorialists and quaternionists here.

The quaternions have two bizarre properties: firstly, that they are not commutative, and secondly that they need four real numbers to describe them, which was rather puzzling when previously everything lived in three dimensions. But Hamilton also writes down what is undeniably a vector operator, $$ \nabla = \mathbf{i} \frac{\partial}{\partial x} + \mathbf{j} \frac{\partial}{\partial y} + \mathbf{k} \frac{\partial}{\partial z} $$ (actually, Hamilton's $\nabla$ is rotated to point left, but it's too fiddly to do that on here). Then $\nabla \mathbf{q}$, where $\mathbf{q}$ is a quaternion, gives as the "scalar part" a quantity that happens to be the negative of the divergence (Maxwell will call it the convergence) and as the "vector part" the curl (as seen on this page of Hamilton's Lectures on Quaternions), so this $\nabla$ is in a sense the only operator you need. On the next page he discusses the action of $\nabla$ on a scalar function (a temperature, or gravitational potential) as describing the vector (of heat flow, or gravitational force acting): in the case of a generic function, a normal vector. So this is certainly a reasonable candidate for the first discussion of a gradient vector (it seems no one else even had the means to write this operator as an object in its own right before Hamilton), although the term gradient is first used much later.

Now, the directional derivative. Hamilton does not connect the directional derivative to $\nabla$ (it should be obvious why: it's missing the scalar part of the quaternion), instead describing everything for Taylor series in terms of differentials (see this page and following). The anticommutativity prevents the production of a gradient vector, since the $dq$s appear interleaved in the expressions. This is one disadvantage of the quaternionic approach. If you want to call this a directional derivative, that's the end of it. The actual phrase appears in A Short Table of Integrals by B. O. Pierce, accompanied by a modern geometric interpretation, but it is clear that the concept is rather older in one form or another. But I think the idea of a directional derivative using an actual vector therefore has to be younger than $\nabla$, since Hamilton is the first to publish anything treating a multi-dimensional object as one thing, rather than a pile of components. It's a matter of opinion rather than absolute fact, though.


It looks like that when think of the gradient as an "arrow" pointing towards a direction (the one of the "steepest ascent"). This can certainly be done, but I think that the gradient was born just as the "list" of the partial derivatives (I have no references for this). Thought like this, it is even more basic that a generic directional derivative.

  • $\begingroup$ Perhaps, but it is a vector by definition. If it were just a list then the components would be arbitrary. I could easily write the y derivative first then the x, in 2d, for example, and get the same list. $\endgroup$ – Michael Stachowsky May 26 '17 at 12:32
  • $\begingroup$ A vector itself can be just a list. It is not mandatory to associate an arrow to, say, $(2,3)$, it might just be thought as a point in the plane, or even just as an abstract couple of numbers $\endgroup$ – lesath82 May 26 '17 at 12:33
  • $\begingroup$ I'm not entirely convinced of that, but I'm willing to hear more about it. Even if you don't associate your vector with Euclidean space, the order of its components matters (or, more accurately, which component is associated with which basis vector in your vector space is important). It's not just a list of elements, it's a list of elements associated with a list of bases, is it not? $\endgroup$ – Michael Stachowsky May 26 '17 at 12:38
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    $\begingroup$ As long as you are happy with your "ordered list", you don't need a basis. $\mathbb R^2$ can be embedded in the plane just as every real number can be thought as the measure of a segment, but it can also just be "ordered couples of numbers" on which you can define operations and other stuff $\endgroup$ – lesath82 May 26 '17 at 12:41
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    $\begingroup$ Honestly I don't think that the idea of a basis is necessary to implement the concept of "ordered list", but they might be shown to be related more than I think @MichaelStachowsky $\endgroup$ – lesath82 May 26 '17 at 13:07

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