# Dx as a small change in x

• I
Can dx be thought of as a sufficiently small change in x? I want to say that dx is the change in x and change in x approaches 0, but that would just be 0.

So I think it might make more sense to just say sufficiently small. Then when we look at something like a derivative dy/dx we can look at the sufficiently small value of x more closely by finding the limit as change in x approaches 0.

So can differentials like dy and dx be thought of as suffiently small changes in y and x?

In calculus I was told that these things had no meaning on their own, but they seemed to be used quite a bit on their own in both calculus and physics. This was never really explained very clearly. I often see answers saying that it’s too complicated or unimportant for early calculus classes, but we have to use these differentials so that’s always a bit irritating.

fresh_42
Mentor
I think it is better to consider them as the basis vectors of the tangent space, e.g. ##dx,dy## in two dimensions. This way ##\frac{dy}{dx}## becomes naturally the slope of a straight, the tangent. Differentiation itself can be viewed from many angles. I once listed some and easily found ten different ways to look at it, and I didn't even mention the term slope. This comes from what you want to take as variable: the function, the point of evaluation, the process of differentiation, the result, or the direction of the linear approximation.

FS98 and jedishrfu
I think it is better to consider them as the basis vectors of the tangent space, e.g. ##dx,dy## in two dimensions.
Can you explain what this means in layman’s terms?

Also, is this a somewhat controversial question? I’ve had a hard time looking for answers, partially because some of the answers I’m seeing seem contradictory. Some mention infitessimals, others seem dismissive of infitessimals or don’t mention them at all.

fresh_42
Mentor
Can you explain what this means in layman’s terms?
If we have a function ##f : \mathbb{R}\longrightarrow \mathbb{R}## then we usually draw it in the ##(x,y)-##plane, i.e. we have a curve ##(x,f(x))##. Now the tangent at some point ##Q## is a straight line, say ##T##.

Lines in general can be written as ##y=mx+b = \dfrac{\Delta P_y}{\Delta P_x}\cdot x + b## with two points ##P_1=(0,b)## and ##P_2=(1,m+b)## on this line. Thus $$T=\left\{ \begin{bmatrix} x \\ y \end{bmatrix}=\begin{bmatrix}0\\b\end{bmatrix}+t\cdot \begin{bmatrix}1\\m\end{bmatrix} = P_1 + T_0\right\}$$ Now ##T=P_1+ T_0## where ##T_0## is a straight line through the origin shifted by ##P_1##. We want to consider our tangent ##T## as an element of a vector space, so we only investigate the direction, the line ##T_0##. As it is simply shifted by ##P_1## to be a certain tangent at a given point, we don't lose information. We only consider the space of tangents as a vector space, whose elements have to be shifted by a certain point vector ##P_1## to get the actual tangent in our graph. But here we're not interested in certain tangents, only in their directions, i.e. ##T_0##.

That might look wrong at first glance, but if we calculate ##f\,'(x)##, we don't have the tangent either, only its slope. E.g. for ##f(x)=x^2##, we get ##f\,'(x)=2x## which is not the tangent at ##x=3##, because we have ##f\,'(3)=6## but the tangent is ##y=6x-9## or ##T_{at \,\,x=3}=(0,-9) + \{y=6x\}##. So differentiation gives us basically the slope and direction part of the tangent, not the actual location in the graph. That's why we are interested to see what ##T_0## does, the tangent shifted such that it goes through the origin, ##T_0=\{y=6x\}## in our example.

Now here's the clue: $$T_0 = \left\{ \begin{bmatrix}x\\y\end{bmatrix}\, : \,0=\left\langle \begin{bmatrix}x\\y\end{bmatrix},\begin{bmatrix}1\\-\frac{1}{m}\end{bmatrix}\right\rangle = x\cdot 1 + y \cdot \left( -\dfrac{1}{m}\right)\right\}$$
where the vector ##(1,-\frac{1}{m})## is the normal vector of our tangent, perpendicular to the tangent direction ##(1,m)##.

I've said, that ##dx## and ##dy## should be regarded as basis vectors of the tangent space, i.e. the vector space where our ##T_0## live. So we should not write ##x,y## anymore, but ##dx,dy## instead. The misleading ##x,y## notation is merely due to school math where both, functions and their tangents are drawn within the same graph with ##(x,y)-##coordinates. Just because we don't distinguish them doesn't mean they are equal. It's only for convenience, and now we pay the price, as things are no longer just drawn on a piece of paper.

Therefore we should write
$$T_0 = \left\{ \begin{bmatrix}dx\\dy\end{bmatrix}\, : \,0=\left\langle \begin{bmatrix}dx\\dy\end{bmatrix},\begin{bmatrix}1\\-\frac{1}{m}\end{bmatrix}\right\rangle = dx\cdot 1 + dy \cdot \left( -\dfrac{1}{m}\right)\right\}$$
which is the same as ##mdx=dy## or ##m=\dfrac{dy}{dx}\,##, the slope ##m=f\,'(Q)## of our tangent at a point ##Q##.
Also, is this a somewhat controversial question?
No. It's a matter of point of view and the degree of abstraction that is needed in certain situations.
I’ve had a hard time looking for answers, partially because some of the answers I’m seeing seem contradictory. Some mention infitessimals, others seem dismissive of infitessimals or don’t mention them at all.
The infinitesimals are only a short hand notation for the actual limits involved: ##\lim_{\Delta x \to 0}\dfrac{\Delta y}{\Delta x}=\dfrac{dy}{dx}##. They are neither of semantic nor syntactic rigor, but historically grown, shorthanded and to some degree convenient. So whether someone speaks of them and someone dismisses it, depends on the kind of rigor he wants to use. E.g. take a closer look at the limit: there is one limit and the denominator is shrunk, whereas the nominator changes only because of that, not by its own. Despite this fact we still write ##dy## with the same small ##d##. As people usually know what they mean, there is no need to explicitly mention all conventions used beforehand. There is simply more than one way to look at it. It's not really controversial, as long as you don't meet a purist or logician, it's more a matter of convenience, habit and intention.

FS98 and K Murty
osilmag
Gold Member
It's usually thought of as an instantaneous change.

PeroK
Homework Helper
Gold Member
2020 Award
Can dx be thought of as a sufficiently small change in x? I want to say that dx is the change in x and change in x approaches 0, but that would just be 0.

So I think it might make more sense to just say sufficiently small. Then when we look at something like a derivative dy/dx we can look at the sufficiently small value of x more closely by finding the limit as change in x approaches 0.

So can differentials like dy and dx be thought of as suffiently small changes in y and x?

In calculus I was told that these things had no meaning on their own, but they seemed to be used quite a bit on their own in both calculus and physics. This was never really explained very clearly. I often see answers saying that it’s too complicated or unimportant for early calculus classes, but we have to use these differentials so that’s always a bit irritating.
http://tutorial.math.lamar.edu/Classes/CalcI/Differentials.aspx