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I'm having trouble understanding the true meaning of $df=f'(x)dx$ here, and I don't have any foundation knowledge in differential geometry. Can anyone explain it more specifically?

Sri-Amirthan Theivendran
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Bubblethan
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    $df$ represents the change in $f$ when given a change $dx$, not necessarily infinitely small (as $df\over dx$ notation suggests). You can use it to calculate the change in $f$ given a change in $x$. – Andrew Li Mar 19 '18 at 16:41
  • In advanced ways of thinking, $dx$ is a tangent vector in $\mathbb R$, the domain of $f.$ $df$ is a tangent vector in the range, also $\mathbb R.$ – Thomas Andrews Mar 19 '18 at 16:59
  • related: https://math.stackexchange.com/questions/2670900/rounding-real-numbers-and-its-effects-on-real-number-line-calculus/2670924#2670924 – Ethan Bolker Mar 19 '18 at 17:20
  • @ThomasAndrews: Differentials are cotangent vectors, actually. –  Mar 21 '18 at 12:06

2 Answers2

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Here is some general abstract nonsense:

Given a function $f:\>{\mathbb R}\to{\mathbb R}$ and a point $x$ in the domain of $f$ one denotes by $df(x)$ the linear map defined by $$df(x).X=f'(x)\, X\ .\tag{1}$$ Here $X$ is a real variable, and can take arbitrary values. In applications $X$ is often assumed to be a small quantity, as in $$f(x+X)-f(x)=f'(x)\,X+\ o\bigl(|X|\bigr)=df(x).X+\ o\bigl(|X|\bigr)\qquad(X\to0)\ .$$ Now the coordinate function $x:\>{\mathbb R}\to{\mathbb R}$ satisfies $$x(x+X)-x(x)\equiv X\ ,$$ which implies $$dx(x).X=X\ .$$ Comparing this with $(1)$ we can write $$df(x)=f'(x)\,dx(x)$$ as an equality between two linear functions, and if we suppress the reference to the point $x$ at which we did this analysis we obtain $$df=f'(x)\,dx$$ with no infinitesimal quantities at stake. What has been explained here can be performed as well in a multivariable environment.

Christian Blatter
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$dx$ represents a very small change in $x$. Whether this change is finite but arbitrarily small or infinitesimal depends on your Calculus book/teacher.

$df$ represents the corresponding change in the resulting function based on the amount of change given by $dx$.

It sounds like you are being taught is that the change is finite but arbitrarily small. What this means is that they are considering $dx$ to always represent a real value, but sufficiently small that its value is close to zero.

Additionally, as $dx$ gets closer and closer to zero, the ratio between $df$ and $dx$ will (generally) converge to a single number.

Personally, I think that the infinitesimal approach (where $dx$ represents a number that is an infinitely small value) is more intuitive and easier to apply, but both methods work.

johnnyb
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