Differential $dx$

Question

I have some trouble understanding a thing. I will reproduce two texts from two different books.

In the first, the author defines the function

$T:\mathbb{R}\longrightarrow \mathbb{R}$, $T(h)=f'(x_0)h$, $\forall h \in \mathbb{R}$,

where $f:I\longrightarrow \mathbb{R}$

is a differentiable function in $x_o$ and he calls it the differential of the function $f$ in $x_0$, denoted $df(x_0)$,

thus $df(x_0)(h)=f'(x_0)h, \forall h \in \mathbb{R}$.

If $f$ is differentiable in $x_0$,

we have the approximation

$f(x)-f(x_0) \cong f'(x_0)(x-x_0)$, for $x$

in a neighbourhood $V$ of $x_0$ (clear for me).

Denoting $h=x-x_0$, we can write:

$f(x)-f(x_0) \cong f'(x_0)h=df(x_0)(h)$.

In the case $f(x)=x$, we have for a point $x_0$: $f'(x_0)=1$,

hence: $df(x_0)(h)=h, \forall h \in \mathbb{R}$.

Denote $dx$ the differential of the function $id(x)$.

Then $dx(h)=h ,\forall h \in \mathbb{R}$

and we get $df(x_0)=f'(x_0)dx$.

If $f$ is differentiable on $I$ (interval),

we have $df(x)=f'(x)dx, \forall x \in I$.

Usually, $dx$ can be understood as an infinitesimal quantity (I don't see a connection with the previous definitions).

Now, in the second text appears the following sentence:

If $u:I\longrightarrow \mathbb{R}$ is a differentiable function, the expression:

$du=u'(x)dx$ is called the formal differential of the function $u$.

My first question: why do they call it "formal"?

Second question: what is the connection between the two texts?

Third: how is this related to $dx$, the infinitesimal quantity?

Answer 1

The first is a good book. The second is a poor book: it "defines" $du$ by means of an undefined symbol $dx$. The author could have written $$ du = u'(x) \spadesuit, $$ and call this a forma differential. The differential is precisely the linear map $T$, and introducing $dx$ as something that we don't want to define is rather old-fashioned.

Finally, infinitesimal quantities are defined only in non-standard analysis. If you study mathematics, please do not speak of infinitesimal quantities. You are allowed only if you teach general physics to students...

PS: of course you can give $dx$ a precise meaning, as the base of the dual space of $\mathbb{R}^1$. This is done in differential geometry courses, but of course you don't really need such an abstract machinery to speak of a linear map from $\mathbb{R}$ to $\mathbb{R}$.

Answer 2

There is no contradiction between the two.

Your difficulty stems from the circumstance that the symbol $x$ is used for two different purposes: (a) as variable for points on ${\Bbb R}$, and (b) as name for a certain function, namely the coordinate function $$\iota:\quad{\Bbb R}\to{\Bbb R},\qquad x\mapsto\iota(x):=x\ .$$ From $\iota(x+h)-\iota(x)=h$ it immediately follows that $$d\iota(x).h=h\qquad\forall h, \ \forall x\ ,$$ or $d\iota.h=h$ for short. Using the common $x$ in place for the fancy $\iota$ we therefore have $$du(x).h=u'(x)\>h=u'(x)\>d\iota.h=u'(x)\>dx.h\qquad\forall h\ .$$ But this says that $du(x)=u'(x)\>dx$. This is not an equation about infinitesimal quantities but the expression of the differential $du$ as a multiple of the coordinate differential $dx$.