Split the differential and differential forms

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In undergraduate dynamics, they do things like this:
--------------------
v = ds/dt
a = dv/dt
Then, from this, they construct: a ds = v dv
And they use that to solve some problems.
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Now I have read that it is NOT wise to treat the derivative like a fraction: it obliterates the meaning.
And that such tricks like the above one, work only in 1D cases. But it is bad policy to get used to it.

I have a FEELING for that, but no PRECISE explanation of why it is unwise to treat the derivative like a fraction.

Can someone please explain this?

And if you can explain it -- and I hope you can -- then I will come back and ask you to discuss that in the context of differential forms where "dx" is a co-vector.

Because with regard to differential forms, one DOES have these bases from the dual space.

Could someone address this for me?
 
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As long as you know what you do, you can treat it as a fraction - as always with abbreviations. A closer inspection, however, gives rise to some questions: Where is the limit? In the "d"? But we have only one limit and two "d", and this is the major risk here ##\displaystyle \lim \frac{\Delta x(t)}{\Delta t} \neq \frac{\lim \Delta x(t)}{\lim \Delta t}##. Another point is, that it marks a derivative, that is ##\displaystyle t_0 \longmapsto \left. \frac{dx(t)}{dt}\right|_{t=t_0}## which is an entire vector field. I recently counted the ways a derivative can be viewed as and found ##10## different interpretations of only one formula, and the fraction wasn't even among them:

$$
D_{x_0}L_g(v)= \left.\frac{d}{d\,x}\right|_{x=x_0}\,L_g(x).v = J_{x_0}(L_g)(v)=J(L_g)(x_0;v)
$$
can be viewed as
  1. first derivative ##L'_g : x \longmapsto \alpha(x)##
  2. differential ##dL_g = \alpha_x \cdot d x##
  3. linear approximation of ##L_g## by ##L_g(x_0+\varepsilon)=L_g(x_0)+J_{x_0}(L_g)\cdot \varepsilon +O(\varepsilon^2) ##
  4. linear mapping (Jacobi matrix) ##J_{x}(L_g) : v \longmapsto \alpha_{x} \cdot v##
  5. vector (tangent) bundle ##(p,\alpha_{p}\;d x) \in (D\times \mathbb{R},\mathbb{R},\pi)##
  6. ##1-##form (Pfaffian form) ##\omega_{p} : v \longmapsto \langle \alpha_{p} , v \rangle ##
  7. cotangent bundle ##(p,\omega_p) \in (D,T^*D,\pi^*)##
  8. section of ##(D\times \mathbb{R},\mathbb{R},\pi)\, : \,\sigma \in \Gamma(D,TD)=\Gamma(D) : p \longmapsto \alpha_{p}##
  9. If ##f,g : D \mapsto \mathbb{R}## are smooth functions, then $$D_xL_y (f\cdot g) = \alpha_x (f\cdot g)' = \alpha_x (f'\cdot g + f \cdot g') = D_xL_y(f)\cdot g + f \cdot D_xL_y(g)$$ and ##D_xL_y## is a derivation on ##C^\infty(\mathbb{R})##.
  10. ##L_x^*(\alpha_y)=\alpha_{xy}## is the pullback section of ##\sigma: p \longmapsto \alpha_p## by ##L_x##.
The question about the dimension is a bit tricky. As long as we consider a single derivative, it is a directional differential, in which direction ever. But - as in the examples above - the entire tangent bundle is often considered, and we get
$$
d\, f = \sum_{i=1}^n \frac{\partial f}{\partial x_i}\,dx_i
$$
and then we have ##n## fractions in ##n## directions, which are reduced to a scalar, the component of the corresponding tangent vector. The fraction is o.k. as long as we consider the whole thing as a slope. If we start using it as a real quotient and calculate with it, we have to keep in mind, that it is merely an abbreviation. If it helps to find a solution, fine, but we should check the answer and especially be careful if we'll deal with functions, that are not smooth.
 
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fresh_42 said:
As long as you know what you do, you can treat it as a fraction - as always with abbreviations. A closer inspection, however, gives rise to some questions: Where is the limit? In the "d"? But we have only one limit and two "d", and this is the major risk here ##\displaystyle \lim \frac{\Delta x(t)}{\Delta t} \neq \frac{\lim \Delta x(t)}{\lim \Delta t}##. Another point is, that it marks a derivative, that is ##\displaystyle t_0 \longmapsto \left. \frac{dx(t)}{dt}\right|_{t=t_0}## which is an entire vector field. I recently counted the ways a derivative can be viewed as and found ##10## different interpretations of only one formula, and the fraction wasn't even among them:

$$
D_{x_0}L_g(v)= \left.\frac{d}{d\,x}\right|_{x=x_0}\,L_g(x).v = J_{x_0}(L_g)(v)=J(L_g)(x_0;v)
$$
can be viewed as
  1. first derivative ##L'_g : x \longmapsto \alpha(x)##
  2. differential ##dL_g = \alpha_x \cdot d x##
  3. linear approximation of ##L_g## by ##L_g(x_0+\varepsilon)=L_g(x_0)+J_{x_0}(L_g)\cdot \varepsilon +O(\varepsilon^2) ##
  4. linear mapping (Jacobi matrix) ##J_{x}(L_g) : v \longmapsto \alpha_{x} \cdot v##
  5. vector (tangent) bundle ##(p,\alpha_{p}\;d x) \in (D\times \mathbb{R},\mathbb{R},\pi)##
  6. ##1-##form (Pfaffian form) ##\omega_{p} : v \longmapsto \langle \alpha_{p} , v \rangle ##
  7. cotangent bundle ##(p,\omega_p) \in (D,T^*D,\pi^*)##
  8. section of ##(D\times \mathbb{R},\mathbb{R},\pi)\, : \,\sigma \in \Gamma(D,TD)=\Gamma(D) : p \longmapsto \alpha_{p}##
  9. If ##f,g : D \mapsto \mathbb{R}## are smooth functions, then $$D_xL_y (f\cdot g) = \alpha_x (f\cdot g)' = \alpha_x (f'\cdot g + f \cdot g') = D_xL_y(f)\cdot g + f \cdot D_xL_y(g)$$ and ##D_xL_y## is a derivation on ##C^\infty(\mathbb{R})##.
  10. ##L_x^*(\alpha_y)=\alpha_{xy}## is the pullback section of ##\sigma: p \longmapsto \alpha_p## by ##L_x##.
The question about the dimension is a bit tricky. As long as we consider a single derivative, it is a directional differential, in which direction ever. But - as in the examples above - the entire tangent bundle is often considered, and we get
$$
d\, f = \sum_{i=1}^n \frac{\partial f}{\partial x_i}\,dx_i
$$
and then we have ##n## fractions in ##n## directions, which are reduced to a scalar, the component of the corresponding tangent vector. The fraction is o.k. as long as we consider the whole thing as a slope. If we start using it as a real quotient and calculate with it, we have to keep in mind, that it is merely an abbreviation. If it helps to find a solution, fine, but we should check the answer and especially be careful if we'll deal with functions, that are not smooth.
Perfect.
Thank you.