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Suppose $x=x(t)$. Let $dx$ and $dt$ be 1-forms.

Does there exist a rigorous theorem for converting $dx=f(t)dt$ into the derivative ("fraction") expression $\frac{dx}{dt}=f(t)$?

Is this possible in general for other possibly multivariate functions (or does it require that $x$ is only a function of $t$). Or is this just false?

My guess is that by the linear independence of the basis differential form dt, it is possible to expand in the following manner and simply equate coefficients of $dt$.

$dx=\frac{dx}{dt}dt$

Therefore, $\frac{dx}{dt}=f(t)$.

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    $\begingroup$ You want to think of a mapping $x\colon\Bbb R\to\Bbb R$. We use $t$ as coordinate on the domain and $u$ as coordinate on the range. You pull back $du$ by the mapping $u=x(t)$ and you get, by definition, $dx=x^*du = x'(t)\,dt$. You now set this equal to your expression, and you have $x'(t)\,dt = f(t)\,dt$, whence $x'(t)=f(t)$. $\endgroup$ Feb 9, 2021 at 18:00
  • $\begingroup$ You haven't responded to this. If this ends the question, perhaps I should post it as an answer for you to accept. $\endgroup$ Feb 11, 2021 at 19:46
  • $\begingroup$ @TedShifrin Thanks, I believe that does answer the question! If you could post it as an answer, I'll accept that. $\endgroup$
    – Kcronix
    Feb 11, 2021 at 21:00

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As I commented, you want to think of a mapping $x\colon\Bbb R\to\Bbb R$. Using $t$ as a coordinate on the domain and $u$ on the range, you pull back $du$ by the mapping $u=x(t)$. Then you get $dx = x^*du = x'(t)\,dt$. Setting this equal to $f(t)\,dt$, you have $x'(t)\,dt = f(t)\,dt$, whence $x'(t)=f(t)$.

You asked about the multivariate situation. Consider, for example, $$dx = P(u,v)\,du + Q(u,v)\,dv.$$ Then this reduces likewise to the equations $$\frac{\partial x}{\partial u} = P(u,v) \quad\text{and}\quad \frac{\partial x}{\partial v} = Q(u,v).$$ Can you make the analogous argument?

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