Is this statement correct : $\frac{dv}{dt}\times dx=\frac{dx}{dt}\times dv$

Question

So I just came across a physics derivation where the treat the $dv$ and $dx$ operators like fractions while I have always heard it's a mistake. But so far what I came up with:

$$\begin{align*} \frac{dv}{dt}\times dx &= \lim_{h\to0} \frac{v(t+h)-v(t)}h \times (x(t+h)-x(t)) \\ &= \lim_{h\to0} \frac{x(t+h)-x(t)}h \times (v(t+h)-v(t)) \\ &= \frac{dx}{dt}\times dv \end{align*}$$

Is my reasoning correct? If not please explain what is the correct way to think about this.

Answer 1

Just before a differential disappears in the limit we are allowed to view them as small segments of length or small packets of velocity. Even after disappearance their ratio continues to make sense.

$$ \frac{\Delta y}{\Delta x} \to\frac{dy}{dx} =y' = \tan \phi$$ $$ \frac{\Delta x}{\Delta s} \to\frac{dx}{ds} =\cos \phi$$ $$ \frac{\Delta y}{\Delta s} \to\frac{dy}{ds} =\sin \phi. $$

Similarly packets of mass or velocity in a packet (interval) of time made.. even now makes sense in the Newtonian formulation of force :

$$ F =\frac{\Delta(m v)}{\Delta t }= \frac{v\Delta m}{\Delta t }+\frac{m\Delta v}{\Delta t }$$

$$\to v\frac{dm}{dt }+m \frac{ dv}{dt } $$

Even after the disappearance of finite differential form their product continues to make sense.

Answer 2

As stated in the comments, taking the limit doesn't work because technically the limit goes to $0$ as $h\rightarrow0$. As stated in Narasimham's answer, you think of differentials as representing very small (infinitesimally small), but non-zero, changes.

The rigorous way to derive the statement you mentioned above is:

$$\cfrac{\text{d}x}{\text{d}t}\text{d}v=\cfrac{\text{d}x}{\text{d}t}\left(\cfrac{\text{d}v}{\text{d}x}\text{d}x\right)=\cfrac{\text{d}x}{\text{d}t}\left(\cfrac{\cfrac{\text{d}v}{\text{d}t}}{\cfrac{\text{d}x}{\text{d}t}}\text{d}x\right)=\cfrac{\text{d}v}{\text{d}t}\text{d}x\ ,$$ where I used the fact that $\cfrac{\text{d}v}{\text{d}t}=\cfrac{\text{d}v}{\text{d}x}\cfrac{\text{d}x}{\text{d}t}$, which is the chain rule.

I think it's good practice to do things more rigorously like this if you ever take (or have taken) multi-variable calculus. Because then differential cancellations are less straightforward.