I've been doing some reading, and even though many people say different things, i think i'm pretty confident in saying that we can't treat differentials as fractions. In some scenarios it works out (chain rule), but in others it will not. Leibniz notation is still notation. It's just very suggestive of the math going on.
Im trying to prove the work-energy theorem, and this is my progress so far: $$\int F*ds => \int ma*ds => m\int a*ds$$
If acceleration is integrated w.r.t ds, then it must be a function of ds too. We also know that ds, the position, must be a function of time. $$m\int a(s(t))*ds(t) => m\int \frac{d}{dt}[v(s(t))]*ds(t) $$
With the chain rule, this gives me: $$m\int v'(s(t)) *v(t)*ds(t)$$
And this is where i'm stuck. I can't seem to find a way to integrate this, especially since i haven't really seen an integral like this before. Time t seems to be the independent variable, but it is still integrated w.r.t ds. All proofs ive found do the following: $$m\int \frac{dv}{ds}*\frac{ds}{dt}*ds => m\int v*dv => m\frac{v^2}{2}$$ And since this integral was defined to be evaluated from a point A, to point B, we measure a difference in kinetic energy.
But i'm not okay with that proof. We can't just treat differentials as fractions whenever we feel like it. We can say that $$ \frac{df}{dx}=\frac{df}{dv} * \frac{dv}{dx}$$ not because the $dv's$ cancel out, but because of the chain rule (since this is just the chain rule written in Leibniz notation). This proof of the work energy-theroum clearly is correct. What is it that allows us to cancel out these fractions in this scenario? What special rule or derivation makes us certain that canceling out $ds$ is a valid step.
Any help is greatly appreciated. Have a good day!