Why $F = m\frac{dv}{dt} = m\frac{dv}{dx}\frac{dx}{dt}$

Question

Why $F = m\frac{dv}{dt} = m\frac{dv}{dx}\frac{dx}{dt}$

I think differentials don't behave like fractions so why is that the previous equation holds, seems like is related to the definition of force where P is momentum using the chain rule:

$$P = mv$$ $$\frac{dP}{dx} = m \frac{dv}{dx}\frac{dx}{dt} $$

I would like to see a detailed explanation in the chain rule step

Answer 1

As noticed in the comments this is simply the chain rule applied at $x(t)$ and $v(x(t))$ that is

$$F = m\frac{dv(x(t))}{dt} = m\frac{dv(x(t))}{dx(t)}\frac{dx(t)}{dt}$$

Starting from the momentum equation, with $m$ constant, we have

$$p=mv \implies F=\frac{dp}{dt}=\frac{dp}{dx}\frac{dx}{dt}=m\frac{dv}{dx}\frac{dx}{dt}$$

Answer 2

For slightly more precision you should consider $m$ to be a function of $t$ as well, as for instance is the case with a rocket, loosing its own mass in order to move forward.

In that case you would have to consider the product rule instead, providing $$F(t)=\frac{d}{dt}p(t)=\frac{d}{dt}\left(m(t)\cdot v(t)\right)=\frac{dm}{dt}(t)\cdot v(t)+m(t)\cdot \frac{dv}{dt}(t)$$

However, if, as in your quest, $p$ is not directly considered to be a function of $t$, but to be a function of $x$, which in turn might be some function of $t$, then you would have to consider instead $$F(x(t))=\frac{d}{dt}p(x(t))=\frac{d}{dt}\left(m(x(t))\cdot v(x(t))\right)=\\ =\left(\frac{d}{dt}m(x(t))\right)\cdot v(x(t))+m(x(t))\cdot\left(\frac{d}{dt}v(x(t))\right)=\\ =\left(\left(\frac{d}{dx}m(x)\right)\cdot\left(\frac{d}{dt}x(t)\right)\right)\cdot v(x(t))+m(x(t))\cdot\left(\left(\frac{d}{dx}v(x)\right)\cdot\left(\frac{d}{dt}x(t)\right)\right)$$ which could have been derived also by applying the chain rule on $p(x(t))$ directly $$F(x(t))=\frac{d}{dt}p(x(t))=\left(\frac{d}{dx}p(x)\right)\cdot\left(\frac{d}{dt}x(t)\right)=\left(\frac{d}{dx}\left(m(x)\cdot v(x)\right)\right)\cdot\left(\frac{d}{dt}x(t)\right)=\\ =\left(\left(\frac{d}{dx}m(x)\right)\cdot v(x)+m(x)\cdot\left(\frac{d}{dx}v(x)\right)\right)\cdot\left(\frac{d}{dt}x(t)\right)$$

--- rk