What exactly is meant by $\frac {dv}{dt} = \frac {dv}{dx}\frac {dx}{dt}$? I'm a bit embarrased to have to ask this, as I guess I'm missing something completely basic: I've seen various physics problems solved using $\frac {dv}{dt} ``=" \frac {dv}{dx}\frac {dx}{dt}$, but I'm not clear on what this is saying.
To be precise, I suppose $x,v$ are functions $x: \Bbb R\to \Bbb R$, with $v:=x'$. Furthermore, I would suspect that $\frac{dv}{dx}:= v'\circ x$, and $\frac {dv}{dt}:=v'$.
With that interpretation, the equality is simply false, as it reads $v'=v'\circ x \cdot v$, which fails for example if $x: k\mapsto k^2$.
Can someone clarify this situation?
 A: Your interpretation of $\frac{dv}{dx}$ is wrong, since $v'$ is the derivative of $v$ "with respect to $t$", not "with respect to $x$".  To take the derivative with respect to $x$, you need to "make $v$ a function of $x$": that is, you need to consider the function $v\circ x^{-1}$.  So $\frac{dv}{dx}$ actually refers to $(v\circ x^{-1})'\circ x$. So the equation says $$v'=((v\circ x^{-1})'\circ x)\cdot v.$$ This equation follows from the chain rule.  Indeed, if you write $w=v\circ x^{-1}$ so $v=w\circ x$, this is just the chain rule for $w\circ x$: $(w\circ x)'=(w'\circ x)\cdot x'$.
A: This notation makes more sense when working with related variables rather than functions. If you want to insist on using functions, then...
Let $f$, $g$, and $h$ be functions such that:


*

*$v = f(t)$

*$x = g(t)$

*$v = h(x)$


Then in terms of functions, the asserted equation is
$$ f'(t) = h'(x) g'(t) $$
or, if you want the whole thing expressed in terms of $t$,
$$ f'(t) = h'(g(t)) g'(t) $$

That said, what is meant by the notation is, for example, that $\frac{\mathrm{d}v}{\mathrm{d}x}$ is the ratio of the two differentials $\mathrm{d}v$ and $\mathrm{d}x$.  And, naturally, the equation
$$\mathrm{d}v = h'(x) \mathrm{d} x$$
is true (with $h$ as defined above), so the ratio $\frac{\mathrm{d}v}{\mathrm{d}x}$ is indeed $h'(x)$.
