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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?

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    $\begingroup$ It's not composite, just the usual product. $\endgroup$ – Vim Jun 12 '17 at 2:33
  • $\begingroup$ Possible duplicate of Is $\frac{\textrm{d}y}{\textrm{d}x}$ not a ratio? $\endgroup$ – Masacroso Jun 12 '17 at 3:32
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    $\begingroup$ Notice that two different functions are both being called by the same name $v$. Really, we should define $\hat v(t) = v(x(t))$, and write this equation as $\hat v'(t) = v'(x(t)) x'(t)$. Conflating $\hat v$ and $v$ is a common abuse of notation but it's one of my pet peeves, and I think it causes a lot of confusion. $\endgroup$ – littleO Jun 12 '17 at 3:49
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    $\begingroup$ @Masacroso: I don't think it's a duplicate, since this question is not about the “quotient of differentals”, but rather about the double meanings of the symbol $v$. $\endgroup$ – Hans Lundmark Jun 12 '17 at 6:02
  • $\begingroup$ This is typical equation used to derive the formula for kinetic energy as $mv^{2}/2$. Note that $x, v$ are functions of $t$ and $dv/dx$ is nothing but $(dv/dt) /(dx/dt) $. This is nothing but chain rule. You say that $v=x'$ where primes denote differentiation with respect to $t$ then $dv/dx=dx'/dx=(dx'/dt)/(dx/dt) = x''/x'=a/v$ where $a$ is acceleration. $\endgroup$ – Paramanand Singh Jun 12 '17 at 14:12
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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 $w\circ x$: $(w\circ x)'=(w'\circ x)\cdot x'$.

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  • $\begingroup$ I must say WHAT. Why would anybody use that notation for $(v\circ x)'$?! Everyday I hate physicists' sloppiness more and more. $\endgroup$ – YoTengoUnLCD Jun 12 '17 at 2:39
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    $\begingroup$ @YoTengoUnLCD: the chain rule is mathematics, not physics. That notation is commonplace, and it is what one uses to do integrals by substitution. $\endgroup$ – Martin Argerami Jun 12 '17 at 2:41
  • $\begingroup$ @MartinArgerami Yes of course, using proper notation $(v\circ x)'=v'\circ x \cdot v $ is completely obvious. $\endgroup$ – YoTengoUnLCD Jun 12 '17 at 2:42
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    $\begingroup$ @YoTengoUnLCD: Here's an attempt to answer your question “why”: to physicists, the symbols $v$, $x$, $t$ denote the values of the physical quantities velocity, position, time, and they don't like wasting letters by giving names to the functions that describe how these quantities depend on each other. A mathematician might write $v=f(x)$ and $v=g(t)$, and then the notations $f'=df/dx$ and $g'=dg/dt$ are unambiguos, but physists (and mathematicians too, in some context) prefer to write $dv/dx$ and $dv/dt$ instead. $\endgroup$ – Hans Lundmark Jun 12 '17 at 5:57
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    $\begingroup$ @YoTengoUnLCD: Giving names to these functions usually clears up any confusion immediately, and I often give that advice to students. In multivariate calculus it's even worse. See this answer for another example: math.stackexchange.com/a/942543/1242 $\endgroup$ – Hans Lundmark Jun 12 '17 at 6:00
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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 $h'(x)$.

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  • $\begingroup$ What do you mean by "making more sense when working with related variables than functions"? Both $x$ and $v$ are functions by definition in classical mechanics (from some interval $I$ to some $\Bbb R^d$). And they are functions related by $x'=v$. $\endgroup$ – YoTengoUnLCD Jun 12 '17 at 6:49
  • $\begingroup$ @YoTengoUnLCD: That's because introductory math classes tend not to explicitly teach related variables, so people learn to translate in terms of functions, because that's all they know. The usual translation style (use $v$ both for the variable and for a function that expresses the dependence of $v$ on some other variable) is problematic -- the OP, in fact, appears to demonstrate one of the reasons why. The function expressing the dependence of $v$ on $x$ is different than the function expressing the dependence of $v$ on $t$... which tends to be overlooked when thinking of $v$ as a function $\endgroup$ – user14972 Jun 12 '17 at 7:06

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