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I have a hopefully simple question. I am guessing that this has been asked before but I wasn't sure how to search for it.

Suppose we have two functions: $x(t)$, $y(t)$.

What is $\frac{dx}{dt}/\frac{dy}{dt}$? And why?

Note that this is not the setup for the standard chain rule (or at least its not obvious to me that it is). I talked to some friends and some of them say "just cancel out the $dt$s." Obviously, $\frac{dx}{dt}$, for example, is just the notation for a limit and not a ratio of numbers and so there is no reason as such that we should be able to "just cancel out."

Working with the limits directly, I can get up to:

$$ \lim_{\triangle t \to 0} \frac{x(t+ \triangle t)-x(t)}{y(t+\triangle t)-y(t)} $$

Thinking in terms of infinitesimals, this "feels" a lot like $dx/dy$ but I don't see any formal reason for why it is.

Thanks for your time.

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3 Answers 3

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The ratio you post is a consequence of the following, where we have $f(g(t))$, with $x = f(t)$, $y= g(t)$. Then by the chain-rule,

$$\dfrac{dx}{dt} = \dfrac{dx}{dy}\cdot \dfrac{dy}{dt} \iff \dfrac{dx}{dy} = \dfrac{\frac{dx}{dt}}{\frac{dy}{dt}}$$

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  • $\begingroup$ I think you are using a setup where $y$ is a function of $x$ and $t$ is a function of y? I know the chain rule applies in this situation but this not the setup I had. Am I missing something? $\endgroup$
    – user81164
    Commented Nov 11, 2013 at 18:05
  • $\begingroup$ Both are functions of $t$, but $x$ is also a function of $y$, like in $x = e^{2t + 1}$, $y = 2t + 1$, so by the chain rule, $dx/dt = dx/dy \cdot dy/dt$ $$x' = \underbrace{\frac{d}{dt}(2t + 1)}_{dy/dt}\underbrace{e^{2t+1}}_{dx/dy}$$ $\endgroup$
    – amWhy
    Commented Nov 11, 2013 at 18:11
  • $\begingroup$ If both are functions of $t$, then $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$, whereas $\dfrac{dx}{dy} = \dfrac{dx/dt}{dy/dt}$. $\endgroup$
    – amWhy
    Commented Nov 11, 2013 at 18:15
  • $\begingroup$ Both are functions of $t$. My question is precisely that: why is $dy/dx = \frac{dy/dt}{dx/dt}$ (for example)? $\endgroup$
    – user81164
    Commented Nov 11, 2013 at 18:40
  • $\begingroup$ And my explanation is that it follows from the chain rule. $$dy/dt = (dy/dx)\cdot(dx/dt)$$ Now we solve for $dy/dx$ $$dy/dx = \dfrac{dy/dt}{dx/dt}$$ $\endgroup$
    – amWhy
    Commented Nov 11, 2013 at 18:45
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If you ask me what $$\frac{\frac{dx}{dt}}{\frac{dy}{dt}}$$ is, my answer is simple: it is $\frac{x'(t)}{y'(t)}$. As your friends say, $\frac{d}{dt}$ is the operator that acts as the derivative (with respect to $t$). All other interpretations require additional assumptions: if you assume that $t$ can be expressed as a differentiable function of $x$, then $$ \frac{dy}{dx} = \frac{dy}{dt}\frac{dt}{dx}, $$ and if $t=t(x)$ is invertible, then $\frac{dt}{dx} = \left(\frac{dx}{dt} \right)^{-1}$, and therefore $$ \frac{dy}{dx} = \frac{dy}{dt} \left(\frac{dx}{dt} \right)^{-1}. $$ But you need to know that you can invert a function, at least.

Your idea about taking a limit as $\Delta t \to 0$ is very dangerous, since $y(t+\Delta t)=y(t)$ also for $\Delta t \neq 0$, and you are not allowed to divide by zero. This is the popular but wrong proof of the chain rule.

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  • $\begingroup$ We're working with $\dfrac{dx}{dy}$, not $dy/dx$. $\endgroup$
    – amWhy
    Commented Nov 11, 2013 at 17:48
  • $\begingroup$ Oh my God, can't you just swap $y$ and $x$? $\endgroup$
    – Siminore
    Commented Nov 11, 2013 at 17:49
  • $\begingroup$ Siminore: Thanks for the reply. The limit expression is just based on the definitions of $dx/dt$ and $dy/dt$ (this is where the $\triangle t \to 0$ comes from). I can take the limit operator out because the limit defined by $dy/dt$ exists by assumption. $\endgroup$
    – user81164
    Commented Nov 11, 2013 at 17:57
  • $\begingroup$ Good answer, although I think you mean that $x(t)$ has to be invertible (otherwise $t(x)$ doesn't make sense). $\endgroup$ Commented Nov 11, 2013 at 18:00
  • $\begingroup$ Well, actually, $dt/dx$ implies that $t$ is a function of $x$. If this function can be inverted as $x=x(t)$, then you can apply the rule to differentiate the inverse function. $\endgroup$
    – Siminore
    Commented Nov 11, 2013 at 18:02
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The formal reason for why $$\lim_{\triangle t \to 0} \frac{x(t+ \triangle t)-x(t)}{y(t+\triangle t)-y(t)} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$ is because the very definition of dy/dt is $$\frac{dy}{dt} = \lim\limits_{\Delta t\rightarrow\infty} \frac{y(t+\Delta t) - y(t)}{\Delta t} $$.

Basically, $$\lim_{\triangle t \to 0} \frac{x(t+ \triangle t)-x(t)}{y(t+\triangle t)-y(t)} =\lim_{\triangle t \to 0} \frac{\frac{x(t+ \triangle t)-x(t)}{\Delta t}}{\frac{y(t+\triangle t)-y(t)}{\Delta t}} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{dy}{dx}$$

To clarify the final step, If we we want to find a tangent line on to a curve where y is a differentiable function of x, then the Chain Rule gives $\frac{dy}{dt} = \frac{dy}{dt}\cdot\frac{dx}{dt}$, solving for $\frac{dy}{dx}$ gives us $\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

If you want a better way to think about it rather than the $dt$'s cancelling out, think of it this way: $$ \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{dy}{dt}\cdot\frac{dt}{dx} = \frac{dy}{dx}\cdot\frac{dt}{dt} = \frac{dy}{dx}\cdot 1 = \frac{dy}{dx} $$

As for $\frac{dt}{dt} = 1$, even though every one on this site is about to kill me for this, the above expression can be thought of as:

$$ \frac{dt}{dt} \rightarrow \frac{\frac{1}{\infty}}{\frac{1}{\infty}} = \frac{\infty}{\infty} = \frac{1}{1} $$

Afterthought

The equation you are trying to prove assumes the relation that you are having fault with. This is assumed because it is first assumed that both f and g are differentiable functions and that y is differentiable with respect to x. It isn't an algebraic manipulation, but rather an equality made by the construction of the functions themselves. Since formulation of $\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ assumes that $ \frac{dy}{dx}$, the relation is true by construction.

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  • $\begingroup$ In the expression after the "Basically", how do you get the last equality? That is my original question. Also, in re-arranging $\frac{dy}{dt} \frac{dt}{dx}$, I guess you are treating these derivatives as ratios? I don't see why this is justified. $\endgroup$
    – user81164
    Commented Nov 11, 2013 at 18:06
  • $\begingroup$ I updated the OP, and in response to the justification of treating them as ratios, you're actually right about it not being justified. Its more of a trick of notation, and there probably is a formal proof of it, but I state that it is only a way to think about it before using it. I'm actually going to try and justify it using limit laws, but there probably isn't one that I can derive at my level. $\endgroup$ Commented Nov 11, 2013 at 18:29
  • $\begingroup$ The proof of the standard chain rule you are using assumes that y is a y function of x (as you mention) and x is a function of t. I am considering a setup where y and x are both functions of t. Maybe there is a way to rewrite the my setup in terms of the standard setup (perhaps with some additional assumptions) but they are not obviously the same. $\endgroup$
    – user81164
    Commented Nov 11, 2013 at 18:45
  • $\begingroup$ Of course we assume that, y and x were parameterized as functions of t assuming dependence on each other. The expression you are trying to prove assumes mutual dependence, or else it would not make any sense. Infact, it would just be evaluated the quotient rule. $\endgroup$ Commented Nov 11, 2013 at 18:47
  • $\begingroup$ @user81164 I updated the answer with a definitive answer. I've checked with 3 textbooks, and they all agree, it isn't a manipulation, but instead the formulation of $\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$ assumes that y is differentiable with respect to x, hence: $ \frac{dy}{dx}$. $\endgroup$ Commented Nov 11, 2013 at 18:58

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