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If you have two derivatives, say $dx/dt$ and $dy/dt$, can you divide the two to get $dx/dy$?

I have looked around and the answer seems to be yes, but I have further questions. For example, what happens if $dy/dt=0$, so the denominator is $0$? I know that derivatives are limits, but we don't have as much to work with here, algebraically speaking. What if you divide two independent derivatives, say $(dw/dx)/(dy/dz)$ (including the case where they are all part of the same system, for example, if $dx/dy=1$)? How else could we algebraically manipulate derivatives (e.g. what happens if we multiply them? If $dx/dt$ is added to $dy/dt$, is the result $(dx+dy)/dt$? Essentially, how can derivatives be manipulated with and/or by each other?

I already tried searching around online with little success with my specific question.

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  • $\begingroup$ Um... this should have been covered in a basic calculus course. $(f + g)' = f' + g'$ so $\frac {d(x+y)}{dt} = \frac {dx}{dt} + \frac{dy} {dt}$ (the expression $dx+ dy$ is meaningless gobbledeegook). $\endgroup$
    – fleablood
    Sep 28 '20 at 19:27
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Well you don't. Not really.

Derivatives are limits. If we assume differentiability and limits behaving "nicely" then

Then if $x = x(y)$ and $y = y(t)$ then

Let $t_1 = t+h$ and $h = t_1 - t$.

Let $y_1=y(t_1) = y(t+h)$ and $j = y_1 - y = y(t+h) - y(t)$.

Let $x_2=x(y_1)=x(y(t+h))$ and $k = x_1 - x = x(y(t+h)) - x(y(t))$.

Then

$\frac {dx}{dy} = \lim_{k\to 0} \frac {x(y+k) - x(y)}k=\lim_{y_1\to y}\frac {x(y_1) - x(y)}{y_1 - y}=\lim_{h\to 0}\frac {x(y(t+h)) - x(y(t))}{y(t+h) - y(h)} $. As

And $\frac {dy}{dt} = \lim_{h\to 0} \frac {y(t+h) - y(t)}h$.

And $\frac {dx}{dt} =\lim_{h\to 0}\frac {x(y(t+h)) - x(y(t))}h$.

So that means

$\frac {dx}{dy}=\lim_{h\to 0}\frac {x(y(t+h)) - x(y(t))}{y(t+h) - y(h)}\cdot \lim_{h\to 0} \frac {y(t+h) - y(t)}h= \lim_{h\to 0}\frac {x(y(t+h)) - x(y(t))}{h} = \frac {dx}{dy}$.

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  • $\begingroup$ You need to assume that $y$ is is one-to-one in a neighborhood of $t$ for this to work (or else use the more complicated and less transparent fix used in , for exmple, Spivak's proof of the Chain Rule). $\endgroup$ Sep 28 '20 at 19:25
  • $\begingroup$ Yeah... well..... that goes in the "assume everything is nice" clause. $\endgroup$
    – fleablood
    Sep 28 '20 at 19:27
  • $\begingroup$ Well, it's more than just the limits behaving nicely... you need $y$ to behave nicely, too. $\endgroup$ Sep 28 '20 at 19:30
  • $\begingroup$ @ArturoMagidin what do you mean by one-to-one? Do you mean that it is a part of the same system (I just assumed that that was implicit in my question)? $\endgroup$ Sep 29 '20 at 13:58
  • $\begingroup$ Also, what I'm getting from this is that you can divide limits like I described, but it's a bit different than simple algebraic manipulation of, say, dx. This is, of course, assuming that the derivatives are "nice". $\endgroup$ Sep 29 '20 at 14:00
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The problem you are encountering is a clash between the notion we are taught of the derivative as the ratio between increasingly small (eventually infinitesimally small) amounts, but the approach of calling $dx$ "a small step in $x$" is a bit flawed. That's where the limits you talk about enter the question. $\frac{dy}{dx}$ is not really a fraction, but a function in its own sense.

"Manipulating derivatives algebraically" doesn't have much sense in that regard, $\frac{dy+dx}{dt}$ is wrong in the sense that this is not a fraction to begin with. It's just notation we've stuck with for ages.

Now, a loooot of the time treating this as fractions will work, and it turns out to be the case in this situation. Your problem is usually framed multiplying by the denominator in this way $$\frac{dz}{dx} = \frac{dz}{dy}\frac{dy}{dx}$$ This is known as the chain rule, and it is a basic result in Differential Calculus. It only requires the derivative of z to exist at y(x) and the derivative of y to exist at x, which probably holds as you wouldn't be checking if it wasn't the case. There are other cases in which it seems as this notation really is a fraction. It is standard practice among Physics literature to make an argument of the integration of a function by "moving the differential in the denominator to the other side and integrating the numerator", but notation for derivatives is just notation. Don't lose track of that

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  • $\begingroup$ Wow, that actually clears up some of the confusion. I've just thought of them as 'small changes' so far. So, this being the case, would, say, 1/dx/dy not be the same as dy/dx? Also, why wouldn't (dy+dx)/dt make sense? Finally, what would happen to the dz/dx equation if dy/dx=0? Thanks again for your answer, by the way. $\endgroup$ Sep 29 '20 at 14:05
  • $\begingroup$ The thing is that it ALMOST always works to think of this as a fraction, and probably in school you will be taught that way, because even Leibniz (the dude that made the whole thing run) thought it worked. For example, if you, as you said, did 1/(dy/dx), you would get (if it exists) the derivative of the inverse function. But it's just notation and it just makes no sense to have (dx+dy)/dz because it just isn't written that way. With dy/dt=0 probably dx/dt=0 as well and you would have a nice 0/0 limit, as x is a function of y and y would increase really slowly there $\endgroup$
    – Evaristo
    Sep 29 '20 at 16:31
  • $\begingroup$ Alright, so when doesn't this analogy work? Also, why would a 0/0 limit be "nice"? I've always found those to be annoying because you have to do more work and it doesn't otherwise make sense... And if dy/dx=0/0 there, why would that mean that the function increases really slowly rather than showing a discontinuity? Finally, to clarify, let's just day that dx/dy=a and dx/dz is a*b. Would dz/dy thus inherently have to be something along the lines of 1/b? $\endgroup$ Sep 30 '20 at 13:53
  • $\begingroup$ This analogy works if you don't make weird stuff like moving the denominator to multiply on the other side (differentials aren't numbers!) and when all derivatives involved work on their own. A 0/0 limit would be nice in the sense that your problem isn't finished and you still have "fun" left, and for the slow rate, think that y is a function on x and as dx/dt=0, x is almost constant. If y isn't too chaotic (which isn't expected as it is differentiable) you expect it to not change much with respect to t, as its parameter doesn't change much. The last one is true if all derivatives exist $\endgroup$
    – Evaristo
    Oct 1 '20 at 8:36
  • $\begingroup$ That last example does make sense. So what manipulations that are similar to algebraic manipulation (multiplication, division, etc.) can you perform on derivatives and dx? $\endgroup$ Oct 1 '20 at 12:11
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The derivative, $\frac{dy}{dx}= \lim_{\Delta x\to 0}\frac{y(x+ \Delta x)- f(x)}{\Delta x}$ is NOT a fraction, but it is the limit of one. That's why $\frac{\frac{dy}{du}}{\frac{du}{dx}}= \frac{dy}{dx}$.

To prove that, go back before the limit. we can write that $\frac{y(u+ \Delta u)}{\Delta u}\frac{\Delta u}{\Delta x}= \frac{y(u+\Delta u)}{\Delta x}$ and as now take the limit as $\Delta x$ goes to 0, so that $\Delta u$ does also.

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  • $\begingroup$ Well, that makes sense. However, I think you accidentally mixed up du and dx in your division equation. At any rate, The delta u/delta x part of your last function seems to not fit, as it isn't a derivative function (i.e. a limit). Is it actually different or am I missing something? $\endgroup$ Sep 29 '20 at 14:08

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