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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$\begingroup$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
$\frac {dx}{dy}=\lim_{h\to 0}\frac {x(y(t+h)) - x(y(t))}{y(t+h) - y(h)}\cdot \lim_{h\to 0} \frac {y(t+h) - y(t)}h= \lim_{h\to 0}\frac {x(y(t+h)) - x(y(t))}{h} = \frac {dx}{dy}$.
$\endgroup$ 10 $\begingroup$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
$\endgroup$ 5 $\begingroup$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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