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I know it's possible to do this:

$$\frac{dy}{dx} \frac{dt}{dt} = \frac{dy}{dt} \frac{dt}{dx}$$

but I wonder if this makes sense?

$$\frac{d}{dx}\left(\frac{dt}{dt}\right) = \frac{d}{dt} \left(\frac{dt}{dx}\right)$$

so if $t=x^4$ then $\displaystyle\frac{dt}{dx} = 4x^3$ and

$$\frac{d}{dx}\left(\frac{dt}{dt}\right) = \frac{d}{dt} \left(4x^3\right)$$

but this is like saying

$$\frac{d}{dx} \left(1\right) = \frac{d}{dt} \left(4x^3\right)$$

and $0=0$.

so it makes sense in that instance at least... I suppose at time this notation is still mysterious.

t.b.
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Noteventhetutorknows
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2 Answers2

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The notation itself is just that, a notation. While it may seem 'clean and safe' to move differentials around as though they're just another variable, technically speaking it isn't a valid way to do mathematics.

Now, provided you understand and follow the rules, some manipulations along those lines can still arrive at a correct result albeit through potentially dubious means.

A good example of this shows up commonly in differential equations texts:

$$f(x,y)dx + g(x,y)dy = 0$$

... which, when written in proper form is:

$$\frac{dy}{dx} = -\frac{f(x,y)}{g(x,y)}$$

However, when used as a mnemonic device the former is a good way to help a student remember how to find the adjoint of the ODE and ultimately arrive at a general solution.

Under the hood, the former can be re-written as:

$$f(x,y)\frac{dx}{dt} + g(x,y)\frac{dy}{dt} = 0$$

... due to the chain rule:

$$\frac{dy}{dt}\frac{dt}{dx} = -\frac{f(x,y)}{g(x,y)}$$

... where here it must be assumed that $x(t)$ has an inverse $x^{-1}(t) = t(x)$ such that $\displaystyle\frac{1}{\frac{dt}{dx}} = \frac{dx}{dt}$ in order to arrive at the homogeneous equation. I'm probably missing some other pertinent details, but this is closer to a more correct way to work with differentials than the ad-hoc methods usually taught in differential equations texts.

t.b.
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Brian Vandenberg
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    I was tutoring a polytechnic institute student for his differential equations exam, and my head hurt when I saw how they move $dx,dy$ from one side to another, just like usual numbers and integrate two sides with different variables. It maybe faster, but as a mathematician I can't stand these methods. :) – Beni Bogosel Jun 22 '11 at 17:21
  • What exactly is dubious in what I did and how can it fail? – Noteventhetutorknows Jun 22 '11 at 17:36
  • Agreed. When I took my ODE class I just went with it, but as I went back and worked through some of that material years later it occurred to me that the homogeneous equation above is actually a partial differential equation in disguise, of sorts -- hence my addendum with $\frac{dy}{dt}$ (etc). Adding $t$ explicitly seemed to make it easier to 'guess' the solution for those problems. – Brian Vandenberg Jun 22 '11 at 17:36
  • @Brian: I tried to improve the readability of your text by using $$...$$ instead of $...$ to make the formulas larger. I hope you don't mind. – t.b. Jun 22 '11 at 17:38
  • @Noteventhetutorknows - re: dubiousness, I was specifically referring to the homogeneous equation I wrote, but what you wrote is fraught with what I would consider to be inappropriate manipulations. The chain rule gives: $\frac{dy}{dx} = \frac{dy}{dt}\frac{dt}{dx}$. Nothing about the chain rule says it's safe or valid to swap denominators as though $\frac{dy}{dt}$ is a simple fraction. – Brian Vandenberg Jun 22 '11 at 17:39
  • @Theo - That's fine. What affect does $$ have? – Brian Vandenberg Jun 22 '11 at 17:41
  • It gives you a displayed formula which looks like $$\frac{df}{dx}$$ instead of the inline $\frac{df}{dx}$ (so in the center, and larger font). – t.b. Jun 22 '11 at 17:43
  • My understanding is that it is not like a fraction but rather a change in rates. I'll have a look at the chain rule again. – Noteventhetutorknows Jun 22 '11 at 17:50
  • @Noteventhetutorknows - The fractional notation helps give the notion of rates, but it also gives the impression you can just do normal algebraic manipulations on them -- which is definitely not true. It can be done, as Americo Tavares points out, but it must be done very carefully and rigorously. – Brian Vandenberg Jun 22 '11 at 21:35
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The differential of a differentiable function $f(x)$ at $x_0$ is the expression $f'(x_0)dx$. If $f(x)=x$, then $f'(x_0)dx=1\cdot dx=dx$. The algebraic manipulation of differentials may be intuitive in certain cases but it has to be checked, i. e. proved or disproved rigorously.

Américo Tavares
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