Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
+1 because it has the potential to generate answers to help explain proper use of differentials. –  Brian Vandenberg Jun 22 '11 at 17:45
    
See also math.stackexchange.com/q/8961/1242. –  Hans Lundmark Jun 22 '11 at 21:04
add comment

2 Answers

up vote 2 down vote accepted

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.

share|improve this answer
1  
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
show 4 more comments

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.