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Homework Help: Rigorousness of derivatives as ratio of differentials

  1. Jan 24, 2010 #1
    1. The problem statement, all variables and given/known data

    I have often heard that given say, ds/dt=k, that it is not entirely rigorous to say that ds=kdt. Why is that? If I view the derivative as nothing more than a ratio of the differentials, that seems perfectly reasonable. Also I see this done all the time, with acknowledgment that it is not perfectly mathematically correct. Can anyone shed some light on this please?
  2. jcsd
  3. Jan 24, 2010 #2
    It is not correct to say that ds/dt=k => ds = k dt if you just think about it as multiplication. This is in fact how the differential ds is defined. So although ds/dt isn't actually a fraction, we've defined differentials so that it acts somewhat like one. A lot of other times it seems like you're manipulating derivatives as if they were fractions, you can rigorously justify the steps using the chain rule.
  4. Jan 24, 2010 #3
    But why? If ds is an infinitesimal change in s, and dt likewise for t, then the ratio will be the rise over run if you define a function s(t) i.e. the slope of s(t), which is the derivative.

    I think I kind of understand now actually, maybe someone can put the following into more concrete language. By themselves ds and dt are not exactly defined relative to each other. An infinitesimal change in s could be larger than an infinitesimal change in t, even if ds/dt were less than 1. If you look at it like that then they are not well defined until you define one as being proportional to another. But what is the advantage to that view, besides perhaps some gained generality if you are doing some thing I don't care about.

    Except that if I say ds=k*dt and dt=ds/k then ds/dt=(k^2 dt)/ds !!! What in the world is that about?
  5. Jan 24, 2010 #4
    You're intuitive way of thinking is wonderful, but it is intuitive. We define differentials so that they agree with this multiplicative intuition, so it's not a bad way to think about them. Your thread just has the word "rigorousness" in the title, so I was pointing out the rigorous interpretation, which is not as simple as multiplication.
  6. Jan 24, 2010 #5
    I believe you but can you explain why?
  7. Jan 24, 2010 #6
    Was I correct in saying that ds by itself is not defined relative to dt until you explicitly state the proportionality?
  8. Jan 24, 2010 #7
    Because the derivative is not a fraction, it is a limit of fractions!

    Yes, in a way. If s = f(t) is a function of t, then we define ds = f'(t) dt, or, if you want it to look like multiplication (even though it isn't technically multiplication), [tex]ds = \frac{ds}{dt} dt[/tex]. So you're "explicitly stating the proportionality" when you write s as a function of t and then differentiate. I'm just trying to show how your (excellent) intuition fits into the precise framework of differentials.
  9. Jan 24, 2010 #8
    Ok I think I understand. But in what situations will treating the derivative as a fraction actually get me into serious trouble?

    Edit: This question came out of me doing my thermal physics homework. So for example, for manipulations of the thermodynamic identity.
    Last edited: Jan 24, 2010
  10. Jan 24, 2010 #9
    You could get into serious trouble when trying to find the second derivative of a parametric curve in the plane. Say x and y are functions of t. You can use the chain rule to rigorously show that [tex]\frac{dy}{dx} = \frac{ \frac{ dy }{ dt } }{ \frac{ dx }{ dt } }[/tex], but it is emphatically not true that [tex]\frac{d^2y}{dx^2} = \frac{ \frac{ d^2y }{ dt^2 } }{ \frac{ d^2x }{ dt^2 } }[/tex], which seems plausible if you're treating the derivatives as fractions. It is an instructive exercise in the chain rule to compute the second derivative correctly.
  11. Jan 24, 2010 #10


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    That's a big if. The only infinitessimal real number is zero, so we run into serious problems if we try to think this way.

    The way (standard) analysis works is that we use the derivative to give meaning to infinitessimals, not the other way around.

    Once you have defined the notion of differential form, you can talk about things like ds and dt. And if it so happens that there is a unique way to write ds as a multiple of dt:
    ds = f dt​
    then it would be fair to define ds/dt to mean the value f.

    There is no a priori reason why this should have anything to do with the derivative, but happily it turns out that this notation agrees with Leibniz notation.

    Incidentally, nonstandard analysis does use infinitessimals to define the derivative, but it's still wrong to view ds/dt as an infinintessimal variation in s over an infinitessimal variation in t. NSA still goes about defining a differential form-like thing. Once you define the derivative and have settled upon what you mean by dx where x is the chosen variable upon which other things are dependent, then df(x) is then defined to be f'(x) dx.

    *: I mean differential functions, not just functions. But I don't know the right adjetive here
  12. Jan 24, 2010 #11
    Man this is a blow to my ego to admit but I'm not entirely sure how to do this. And I really really need to iron this out.


    [tex]\frac{dy}{dx}=\frac{\partial y}{\partial t}\frac{dt}{dx}[/tex]

    [tex]\frac{d^{2}y}{dx^{2}}=\frac{\partial}{\partial x}(\frac{\partial y}{\partial t}\frac{d^{2}t}{dx^{2}})[/tex]

    [tex]\frac{d^{2}y}{d x^{2}}=\frac{\partial^{2}y}{\partial t^{2}}\frac{d^{2}t}{dx^{2}}+\frac{\partial y}{\partial t}\frac{\partial}{\partial x}(\frac{d^{2}t}{dx^{2}})[/tex]

    Is that right?
  13. Jan 24, 2010 #12
    No, it is not.
  14. Jan 24, 2010 #13
    Fixed for the sake of my ego. Edit: but still working on it I guess I have a problem w/ partial vs total derivatives.

    [tex]\frac{dy}{dx}=\frac{\partial y}{\partial t}\frac{dt}{dx}[/tex]

    [tex]\frac{d^{2}y}{dx^{2}}=\frac{d}{dx}(\frac{\partial y}{\partial t}\frac{dt}{dx})[/tex]

    [tex]\frac{d^{2}y}{d x^{2}}=\frac{\partial^{2}y}{\partial t^{2}}\frac{d^{2}t}{dx^{2}}+\frac{\partial y}{\partial t}(\frac{d^{2}t}{dx^{2}})[/tex]
  15. Jan 24, 2010 #14
    Still no. Also, no partial derivatives are necessary. You're misusing the product rule.
  16. Jan 24, 2010 #15
    Thank god this is anonymous my advisor would probably bar me from graduating if he saw this.
  17. Jan 24, 2010 #16


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    You just want to use d/dx=(dt/dx)*d/dt=(1/(dx/dt))*d/dt. I.e. the d/dx operation is (1/x'(t)) times the d/dt operation. Apply that twice to y(t). You'll need to use the quotient rule. Try not to go nuts about this.
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