1. PF Insights is off to a great start! Fresh and interesting articles on all things science and math. Here: PF Insights

I have a function

  1. I have a function "f", which is a function of "T" but "T" is a function of small "t".
    Now my question is what is the derivative of "f" with respect to "t"?
     
  2. jcsd
  3. micromass

    micromass 18,723
    Staff Emeritus
    Science Advisor
    Education Advisor

    Re: derivative

    What you're asking simply has no sense. Where did you encounter this?

    Basically, T could be a function [tex]T:\mathbb{R}\rightarrow \mathbb{R}[/tex] and [tex]f:\mathcal{C}(\mathbb{R},\mathbb{R})\rightarrow \mathbb{R}:T\rightarrow f(T)[/tex].

    But now there are two problems
    1) I have no clue how to define a derivative on [tex]\mathcal{C}(\mathbb{R},\mathbb{R})[/tex], I'm certain it can be done, but it's not immediately clear.
    2) f is not a function of t. The best thing you can do is to define a derivative of f w.r.t. T.

    However, you possible can do the following:
    define the function [tex]g:\mathbb{R}\times\mathcal{C}(\mathbb{R},\mathbb{R}):(t,T)\rightarrow T(t)[/tex]
    And you could possible use this to define a derivative w.r.t. t. But I'm quite sure this is not what you mean...


    Where did you encounter this, can you give me the reference??
     
  4. HallsofIvy

    HallsofIvy 40,410
    Staff Emeritus
    Science Advisor

    Re: derivative

    I think saravanan13 is talking about the "chain rule":
    if y= f(T) is a function to the variable T and T itself is a function of the variable t, then we can think of y as a function of t: y= f(T(t)).

    Further, if both functions are differentiable then so is the composite function and
    [tex]\frac{dy}{dt}= \frac{df}{dT}\frac{dT}{dt}[/tex]

    So that, for example, if [itex]y= T^3[/itex] and [itex]T= 3t^2+ 1[/itex] then we can calculate that [itex]y= (3t^2+ 1)^3= 27t^6+ 27t^4+ 9t^2+ 1[itex] so that
    [tex]\frac{dy}{dt}= 182t^5+ 108t^3+ 18t[/tex]

    Or we could calculate that
    [tex]\frac{dy}{dT}= 3T^2[/tex]
    and
    [tex]\frac{dT}{dt}= 6t[/tex]
    so that
    [tex]\frac{dy}{dt}= 3(3t^2+ 1)^2(6t)= 18t(9t^4+ 6t^2+1)= 162t^5+ 108t^2 18t[/tex]
    as before.
     
  5. Re: derivative


    I came across this problem in perturbation analysis formulated by Ablowitz and Kodama.
    In that T is slowly varying time and t is a fast variable.
    Thanks for your kin reply...
     
  6. Re: derivative

    Could you help me out how to type the mathematics formula in this forum.
    After i used some latex that give in the last icon of top left go for a preview it was not shown that i typed.
     
  7. pwsnafu

    pwsnafu 909
    Science Advisor

    Re: derivative

    Just use the Fréchet derivative.
     
  8. Re: derivative

    After you click 'preview', refresh the page - it should now show you what you typed. This is a known issue on these forums.
     
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook

Have something to add?