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"?
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??
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.
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...
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.
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.