I have a function


by saravanan13
Tags: function
saravanan13
saravanan13 is offline
#1
May2-11, 06:48 AM
P: 56
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"?
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micromass
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#2
May2-11, 07:11 AM
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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??
HallsofIvy
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#3
May2-11, 08:15 AM
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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.

saravanan13
saravanan13 is offline
#4
May2-11, 10:50 AM
P: 56

I have a function


Quote Quote by micromass View Post
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??

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...
saravanan13
saravanan13 is offline
#5
May2-11, 10:56 AM
P: 56
Quote Quote by micromass View Post
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??
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.
pwsnafu
pwsnafu is offline
#6
May2-11, 09:16 PM
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P: 778
Quote Quote by micromass View Post
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.
Just use the Fréchet derivative.
plasmoid
plasmoid is offline
#7
May3-11, 12:45 AM
P: 15
Quote Quote by saravanan13 View Post
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.
After you click 'preview', refresh the page - it should now show you what you typed. This is a known issue on these forums.


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