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Derivative of expanded function wrt expanded variable?

  • #1
27
0
Homework Statement
If I have the following expansion
[tex]
f(r,t) \approx g(r) + \varepsilon \delta g(r,t) + O(\varepsilon^2)
[/tex]

This means for other function U(f(r,t))
[tex]
U(f(r,t)) = U( g(r) + \varepsilon \delta g(r,t)) \approx U(g) + \varepsilon \delta g \dfrac{dU}{dg} + O(\varepsilon^2)
[/tex]

Then up to linear order in ε how to calculate
[tex]
\dfrac{dU}{df} = \ldots O(\varepsilon^2)?
[/tex]

The attempt at a solution

No idea how to approach this:
[tex]
\dfrac{dU}{df} = \dfrac {d U}{d g}\dfrac{dg}{df} = \dfrac{dU/dg}{df/dg} = ?
[/tex]

But then again not sure how to calculate this, if I try
[tex]
\dfrac {d U}{d g} = (U(f(r,t)) - U(g))\dfrac{1}{\varepsilon \delta g}
[/tex]
This will lead me again to dU/dg

Additionally how to calculate dg/df term

EDIT: added details to expression
 
Last edited:

Answers and Replies

  • #2
MathematicalPhysicist
Gold Member
4,285
200
How did you get the expression of ##f(r,t)\approx g(r)+\epsilon \delta g +\mathcal{O}(\epsilon^2)##?, if it's a Taylor expansion around ##t=0##, then ##f(r,0)=g(r)##, and then ##df/dg = \epsilon##, from the definition of Taylor expansion.
 

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