Derivative of expanded function wrt expanded variable?

[dpopchev]

Homework Statement
If I have the following expansion
<br /> f(r,t) \approx g(r) + \varepsilon \delta g(r,t) + O(\varepsilon^2)<br />

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

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

The attempt at a solution

No idea how to approach this:
<br /> \dfrac{dU}{df} = \dfrac {d U}{d g}\dfrac{dg}{df} = \dfrac{dU/dg}{df/dg} = ?<br />

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

Additionally how to calculate dg/df term

EDIT: added details to expression

 

[MathematicalPhysicist]

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.