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**Chain rule or Substitution rule???**

## Homework Statement

It appears that a standard result in ODE is the following: if [itex] f(x,t) [/itex] is smooth enough, then the solution [itex]\psi(t)[/itex] to the initial value problem:

[itex] x(t)=x_{0}+\int_{0}^{t}f(x(s),s)ds[/itex]

[itex]x(0)=x_{0}[/itex]

is continuously differentiable with respect to its initial condition (or a parameter in the problem), and that this derivative satisfies the following first variational equation:

[itex] \frac{\partial \psi(t)}{\partial x}=c +\int_{0}^{t}\frac{\partial f(\psi(s,x),s)}{\partial x}\frac{\partial \psi}{\partial x}ds[/itex]

## Homework Equations

To obtain the second equation above one makes use of the chain rule applied to [itex]f(\psi(t,x),t)[/itex]. However, I am not certain we can always use this rule. Consider the following `f' function:

[itex] f(\psi(s),s)=\biggl(\frac{1-\int_{0}^{s}G(\psi(\xi))d\xi}{F(t)F(\psi(t))+\int_{0}^{t}G(\psi(\xi))d\xi}\biggl)^N [/itex]

where [itex]G[/itex] is a sufficiently continuously differentiable function, and [itex]N>2[/itex]. The question is how could I apply the chain rule to obtain a first variational equation as the one I presented above?

## The Attempt at a Solution

No sure yet. I truly need help here as I've trying to understand this for a long time without any success...Thanks!