Ok, may be this is a silly question but consider the following.
Let $\dot x=f(x)$ be an autonomous differential equation with $f$ having enough smoothness (Say $C^2$). Let $\xi:\mathbb R\to[0,\infty)$ with compact support $[-r,r]$ for some $r>0$. Let $\phi_t(x)$ denote the flow of the given differential equation. I wan to find the following derivative.
$$ \frac{d}{dx}(\xi(\phi_t(x)) \tag{A}$$
Here is what I did. By chain rule we get \begin{align} \frac{d}{dx}(\xi(\phi_t(x)) &=\xi'(\phi_t(x)\frac{\partial}{\partial x}(\phi_t(x))) \quad \text{by chain rule}\\ &=\xi'(\phi_t(x)\frac{d}{dt}\left(\phi_t(x) \right)\cdot \frac{dt}{dx} \tag{*}\\ &=\xi'(\phi_t(x)(f(\phi_t(x)) \frac{1}{\dot x} \tag{**} \end{align}
Question? I am not convinced about the lines denoted as (* ) and (**) above. Can anybody please show me how this derivative can be found. May be I am missing some property of the flow of a differential equation. (I know that for a flow $\phi_t(x)$ of a differential equation that $\dot \phi_t =f(\phi_t)$)
Some context I eventually want to be able to compute the derivative of the following integral. $$ \eta (x) =\int_0^\infty e^{at} \xi(\phi_t(x)) dt$$ for a given $f$. Since differentiation under the integral sign can be easily justified this shouldn't be a problem if I can somehow find the derivative in $(A)$
Addendum What if $f(x)=-ax+\xi(x)?$. Does that make any difference as far as $\partial \phi_t/\partial x $ not being able to be simplified?. If $f(x)$ is the given function then I know some properties of $\phi_t$. For example the origin is asymptotically stable and so $\phi_t(x) \to 0$ as $t\to \infty$. I can also assume $\xi(0)=\xi'(0)=0$ as well. By continuous dependance theorem I can also safely say that $\phi_t$ is sufficiently smooth.
Remember that $\phi_t(x)$ is the solution at time $t$ to the system $\dot{y} = f(y)$ with initial condition $y(0) = x$.
The following skips a lot of details, but hopefully gives some idea of one approach to 'computing' the derivative you required.
To compute ${\partial \phi_t(x) \over \partial x}$, consider the system $L(z,x) = 0$, where $L(z,x)(t) = x+ \int_0^t f(z(\tau)) d \tau -z(t)$ and $L : C^n[0,t] \times \mathbb{R}^n \to C^n[0,t]$.
Some work shows that $({ \partial L(z,x) \over \partial z}h) (s) = \int_0^s { \partial f(z(\tau)) \over \partial y} h(\tau) d \tau-h(s)$ (with $ s \in [0,1]$), some more work shows that the map $D h = { \partial L(z,x) \over \partial z}h$ is continuous, and that the equation $D h = b$ is invertible (with $b \in C^n[0,t]$), and so the operator ${ \partial L(z,x) \over \partial z}$ is invertible.
It is straightforward to see that $({ \partial L(z,x) \over \partial x}\delta)(s) = \delta$ (with $ s \in [0,1]$), and applying the implicit function theorem (see Kantorovich & Akilov, "Functional analysis", for example) to $L(z,x) = 0$ gives the existence of some $\zeta$ such that $L(\zeta(a), a) = 0$ for $a$ in some neighbourhood of $x$. Furthermore, ${ \partial L(z,x) \over \partial z} { \partial \zeta(x) \over \partial x} = -{ \partial L(z,x) \over \partial x}$, and so, for any $\delta \in \mathbb{R}^n$, we have $({ \partial L(z,x) \over \partial z} ({ \partial \zeta(x) \over \partial x} \delta))(s) = -\delta$.
Unwinding this, and letting $\eta = { \partial \zeta(x) \over \partial x} \delta $, gives $\eta$ as the solution to the equation $\int_0^s { \partial f(z(\tau)) \over \partial y} \eta(\tau) d \tau-\eta(s) = -\delta$, or in other words, $\eta$ solves $\dot{\eta}(s) = { \partial f(z(s)) \over \partial y} \eta(s)$, with $\eta(0) = \delta$.
Suppose $\Phi(t_1,t_2)$ is the state transition matrix for $\dot{\eta}(s) = { \partial f(z(s)) \over \partial y} \eta(s)$, then we have ${ \partial \phi_t(x) \over \partial x} \delta = \eta(t) = \Phi(t,0) \delta$, the solution of the 'linearised' differential equation at time $t$.