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Let $phi(t, x_0)$ be a solution of the one-dimensional differential equation
$$dot{x}= f(x),$$
with $phi(0, x_0) = x_0$. Show that its derivative $frac{partial}{partial x_0}phi(t, x_0)$ is given by
$$frac{partial}{partial x_0}phi(t, x_0) = exp Big(int_{0}^{t}f'(phi(s, x_0))dsBig)$$

Here in my proposed solution.

By the fundamental theorem of calculus, we have

$$phi(t, x_0)=x_0 + int_{0}^{t}f(phi(s, x_0))ds$$

because $frac{partial{phi}}{partial t}(t, x_0)=f(phi(t, x_0))$ and $phi(0, x_0)=x_0$.

If we differentiate this solution with respect to $x_0$, we obtain via the chain rule that

$$frac{partial{phi}}{partial x_0}(t, x_0)=1 + int_{0}^{t}frac{partial{f}}{partial{x_0}}(phi(s, x_0))
cdot frac{partial{phi}}{partial x_0}(s, x_0)ds$$

Let $$z(t)= frac{partial{phi}}{partial x_0}(t, x_0)$$

Then,

$$z(0)= frac{partial{phi}}{partial x_0}(0, x_0)=1$$

by the analysis above. Therefore, if we differentiate $z$ with respect to $t$, we find that

begin{equation}
begin{split}
z'(t) & = frac{partial{f}}{partial{x_0}}(phi(t, x_0)) cdot frac{partial{phi}}{partial x_0}(t, x_0)\
& = frac{partial{f}}{partial{x_0}}(phi(t, x_0)) cdot z(t)
end{split}
end{equation}

We do not have an explicit solution to $phi(t,x_0)$, but the above equation tells us that $z(t)$ solves the following differential equation,

$$z'(t)= frac{partial{f}}{partial{x_0}}(phi(t, x_0)) cdot z(t)$$

Therefore,

$$frac{dz}{dt}= frac{partial{f}}{partial{x_0}}(phi(t, x_0)) cdot z(t)$$

Rearranging terms produces

$$frac{1}{z(t)}{dz}= frac{partial{f}}{partial{x_0}}(phi(t, x_0))dt$$

Hence,

$$ln(z(t))= int_{0}^{t}frac{partial{f}}{partial{x_0}}(phi(s, x_0))ds$$

And taking the exponential of both sides produces

$$z(t)= exp Big(int_{0}^{t}frac{partial{f}}{partial{x_0}}(phi(s, x_0))dsBig)$$

As $z(t)=frac{partial{phi}}{partial x_0}(t, x_0)=exp Big(int_{0}^{t}f'(phi(s, x_0))dsBig)$, we are done.

I'm not certain if there is a more direct approach. Please let me know if the solution can be improved.

From the comments above, here is a shorter answer:

As $dot{x}=f(x) implies frac{dx}{dt}=f(x)$, we can differentiate both sides with respect to $x_0$ to form

$$frac{d}{dt}frac{partial{x}}{partial{x_0}}=frac{df(x)}{dx}frac{partial{x}}{partial{x_0}}=f'(x)frac{partial{x}}{partial{x_0}}$$

Therefore, as $phi(t,x_0)$ is a solution to the ODE, we know that $x(t)=phi(t,x_0)$. Hence,

$$frac{d}{dt}frac{partialphi}{partial{x_0}}(t,x_0)=f'(phi(t,x_0))frac{partialphi}{partial{x_0}}(t,x_0)$$

Now, let $z(t)=frac{partialphi}{partial{x_0}}(t,x_0)$. Then,

$$frac{d}{dt}z(t)=f'(phi(t,x_0))z(t)$$

So,

$$frac{1}{z(t)}d{z(t)}=f'(phi(t,x_0))dt$$

Hence, by the fundamental theorem of calculus,

$$ln(z(t))=int_0^t{f'(phi(s,x_0))ds}$$

Therefore, if we take the exponential of both sides,

$$z(t)=expBig(int_0^t{f'(phi(s,x_0))ds}Big)$$

So, $frac{partialphi}{partial{x_0}}(t,x_0)=expBig(int_0^t{f'(phi(s,x_0))ds}Big)$.