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Consider the IVP $\dot{z} = f(t,z)$, $z(t_0) = x$. Let $\phi(t, t_0, x)$ be a solution to the IVP, where the dependence on initial conditions is emphasized. Assume $f \in C^1(U,\mathbb{R}^n)$, with $U$ an open subset of $\mathbb{R}^{n+1}$.

Gerald Teschl's book ODE's and dynamical systems claims that if $\phi$ is differentiable w.r.t. x, then so is $\dot{\phi}$. However, he gives no proof of this apparent fact and embarrassingly I cannot see why this necessarily must be the case.

It would be very helpful if someone could provide a proof of this statement!

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Since $\phi(t,t_0,x)$ is a solution, we have that $$\frac{\partial}{\partial t} \phi(t,t_0,x) = f(t,\phi(t,t_0,x)).$$ Since $f$ is differentiable with respect to its second argument, and we have assumed that $\phi$ is differentiable with respect to $x$, then their composition must also be differentiable with respect to $x$. Explicitly, using the chain rule: $$\frac{\partial}{\partial x_i} \dot{\phi} = \frac{\partial}{\partial x_i}f(t,\phi(t,t_0,x)) = \sum_j \frac{\partial f}{\partial z_j}\frac{\partial \phi_j}{\partial x_i}. $$

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    $\begingroup$ I am still a little hung up. We know f is differentiable w.r.t. z, its second argument, but x is our initial condition and f is not a function of x, so how can we say anything about differentiating f with respect to x without knowing a priori whether $\phi$ is? $\endgroup$ Sep 15, 2017 at 14:13
  • $\begingroup$ Thanks for pointing that out, instead of $x$ it should be "its second argument". I have edited my answer. $\endgroup$ Sep 15, 2017 at 14:16

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