Okay I tried to edit stuff:

I have a function $$f:\mathbb{R}\to \mathbb{R}$$I also have the time derivative of V as a function $$\frac{\mathrm{d}V}{\mathrm{d}t}(f(V),V,a,b,c,....):\mathbb{R}\times\mathbb{R}\times A\times B\times C\times\ldots\to\mathbb{R}$$ so V should be a function implicitly depending on $$(f(V),V,a,b,c,....,t)\subseteq\mathbb{R}\times\mathbb{R}\times A\times B\times C\times\ldots\times T\to\mathbb{R}$$ Or maybe $$(f,V(t),a,b,c,....,t)\subseteq\mathcal{F}\times\mathbb{R}\times T\times A\times B\times C\times\ldots\times T\to\mathbb{R}$$

What is df/dV then? The chain rule is confusing me. Is it still the good old df/dV or do I need to take some kind of total derivative of the infinite composition $$\frac{\mathrm{d}f(V(f(V(f(V(f(V(...............)))))...,V,a,b,c,...t)...)}{\mathrm{d}V}$$

  • $\begingroup$ I cannot visualize $f(V)$ as a function, unless $V$ is some parameter, and $f(V): R\rightarrow R$ means $f(V,\cdot): R\rightarrow R$. Could you please explain your notation a bit more? Is $R=\mathbb R?$ What is $V$? $\endgroup$ – Avitus Aug 23 '13 at 13:48
  • $\begingroup$ This is what I really have: A function $$\alpha(\mathbf{V}):\mathbb{R}^n\to \mathbb{R}^n$$ A system of ODE's $$ \begin{cases} \frac{\mathrm{d}}{\mathrm{d}t}\mathbf{x}=f_1(\alpha(\mathbf{V}),\mathbf{x})\\ \frac{\mathrm{d}}{\mathrm{d}t}\mathbf{V}=f_2(\mathbf{x},\mathbf{V},a,b,c,\ldots) \end{cases} $$ and the chain rule seems to tell I need to take derivatives like the one above to compute the jacobian? I don't know if I translated the problem in the correct way above... $\endgroup$ – Emil Aug 23 '13 at 14:32
  • $\begingroup$ Is $V$ a function of $t$? (and only of $t$)? In other words, you have: $x=x(t)$ in $\mathbb R^n$, $V=V(t)$ (another vector in $\mathbb R^n$?) and $\alpha(V):=\alpha\circ V$? Am I correctly translating your problem? $\endgroup$ – Avitus Aug 23 '13 at 16:30
  • $\begingroup$ I don't know. Maybe it should be a partial derivative. I mean, I only have the ODE system. I don't know the analytical solution. But the solution changes depending on the other parameters... So.. Umm... $\endgroup$ – Emil Aug 25 '13 at 12:04

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