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I am trying to understand how the chain rule was applied on page 42 for item 2. https://math.berkeley.edu/~evans/control.course.pdf

When you go from $i(\tau)$ to $i'(\tau)$ I noticed that the author was using $$\frac{\partial}{\partial \tau}F(x(t)+\tau y(t),\dot{x}(t)+\tau \dot{y}(t))=y(t)\frac{\partial F(\cdots)}{\partial x}+\dot{y}(t)\frac{\partial F(\cdots)}{\partial \dot{x}}$$ Why did the partials with respect to $\tau$, $y$ and $\dot{y}$ vanished? I feel embarrassed that I got lost in the basic calculus here. I would be welcoming references from different books.

I asked the author, but maybe someone would help me here as well,

Thank you for your time

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This is just the usual basic chain rule. You have a function $F(x,\dot x)$. It's perhaps confusing that $x$ becomes $x(t)+\tau y(t)$ and $\dot x$ becomes $\dot x(t)+\tau\dot y(t)$. Maybe it would be better to write the original function $F$ as a function of $u$ and $\dot u$. Then we set $u=x(t)+\tau y(t)$ and $\dot u = \dot x(t)+\tau\dot y(t)$. This means that we now have a function of $t$ and $\tau$, and \begin{align*} \frac{\partial}{\partial\tau} F(x(t)+\tau y(t),\dot x(t)+\tau\dot y(t)) &= \frac{\partial F}{\partial u}\frac{\partial}{\partial\tau}(x(t)+\tau y(t)) + \frac{\partial F}{\partial\dot u}\frac{\partial}{\partial\tau}(\dot x(t)+\tau\dot y(t)) \\ &= \frac{\partial F}{\partial u} y(t) + \frac{\partial F}{\partial\dot u}\dot y(t). \end{align*} The $F(\cdots)$ is very misleading. You evaluate $\dfrac{\partial F}{\partial u}$ at $(x(t)+\tau y(t),\dot x(t)+\tau\dot y(t))$. This is not the partial derivative of a composition. This is the usual issue with sloppy notation with the chain rule.

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  • $\begingroup$ Thank you for your detailed answer, @Ted Shifrin! So $\frac{\partial F}{\partial u}\neq \frac{\partial F}{\partial x}$ but the author arrived at $\frac{\partial F}{\partial x}$ once he made $\tau =0$? I just want to make sure this is the follow-up, and I am not mixing it up. $\endgroup$ Jun 26 at 3:26
  • $\begingroup$ No, his notation is bad. As I said, he’s writing $F(x,\dot x)$ to start with. $\endgroup$ Jun 26 at 3:47
  • $\begingroup$ Thank you, @Ted Shifrin! $\endgroup$ Jun 26 at 5:37

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