# Partial Derivatives application

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,

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.
• 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. Jun 26 at 3:26
• No, his notation is bad. As I said, he’s writing $F(x,\dot x)$ to start with. Jun 26 at 3:47