If $foo=f(x,\dot{x})$, what are $\frac{\partial f}{\partial \dot{x}}$ and $\frac{\partial f}{\partial x}$?

Question

Question

Let $x$ be a function of $t$ and $f=f(x,\dot{x})$. What are $\frac{\partial f}{\partial \dot{x}}$ and $\frac{\partial f}{\partial x}$ ?

Example

For example if $f(x(t),\dot{x}(t))=ax(t)+b\dot{x}(t)$: Does $\frac{\partial f}{\partial \dot{x}}=b$ and $\frac{\partial f}{\partial x}=a+b\frac{\partial}{\partial x}\frac{d}{dt}x(t)=a, \forall x(t)?$

Answer 1

In and of itself, the notation $\frac{\partial f}{\partial x}$ doesn't make sense. You take a partial derivative with respect to an index, that is, you just specify which "slot" in the function is being considered a variable. It makes no sense to take a partial derivative of a function with respect to another function. A better notation for partial derivatives is $D_1$ or $D_2$ and so on.

What's implicitly going on here is the following.

1. $f$ is a function of two variables.
2. $x$ is a function of one variable.
3. By $\frac{\partial f}{\partial x}$, we really mean the function $t\to (D_1f)(x(t), x'(t))$. By $\frac{\partial f}{\partial x'}$ we mean $t\to (D_2f)(x(t), x'(t))$

This corresponds to the idea of "pretending $x$ and $x'$ are variables". So for example, if $f(x, x')=ax+bx'$, what we really have is a two-variable function $f(u, v)=au+bv$, which we are composing with the two single variable functions $x(t)$ and $x'(t)$. To get $\frac{\partial f}{\partial x}$ we take $D_1f$, which is a two variable function $D_1f(u, v)=a$. We then compose that with $x'$ and $x$ to get $D_1f(x, x')$, which happens to still equal $a$, of course.

Answer 2

I would say that

$f(x(t),\dot{x}(t))=ax(t)+b\dot{x}(t)\Rightarrow \frac{\partial f}{\partial x}=a,\;\frac{\partial f}{\partial \dot{x}}=b$

and

$\frac{df}{dt}=\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial \dot{x}}\frac{d\dot{x}}{dt}=a\dot{x}+b\ddot{x}$