This may be a simple partial derivative formula but I'm having a hard time to understand it. $$ \left( \frac{\partial }{\partial t} \right)_x f(t,\vec{r}(t,\vec{x})) = \left( \frac{\partial }{\partial t} \right)_r f(t,\vec{r}) + \dot{r} \cdot \left( \frac{\partial}{\partial r} \right)_t f(t,\vec{r}) $$
Since the function "f" explicitly depends on "t", isn't just the partial derivative by "t" enough?
I already know the basic concept of the chain rule, but this formula is tricky for me to understand. What kinds of steps are going on?
The subscripting is used to distinguish the two $\tfrac{\partial f}{\partial t}$ notations by clearly indicating which 'independent variable' are being kept 'constant for purpose of partial differentiation'.
$\left(\dfrac{\partial f(t, r(x,t))}{\partial t}\right)_x$ is the partial derivative of the composite function $f(t,r(x,t))$ with respect to $t$ when $x$ is the independent variable. If we define $g(t,x)=f(t,r(x,t))$ then it is the differentiation of $g$ with respect to its first argument.
$\left(\dfrac{\partial f(t, r)}{\partial t}\right)_r$ is the partial derivative of the function $f(t,r)$ with respect to $t$ and 'independent variable' $r$ . So it is just the differentiation of $f$ with respect to its first argument.
So we often first define $g(t,x)= f(t,r(x,t))$ and present the chain rule as $$\dfrac{\partial g(t,x)}{\partial t}=\left[\dfrac{\partial f(t,y)}{\partial t}+\dfrac{\partial r(t,x)}{\partial t}\cdot\dfrac{\partial f(t,y)}{\partial y}\right]_{y=r(x,t)}$$
Which you will see most commonly shorthanded to: $$\dfrac{\partial g}{\partial t}=\dfrac{\partial f}{\partial t}+\dfrac{\partial r}{\partial t}\cdot\dfrac{\partial f}{\partial r}$$
Alternatively:
$$g^{(1,0)}(t,x) = f^{(1,0)}(t,r(x,t)) + r^{(0,1)}(x,t)\cdot f^{(0,1)}(t,r(x,t))$$