Partial derivative of a total derivative Assume that we have some function $f(x(t),t)$ that is two times differentiable.
The partial derivative $\partial_t$ and $\partial_x$ are defined in the same way as the total derivative, but with $x$, $t$ constant, respectively.
Now assume that $f(x(t),t) = \frac {dx} {dt}$. How do I take $\partial_t \frac {dx} {dt}$? I could say, since $x$ is taken to be constant, the time derivative is zero, thus $\partial_t \frac {dx} {dt} = 0$. Also $\partial_x \frac {dx} {dt} = \frac {d1} {dt} = 0$ where I exhanged the derivatives.
Both statements can't be true, so where am I wrong?
 A: I think that you are heavily mislead by notations. And there is no reason to be ashamed about it: usual notations for derivatives are misleading.
What does an expression like $f(x(t),t)$ contains ? Two important things:


*

*a function $(x,t) \in \mathbb{R}^2 \mapsto f(x,t)$;

*a function $t \in \mathbb{R} \mapsto x(t)$.


Here is the first trap: the former $x$ is a real number whereas the latter refers to a function. Strictly speaking this double use is an inconsistency but it is unfortunately very common with partial derivatives. To avoid this confusion, let us rename our function $t \mapsto \hat{x}(t)$. Provided it is smooth enough, function $\hat{x}$ has a derivative $\hat{x}'$. It is also common to use Leibnitz notation $\frac{d\hat{x}}{dt}$. Similarly function $f$ has two partial derivatives and almost inevitably the Leibnitz notations $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial t}$ are used (or Leibnitz-like notations $\partial_x f$ and $\partial_t f$). $\partial_t f$ means partial derivative of $f$ as a function of variable $t$ with variable $x$ kept constant. Note that we can rewrite your equation like this:
$$\forall t \in \mathbb{R}, f(\hat{x}(t), t) = \hat{x}'(t)$$
Now if we reconsider your reconsider your two expressions: 


*

*$\partial_t \frac{d\hat{x}}{dt}$ can be interpreted (even if the notation is not very conventional) as $\frac{d^2 \hat{x}}{dt^2}$. It's a second derivative, its actual value is unknown and it is perfectly unrelated to the fact that variable $x$ is held constant when $f$ is derivated with respect to variable $t$.

*$\partial_x \frac{d\hat{x}}{dt}$ is in fact almost meaningless because function $\hat{x}$ does not depend on any variable $x$ (it might be possible to give it a meaning if we knew $\hat{x}$ had some specific properties but it is far from current discussion). 


Similarly expression $f(\hat{x}(t), t)$ does not depend on any variable $x$. $t \mapsto f(\hat{x}(t), t)$ is a function with a unique real parameter so $\frac{d}{dt} f(\hat{x}(t), t)$ is a quantity that makes sense and happens to be:
$$\frac{d}{dt} f(\hat{x}(t), t) = \frac{d\hat{x}}{dt}(t) \cdot \frac{\partial f}{\partial x}(\hat{x}(t), t) + \frac{\partial f}{\partial t}(\hat{x}(t), t)$$
Note that some people may abbreviate $\frac{df}{dt}$ for $\frac{d}{dt} f(\hat{x}(t), t)$ which is according to me dangerously misleading because it hides that this quantity depends on function $\hat{x}$.
