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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?

physicsGuy
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  • $x(t)$ is a function of $t$, you can't set it as a constant. This might help you understand : http://tutorial.math.lamar.edu/Classes/CalcIII/ChainRule.aspx – user88595 Dec 09 '13 at 10:40
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    The chain rule yields $\frac{df(x(t),y(t))}{dt} = \frac {\partial f}{\partial x} \frac {dx}{dt} + \frac {\partial f}{\partial y} \frac {dy}{dt}$. Now assume $y(t) = t$. Then we get $\frac{df(x(t),t)}{dt} = \frac {\partial f}{\partial x} \frac {dx}{dt} + \frac {\partial f}{\partial t}$. – physicsGuy Dec 10 '13 at 07:54
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    Next assume that $f(x(t),t) = \frac{dx}{dt} = \dot{x}$. Now we get $\frac{df(x(t),y(t))}{dt} = \ddot{x} = \frac {\partial \dot{x}}{\partial x} \dot{x} + \frac {\partial \dot{x}}{\partial t}$. Following this result I would say that my first argument is wrong and we have $\frac{\partial \dot{x}}{\partial x} = 0$ and $\frac{\partial \dot{x}}{\partial t} = \ddot{x}$. Stil this step is not very nice and I would not call it a proof, but that is what I am looking for. – physicsGuy Dec 10 '13 at 08:02
  • Do you have any idea for a proof? – physicsGuy Dec 10 '13 at 08:04

1 Answers1

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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 $\hat{x}$.

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