Commutativity of differentiation in a special case

[HotMintea]

Problem

I'd like to prove $\frac{d}{dt}[\frac{\partial}{\partial{x}}f(x(t),y(t),t)]=\frac{\partial}{\partial{x}}[\frac{d}{dt}f(x(t),y(t),t)]$.

Attempt
$$\begin{equation*}\begin{split}<br /> \frac{d}{dt}[\frac{\partial}{\partial x}f(x(t),y(t),t)]=\frac{d}{dt}\lim_{\epsilon\to 0}\frac{f(x(t)+\epsilon,y(t),t)-f(x(t),y(t),t)}{\epsilon}\\<br /> =\lim_{\delta\to 0}\lim_{\epsilon\to 0} \frac{[f(x(t+\delta)+\epsilon,y(t+\delta),t+\delta)-f(x(t)+\epsilon,y(t),t)]-[f(x(t+\delta),y(t+\delta),t+\delta)-f(x(t),y(t),t)]}{\epsilon\delta}\end{split}\end{equation*}$$

So if my previous steps are correct, I need to show that the 2 limits are commutative, which I have no idea...

 

[hunt_mat]

Did you know you can express the total derivative as a sum of partial derivatives?

 

[HallsofIvy]

Are you required to go all the way back to the definition of derivative? If not, use the chain rule as hunt mat suggests.

 

[HotMintea]

Thanks for the hints!

$\frac{d}{dt}\partial_{x}f(x(t), y(t), t)=(\sum\dot{x}_k \partial_{x_k}+\partial_t)\partial_{x}f=\sum\dot x_k\partial_{x_k}\partial_{x}f+\partial_t\partial_{x}f$.

At this point, I need to show the "equality of mixed partials". I found the proof of $f_{xy}=f_{yx}$ for $f(x, y)$ here: http://www.sju.edu/~pklingsb/clairaut.pdf

My function is f(x(t), y(t), t) so a bit different.

For proving $f_{x(t)\hspace{1mm}t} = f_{t\hspace{1 mm}x(t)}$ for f(x(t), t), I think the proof in the note works if I substitute c=x(a) and d=x(b).

I also think the theorem can be extended to $f(x_1(t), x_2(t), ..., x_n(t), t)$ if I imagine slicing the n dimensional space by a plane parallel to $x_i x_j$ plane (or $x_i\hspace{1 mm}t$ plane).

So I have $f_{x_i x_j}=f_{x_j x_i}$ and $f_{x_i\hspace{1 mm}t}=f_{t\hspace{1 mm}x_i}$ for $f(x_1(t), ..., x_n(t), t)$. Is that okay?

By the way, I don't understand why the author of the note wrote $f_{xy}(x,y)-f_{yx}(x,y)\ge \frac{h}{2}$. Is that arbitrary? Can I use h/3 instead for example?