Let's say I have a set of $m$ functions depending on $n$ variables $$x^{i} = x^{i}(s^{1}, ... , s^{n})$$ I use latin indices to denote $(1, ..., m)$ and greek letters to denote $(1, ..., n)$.
I know $$\frac{\partial x^{i}}{\partial x^{j}} = \delta^{i}_{j}$$ $$ \implies \frac{\partial^{2} x^{i}}{\partial s^{\alpha} \partial x^{j}} = \frac{\partial (\delta^{i}_{j})}{\partial s^{\alpha}} = 0$$
What I'd like to know is whether commutativity of partial derivatives allows me to do this:
$$\frac{\partial^{2} x^{i}}{\partial x^{j} \partial s^{\alpha}} = \frac{\partial^2 x^{i}}{\partial s^{\alpha} \partial x^{j}} = 0$$
And if I'm not allowed to do so, why?
Thanks a lot, and I'm sorry if it is a silly question.
Edit:
For example, I get troubles when I'm in the simplest case, $m=1, n=1$. Let's say $x=s^2$ $$\frac{dx}{dx}=1$$ $$\implies \frac{d}{ds}\left( \frac{dx}{dx} \right) = 0$$
But when I do it in the opposite order:
$$\frac{dx}{ds}=2s$$ $$\implies \frac{d}{dx}\left( \frac{dx}{ds} \right) = \frac{d}{dx}\left( 2s \right)$$ substituting $s = \sqrt{x}$ $$\frac{d(2s)}{dx} = \frac{d(2\sqrt{x})}{dx} = \frac{1}{\sqrt{x}} = \frac{1}{s}$$
What am I doing wrong?
Thats not necessarily true. Take a counterexample $$ x = r \cos \theta, \quad y= r \sin \theta $$ With $x^1=x$, $x^2=y$ and $s^1=r$, $s^2=\theta$, you'll have $$ \frac{\partial}{\partial y}\frac{\partial x}{\partial r} = \frac{\partial}{\partial y}(\cos \theta) = \Bigg(\frac{\partial r}{\partial y} \frac{\partial}{\partial r}+ \frac{\partial \theta}{\partial y} \frac{\partial}{\partial \theta}\Bigg) (\cos \theta) = 0 + \frac{\partial \theta}{\partial y} (-\sin \theta) $$ whereas $$ \frac{\partial}{\partial r}\frac{\partial x}{\partial y} = 0. $$ By Schwarz's theorem, the commutativity of partial derivatives only apply to variables on the same coordinates.
In your case, when you compare these two expression $$\frac{\partial }{\partial x^{j}} \frac{\partial x^{i}}{\partial s^{\alpha}} \quad \text{and} \quad \frac{\partial}{\partial s^{\alpha}} \frac{\partial x^{i}}{\partial x^{j}}, $$ we know that the second expression will always zero, but the first is not. Because if you try to expand as below \begin{align} \frac{\partial }{\partial x^{j}} \frac{\partial x^{i}}{\partial s^{\alpha}} &= \frac{\partial s^{\beta}}{\partial x^j} \frac{\partial }{\partial s^{\beta}} \Bigg( \frac{\partial x^{i}}{\partial s^{\alpha}} \Bigg) = \frac{\partial s^{\beta}}{\partial x^j} \frac{\partial }{\partial s^{\alpha}} \Bigg( \frac{\partial x^{i}}{\partial s^{\beta}} \Bigg) = \frac{\partial}{\partial s^{\alpha}} \Bigg( \frac{\partial s^{\beta}}{\partial x^j} \frac{\partial x^{i}}{\partial s^{\beta}}\Bigg) - \frac{\partial}{\partial s^{\alpha}} \frac{\partial s^{\beta}}{\partial x^j} \\ &= \frac{\partial}{\partial s^{\alpha}} \frac{\partial x^i}{\partial x^j} - \frac{\partial}{\partial s^{\alpha}} \frac{\partial s^{\beta}}{\partial x^j} , \end{align} the last equality immidiately tells us that the expression is not always equal to $\frac{\partial}{\partial s^{\alpha}} \frac{\partial x^i}{\partial x^j}$.