Are these mixed partial derivatives equal? 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?
 A: 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 $\frac{\partial}{\partial s^{\alpha}} \frac{\partial x^i}{\partial x^j}$.
A: As long as $x_i$ varies with $s_i$ you cannot switch the partial derivative variables. That is why the result does not make sense to you. According to Schwarz's theorem the commutative property of partial derivative is valid if $x_i$ are independent variables (e.g. $x,y,z$ in Cartesian coordinate systems).
