If $x=\pi a y^{1/2}$ then why is $\frac{\partial^n}{\partial x^n}=-2\left(\frac{y^{3/2}}{\pi a}\right)^n \frac{\partial^n}{\partial y^n}$? While I was reading this question, I was surprised that the transformation of a 'simple' differential operator $\displaystyle \frac{\partial^n}{\partial x^n}$ by substituting $x=\pi a y^{\frac{1}{2}} $ so that $$\displaystyle\frac{\partial^n}{\partial x^n}=-2\left(\frac{y^{\frac{3}{2}}}{\pi a}\right)^n \frac{\partial^n}{\partial y^n}$$ stumped me; I'm not sure how he derived it.
I actually have a simpler question: given $\displaystyle\frac{d^n}{dx^n}$, how will it transform if $x=cu$ where $c$ is a constant?
Edit: Here's a very relevant example from wikipedia, I think that the example in the article is essentially the same kind of thing that the OP in the reference link above did. The problem is that I don't really know how to manipulate operators, if I'm allowed to use a "physicist's" argument I'd say that for some differentiable function $f$ and if $u=x/c$:
$$
\frac{df}{dx}=\frac{du}{dx}\frac{df}{du}=\frac{1}{c}\frac{df}{du}
$$ 
$$
\therefore \frac{d}{dx}=\frac{1}{c}\frac{d}{du}.
$$
Is this even remotely correct? This only addresses the case $n=1$ though, doing the same for latter values seems harder.
Edit #2:
The final answer from the original link seems incorrect, the OP's result is:
$$
\frac{\partial^n}{\partial y^n}e^{-\frac{\pi^2a^2}{y}}=\frac{1}{2}e^{-\frac{\pi^2a^2}{y}}(-1)^{n+1}H_n(\pi a y^{-\frac{1}{2}})\left(\frac{y^{\frac{3}{2}}}{\pi a}\right)^n
$$
which is obviously wrong for the smaller values $n=0,1$ (since $H_1(x)=2x)$.
 A: As you noted, that expression is incorrect. However, we can still figure out things like this with the chain rule, in a manner similar to your calculation for $u=x/c$. Although I will be mixing notations, I think it might make things clearer than to try to use Leibniz notation alone. Also, while the expressions in your problem use $\partial$, I will use $\mathrm d$ since all variables other than $x$ and $y$ (which depends only on $x$) are effectively being held constant. (Things are more complicated in multivariable calculus.) Let $f(x)$ be an arbitrary function. Two operators are the same if they do the same thing to $f(x)$. Finally, since $\pi a$ is just a constant, I'll replace it with $c$ for convenience.

First, note that $\dfrac{\mathrm d}{\mathrm dx}f(x)=f'(x)$. Then $\dfrac{\mathrm d}{\mathrm dy}f(x)=\dfrac{\mathrm d}{\mathrm dy}f(c y^{1/2})=\left(f'(c y^{1/2})\right)*\left(\dfrac{c}{2y^{1/2}}\right)$ by the chain rule. But this can be rewritten as $\left(f'(x)\right)*\left(\dfrac{c}{2y^{1/2}}\right)=\left(\dfrac{\mathrm d}{\mathrm dx}f(x)\right)\left(\dfrac{c}{2y^{1/2}}\right)$. Dividing, we have $$\dfrac{\mathrm d}{\mathrm dx}f(x)=\frac{2y^{1/2}}{c}\dfrac{\mathrm d}{\mathrm dy}f(x)\text{, so }\dfrac{\mathrm d}{\mathrm dx}=\frac{2y^{1/2}}{c}\dfrac{\mathrm d}{\mathrm dy}$$

For $n=2$, things get a bit more complicated. First, the product rule and chain rule give us:
$$\frac{\mathrm{d}^{2}}{\mathrm{d}y^{2}}f\left(x\right)=\frac{\mathrm{d}}{\mathrm{d}y}\left(\frac{c}{2y^{1/2}}\frac{\mathrm{d}}{\mathrm{d}x}f\left(x\right)\right)=-\frac{c}{4y^{3/2}}\frac{\mathrm{d}}{\mathrm{d}x}f\left(x\right)+\left(\frac{c}{2y^{1/2}}\right)^{2}\frac{\mathrm{d}^{2}}{\mathrm{d}x^{2}}f\left(x\right)$$
Next, we can substitute our formula for $\dfrac{\mathrm d}{\mathrm dx}f(x)$:
$$\frac{\mathrm{d}^{2}}{\mathrm{d}y^{2}}f\left(x\right)=-\frac{c}{4y^{3/2}}\left(\left(\dfrac{2y^{1/2}}{c}\right)\frac{\mathrm{d}}{\mathrm{d}y}f\left(x\right)\right)+\left(\frac{c}{2y^{1/2}}\right)^{2}\frac{\mathrm{d}^{2}}{\mathrm{d}x^{2}}f\left(x\right)$$
Simplifying and adding the $\dfrac{\mathrm d}{\mathrm dy}$ term to both sides yields:
$$\frac{\mathrm{d}^{2}}{\mathrm{d}y^{2}}f\left(x\right)+\dfrac{1}{2y}\frac{\mathrm{d}}{\mathrm{d}y}f\left(x\right)=\left(\frac{c}{2y^{1/2}}\right)^{2}\frac{\mathrm{d}^{2}}{\mathrm{d}x^{2}}f\left(x\right)$$
Finally, division gives us a formula for $\frac{\mathrm{d}^{2}}{\mathrm{d}x^{2}}$ in terms of $\frac{\mathrm{d}^{2}}{\mathrm{d}y^{2}}$ and $\frac{\mathrm{d}}{\mathrm{d}y}$:
$$\dfrac{4y}{c^2}\frac{\mathrm{d}^{2}}{\mathrm{d}y^{2}}f\left(x\right)+\dfrac{2}{c^2}\frac{\mathrm{d}}{\mathrm{d}y}f\left(x\right)=\frac{\mathrm{d}^{2}}{\mathrm{d}x^{2}}f\left(x\right)$$

These formulas have been a pain to derive this way because everything is backwards. It's much easier to use $y=\dfrac{x^2}{c^2}$ and let the arbitrary function be $f(y)$ (and substitute $x=c*y^{1/2}$ at the end).
The $n=1$ calculation then becomes $$\frac{\mathrm{d}}{\mathrm{d}x}f\left(y\right)=\frac{\mathrm{d}}{\mathrm{d}x}f\left(\frac{x^{2}}{c^{2}}\right)=\frac{2x}{c^{2}}f'\left(\frac{x^{2}}{c^{2}}\right)=\frac{2x}{c^{2}}\frac{\mathrm{d}}{\mathrm{d}y}f\left(y\right)=\boxed{\left(\dfrac{2y^{1/2}}{c}\dfrac{\mathrm{d}}{\mathrm{d}y}\right)f\left(y\right)}$$
$n=2$ is as straightforward as $$\frac{\mathrm{d}^{2}}{\mathrm{d}x^{2}}f\left(y\right)=\frac{\mathrm{d}}{\mathrm{d}x}\left(\frac{2x}{c^{2}}f'\left(\frac{x^{2}}{c^{2}}\right)\right)=\frac{2}{c^{2}}f'\left(\frac{x^{2}}{c^{2}}\right)+\left(\frac{2x}{c^{2}}\right)^{2}f''\left(\frac{x^{2}}{c^{2}}\right)$$
$$=\frac{2}{c^{2}}\frac{\mathrm{d}}{\mathrm{d}y}f\left(y\right)+\left(\frac{2y^{1/2}}{c}\right)^{2}\frac{\mathrm{d}^{2}}{\mathrm{d}y^{2}}f\left(y\right)=\boxed{\left(\dfrac{2}{c^{2}}\dfrac{\mathrm{d}}{\mathrm{d}y}+\dfrac{4y}{c^{2}}\dfrac{\mathrm{d}^{2}}{\mathrm{d}y^{2}}\right)f\left(y\right)}$$
To go further than this. You may as well use a calculating device. Wolfram|Alpha can verify n=1, n=2, etc. For example, n=7 yields: 
$$\frac{\mathrm{d}^{7}}{\mathrm{d}x^{7}}=\frac{1680 y^{1/2}}{c^7}\frac{\mathrm{d}^{4}}{\mathrm{d}y^{4}}+\frac{3360
   y^{3/2} }{c^7}\frac{\mathrm{d}^{5}}{\mathrm{d}y^{5}}+\frac{1344 y^{5/2}}{c^7}\frac{\mathrm{d}^{6}}{\mathrm{d}y^{6}}+\frac{128 y^{7/2}}{c^7}\frac{\mathrm{d}^{7}}{\mathrm{d}y^{7}}$$

What about a general formula? Well, asking OEIS for 0,0,0,0,1680,3360,1344,128 finds the sequence of coefficients for these formulas, and a lot of facts/formulas for the sequence. In particular, if we define $$a_{n,k}=\begin{cases}\dfrac{n!}{2^{n-2k}\left(\left(n-k\right)!\right)\left(\left(2k-n\right)!\right)} & \text{ if }n/2\le k\le n\\0 & \text{ otherwise}\end{cases}$$
Then we have $$\frac{\mathrm{d}^{n}}{\mathrm{d}x^{n}}=\dfrac1{c^n}\sum_{k=0}^na_{n,k}*y^{k/2}\frac{\mathrm{d}^{k}}{\mathrm{d}y^{k}}$$
I didn't actually prove that this works, but it shouldn't be too hard to show that the coefficients of the operators satisfy the same recurrence relation as $a_{n,k}$, as given on the OEIS page.
A: Hint: I've added an  answer to the original question which might be helpful.
The questions you are addressing here are analysed in some detail and a general formula for the $n$-th derivative of $D_x^ne^{-\frac{c}{x}}$

\begin{align*}
D_x^ne^{-\frac{c}{x}}=e^{-\frac{c}{x}}\sum_{k=0}^{n-1}(-1)^k\binom{n}{k}\binom{n-1}{k}k!c^{n-k}x^{-2n+k}
\end{align*}

is provided.
