Finding $\frac{\partial^2 U}{\partial y \,\partial x}$ Let $x=2r-s$ and $y=r+2s$. Let $U:\mathbb{R}^2\to\mathbb{R}$ be a smooth function. Find $\frac{\partial^2 U}{\partial y \,\partial x}$ in terms of derivatives of only $U$ with respect to $r$ and $s$
I know the chain rule so this is my attempted solution and I wanted to know if I did this correctly? I'm confused about the change of variables.
$$\frac{\partial U}{\partial x} = \frac{\partial U}{\partial r}\frac{\partial r}{\partial x} + \frac{\partial U}{\partial s} \frac{\partial s}{\partial x}$$
$$x = 2r - s, y = r+2s \implies x-2y = -5s \implies s = \frac{-1}{5}(x-2y)$$
also $$2x+y = 5r \implies r = \frac{1}{5}(2x+y)$$
$$\frac{\partial r}{\partial x} = \frac{2}{5} 
\qquad \frac{\partial r}{\partial y} = \frac{1}{5}
\qquad \frac{\partial s}{\partial x} = \frac{-1}{5}
\qquad \frac{\partial s}{\partial y} = \frac{2}{5}$$
$$\frac{\partial U}{\partial x} = \frac{\partial U}{\partial r}\frac{2}{5} + \frac{\partial U}{\partial s}\frac{-1}{5}$$
\begin{align}
\frac{\partial^2 U}{\partial y \, \partial x} &= \frac{\partial}{\partial y} \left[\frac{\partial U}{\partial x}\right] \\[8pt]
&= \frac{\partial}{\partial y} \left[\frac{\partial U}{\partial r}\frac{2}{5} +\frac{\partial U}{\partial s} \left(\frac{-1}{5}\right)\right] = \frac{2}{5} \left[\frac \partial {\partial y} \frac{\partial U}{\partial r} \right] + \frac{-1}{5} \left[\frac{\partial}{\partial y}\frac{\partial U}{\partial s} \right]  \\[8pt]
&= \frac{2}{5} \left[\frac{\partial^2 U}{\partial r^2} \frac{\partial r}{\partial x} + \frac{\partial^2 U}{\partial s \, \partial r} \frac{\partial U}{\partial s}\right]+\frac{-1}{5} \left[\frac{\partial^2 U}{\partial s\,\partial r} \frac{\partial r}{\partial x} + \frac{\partial^2 U}{\partial s^2}\frac{\partial s}{\partial x}\right] \\[8pt]
&= \frac{2}{5}\left[\frac{\partial^2 U}{\partial r^2} \frac{2}{5} + \frac{\partial^2 U}{\partial s \,\partial r}\frac{-1}{5}\right]+\left(\frac{-1}{5} \right) \left[\frac{\partial^2 U}{\partial s\, \partial r} \frac{2}{5} + \frac{\partial^2 U}{\partial s^2}\frac{-1}{5}\right] \\[8pt]
 &= \frac{4}{25}\left(\frac{\partial^2 U}{\partial r^2}\right)+\frac{1}{25} \left(\frac{\partial^2 U}{\partial s^2}\right)
\end{align}
I think I solved this problem correctly but I am not sure, would anyone be able to verify please?
 A: The step
$$ \frac{2}{5}\left[\frac{\partial}{\partial{y}}\frac{\partial{U}}{\partial{r}}\right]+\frac{-1}{5}\left[\frac{\partial}{\partial{y}}\frac{\partial{U}}{\partial{s}}\right]= \frac{2}{5}\left[\frac{\partial^2{U}}{\partial{r^2}} \frac{\partial{r}}{\partial{x}} + \frac{\partial^2{U}}{\partial{s}\partial{r}}\frac{\partial{U}}{\partial{s}}\right]+\frac{-1}{5}\left[\frac{\partial^2{U}}{\partial{s}\partial{r}} \frac{\partial{r}}{\partial{x}} + \frac{\partial^2{U}}{\partial{s^2}}\frac{\partial{s}}{\partial{x}}\right] $$
has a few errors. Why do you go from the derivative with respect to $y$ to the derivative with respect to $x$? All the terms on the right hand side should contain $\frac{\partial r}{\partial y}$ or $\frac{\partial s}{\partial y}$.
Even if this would be fine, the $\frac{\partial^2{U}}{\partial{s}\partial{r}}$ terms do not cancel (they might if you do the previous step right). The coefficients are both $-2/25$, so there should be a cross term with coefficient $-4/25$
A: The first derivative looks okay. The second derivative isn't: it looks like your problem is $\partial/\partial y$. We have
$$ \frac{\partial F}{\partial y} = \frac{\partial r}{\partial y}\frac{\partial F}{\partial r} + \frac{\partial s}{\partial y} \frac{\partial F}{\partial s} = \frac{1}{5} \frac{\partial F}{\partial r} + \frac{2}{5} \frac{\partial F}{\partial s} $$
for any $F$. Applying this to the terms in $\partial U/\partial x$,
$$ \frac{\partial^2 U}{\partial y \partial x} = \frac{1}{5} \frac{\partial }{\partial r} \left( \frac{2}{5} \frac{\partial U}{\partial r} - \frac{1}{5} \frac{\partial U}{\partial s} \right) + \frac{2}{5} \frac{\partial }{\partial s} \left( \frac{2}{5} \frac{\partial U}{\partial r} - \frac{1}{5} \frac{\partial U}{\partial s} \right) \\
= \frac{2}{25} \frac{\partial^2 U}{\partial r^2} + \frac{3}{25} \frac{\partial^2 U}{\partial r \partial s} - \frac{2}{25} \frac{\partial^2 U}{\partial s^2}, $$
since the $r$ and $s$ derivatives commute (if they don't, the middle term is replaced by two terms that you can work out yourself from the previous line).
