Rotational invariance of Laplacian operator I was reading in Wikipedia about Rotational invariance and noticed that the two-dimensional Laplacian operator $\nabla^2 = \frac{\partial^2 }{\partial x^2} + \frac{\partial^2 }{\partial y^2}$ is thought to be invariant under rotations. I was trying to prove this for a given function $f\in \mathbb{R}^2$ but I couldn't find a way.
Let's assume we have a 2D given function $f(x,y)$ in Cartesian coordinates. I am trying to show that the Laplacian operator is rotational invariant, which means that:
$$\nabla^{2}_{xy} f = f_{xx}+f_{yy}=f_{x^\prime x^\prime}+f_{y^\prime y^\prime }=\nabla^{2}_{x^\prime y^\prime} f$$
Which is the right way to approach this? 
 A: In this 2-dimensional case, everything is much simpler, I agree. In fact you can even write down explicitly what a general rotation looks like. So, suppose you have two sets of coordinates; $(x,y)$ and $(u,v)$, where one is obtained from another by a rotation, say of angle $\phi$:
\begin{align}
\begin{cases}
u &= x\cos \phi - y \sin \phi \\
v &= x \sin \phi + y \cos \phi
\end{cases}
\end{align}
Now, using the chain rule, we find that
\begin{align}
\dfrac{\partial}{\partial x} &= \dfrac{\partial u}{\partial x} \dfrac{\partial }{\partial u}  + \dfrac{\partial v}{\partial x} \dfrac{\partial}{\partial v} \\
&= \cos \phi \dfrac{\partial}{\partial u} + \sin \phi \dfrac{\partial}{\partial v}
\end{align}
and similarly,
\begin{align}
\dfrac{\partial}{\partial y} &= -\sin \phi \dfrac{\partial}{\partial u} + \cos \phi \dfrac{\partial}{\partial v}
\end{align}
Now, try to calculate $\dfrac{\partial^2}{\partial x^2}$ and $\dfrac{\partial^2}{\partial y^2}$ similarly, and then add them up. You should find in a few lines of algebra (after using $\sin^2 + \cos ^2 = 1$ a couple of times) that
\begin{align}
\dfrac{\partial^2}{\partial x^2} + \dfrac{\partial^2}{\partial y^2} = \dfrac{\partial^2}{\partial u^2} + \dfrac{\partial^2}{\partial v^2}
\end{align}

Edit: Answering Question in comments
We have
\begin{align}
\dfrac{\partial^2 f}{\partial x^2} &=\dfrac{\partial}{\partial x} \left(\dfrac{\partial f}{\partial x} \right) 
\end{align}
Now, temporarily define $g$ as 
\begin{align}
g:= \dfrac{\partial f}{\partial x} = \dfrac{\partial u}{\partial x} \dfrac{\partial f }{\partial u}  + \dfrac{\partial v}{\partial x} \dfrac{\partial f}{\partial v} = \cos \phi \dfrac{\partial f}{\partial u} + \sin \phi \dfrac{\partial f}{\partial v}
\end{align}
So,
\begin{align}
\dfrac{\partial ^2 f}{\partial x^2} &= \dfrac{\partial g}{\partial x} \\
&= \dfrac{\partial u}{\partial x} \cdot \dfrac{\partial g}{\partial u} + \dfrac{\partial v}{\partial x} \cdot \dfrac{\partial g}{\partial v} \\
&= \cos \phi \dfrac{\partial g}{\partial u} + \sin \phi \dfrac{\partial g}{\partial v} \\
&= \cos \phi \dfrac{\partial }{\partial u} \left( \cos \phi \dfrac{\partial f}{\partial u} + \sin \phi \dfrac{\partial f}{\partial v}\right) + \sin \phi \dfrac{\partial }{\partial v} \left( \cos \phi \dfrac{\partial f}{\partial u} + \sin \phi \dfrac{\partial f}{\partial v} \right) \\
&= \cos^2 \phi \dfrac{\partial ^2 f}{\partial u^2} + 2\cos \phi \sin \phi \dfrac{\partial ^2 f}{\partial u \partial v} + \sin^2 \phi \dfrac{\partial ^2 f}{\partial v^2}
\end{align}
where in the last line, I expanded everything, and used equality of mixed partials. If you do a similar thing with $y$, you'll get a $-2 \sin \phi \cos \phi$ term instead.
A: A rotation from one system of Cartesian coordinates $x_i$ to another with coordinates $y_J$ satisfies $x_i=R_{iJ}y_J$, and hence a chain rule of the form $dx_i=R_{iJ}dy_J$, where we sum over repeated indices and the orthogonal matrix $R$ satisfies $RR^T=I$, or in terms of the Kronecker delta$R_{iJ}R_{kJ}=\delta_{ik}$. First derivatives obey$$R_{iJ}\partial_ifdy^J=\partial_i fdx^i=df=\partial_Jfdy^J\implies R_{ij}\partial_i=\partial_J.$$So$$\partial_J\partial_L=R_{iJ}R_{kL}\partial_i\partial_k\implies\nabla^{\prime2}=\partial_J\partial_J=R_{iJ}R_{kJ}\partial_i\partial_k=\partial_i\partial_i=\nabla^2.$$
