I'm trying to show that the Laplace operator is rotationally invariant. It's similar to this topic, but I can't seem to figure my mistake. I want to prove that: $$ \frac{\partial^{2}f}{\partial x^{2}}+\frac{\partial^{2}f}{\partial y^{2}}=\frac{\partial^{2}f}{\partial u^{2}}+\frac{\partial^{2}f}{\partial v^{2}} $$ where: $$ \begin{cases} u=x\cos\theta-y\sin\theta\\ v=x\sin\theta+y\cos\theta \end{cases} $$ On one hand we get: $$ \frac{\partial f}{\partial x}=\frac{\partial f}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial x}=\frac{\partial f}{\partial u}\cos\theta+\frac{\partial f}{\partial v}\sin\theta $$ So we get: $$ \frac{\partial^{2}f}{\partial x^{2}}=\frac{\partial}{\partial x}\left(\frac{\partial f}{\partial x}\right)=\frac{\partial}{\partial x}\left(\frac{\partial f}{\partial u}\cos\theta+\frac{\partial f}{\partial v}\sin\theta\right)=\frac{\partial}{\partial x}\frac{\partial f}{\partial u}\cos\theta+\frac{\partial}{\partial x}\frac{\partial f}{\partial v}\sin\theta $$ Then: $$ \frac{\partial}{\partial x}\frac{\partial f}{\partial u}=\frac{\partial}{\partial u}\frac{\partial f}{\partial x}=\frac{\partial}{\partial u}\left(\frac{\partial f}{\partial u}\cos\theta+\frac{\partial f}{\partial v}\sin\theta\right)=\frac{\partial^{2}f}{\partial u^{2}}\cos\theta+\frac{\partial^{2}f}{\partial u\partial v}\sin\theta\\ \frac{\partial}{\partial x}\frac{\partial f}{\partial v}=\frac{\partial}{\partial v}\frac{\partial f}{\partial x}=\frac{\partial}{\partial v}\left(\frac{\partial f}{\partial u}\cos\theta+\frac{\partial f}{\partial v}\sin\theta\right)=\frac{\partial^{2}f}{\partial u\partial v}\cos\theta+\frac{\partial^{2}f}{\partial v^{2}}\sin\theta $$ Lets put it back together: $$ \frac{\partial^{2}f}{\partial x^{2}}=\left(\frac{\partial^{2}f}{\partial u^{2}}\cos\theta+\frac{\partial^{2}f}{\partial u\partial v}\sin\theta\right)\cos\theta+\left(\frac{\partial^{2}f}{\partial u\partial v}\cos\theta+\frac{\partial^{2}f}{\partial v^{2}}\sin\theta\right)\sin\theta $$
Now I want to do the same for $\frac{\partial f}{\partial y}$: $$ \frac{\partial f}{\partial y}=\frac{\partial f}{\partial v}\frac{\partial v}{\partial y}+\frac{\partial f}{\partial u}\frac{\partial u}{\partial y}=-\frac{\partial f}{\partial u}\sin\theta+\frac{\partial f}{\partial v}\cos\theta $$ So we get: $$ \frac{\partial^{2}f}{\partial y^{2}}=\frac{\partial}{y}\left(\frac{\partial f}{\partial y}\right)=\frac{\partial}{\partial y}\left(-\frac{\partial f}{\partial u}\sin\theta+\frac{\partial f}{\partial v}\cos\theta\right)=-\frac{\partial}{\partial y}\frac{\partial f}{\partial u}\sin\theta+\frac{\partial}{\partial y}\frac{\partial f}{\partial v}\cos\theta $$ Then we get: $$ \frac{\partial}{\partial y}\frac{\partial f}{\partial u}=\frac{\partial}{\partial u}\frac{\partial f}{\partial y}=\frac{\partial}{\partial u}\left(-\frac{\partial f}{\partial u}\sin\theta+\frac{\partial f}{\partial v}\cos\theta\right)=-\frac{\partial^{2}f}{\partial u^{2}}\sin\theta+\frac{\partial^{2}f}{\partial u\partial v}\cos\theta\\ \frac{\partial}{\partial y}\frac{\partial f}{\partial v}=\frac{\partial}{\partial v}\frac{\partial f}{\partial y}=\frac{\partial}{\partial v}\left(-\frac{\partial f}{\partial u}\sin\theta+\frac{\partial f}{\partial v}\cos\theta\right)=-\frac{\partial^{2}f}{\partial u\partial v}\sin\theta+\frac{\partial^{2}f}{\partial v^{2}}\cos\theta $$ Lets put it together: $$ \frac{\partial^{2}f}{\partial y^{2}}=\left(-\frac{\partial^{2}f}{\partial u^{2}}\sin\theta+\frac{\partial^{2}f}{\partial u\partial v}\cos\theta\right)\sin\theta+\left(-\frac{\partial^{2}f}{\partial u\partial v}\sin\theta+\frac{\partial^{2}f}{\partial v^{2}}\cos\theta\right)\cos\theta $$
But, as I understand, in order to show $\frac{\partial^{2}f}{\partial x^{2}}+\frac{\partial^{2}f}{\partial y^{2}}=\frac{\partial^{2}f}{\partial u^{2}}+\frac{\partial^{2}f}{\partial v^{2}}$, I need to have: $$ \frac{\partial^{2}f}{\partial y^{2}}=\left(\frac{\partial^{2}f}{\partial u^{2}}\sin\theta-\frac{\partial^{2}f}{\partial u\partial v}\cos\theta\right)\sin\theta+\left(-\frac{\partial^{2}f}{\partial u\partial v}\sin\theta+\frac{\partial^{2}f}{\partial v^{2}}\cos\theta\right)\cos\theta $$
Because then I get: $$ \begin{align*} \frac{\partial^{2}f}{\partial x^{2}}+\frac{\partial^{2}f}{\partial y^{2}}&=\left(\frac{\partial^{2}f}{\partial u^{2}}\cos\theta+\frac{\partial^{2}f}{\partial u\partial v}\sin\theta-\frac{\partial^{2}f}{\partial u\partial v}\sin\theta+\frac{\partial^{2}f}{\partial v^{2}}\cos\theta\right)\cos\theta+\left(\frac{\partial^{2}f}{\partial u\partial v}\cos\theta+\frac{\partial^{2}f}{\partial v^{2}}\sin\theta+\frac{\partial^{2}f}{\partial u^{2}}\sin\theta-\frac{\partial^{2}f}{\partial u\partial v}\cos\theta\right)\sin\theta\\&=\left(\frac{\partial^{2}f}{\partial u^{2}}+\frac{\partial^{2}f}{\partial v^{2}}\right)\cos^{2}\theta+\left(\frac{\partial^{2}f}{\partial u^{2}}+\frac{\partial^{2}f}{\partial v^{2}}\right)\sin^{2}\theta\\&=\left(\frac{\partial^{2}f}{\partial u^{2}}+\frac{\partial^{2}f}{\partial v^{2}}\right)\left(\cos^{2}\theta+\sin^{2}\theta\right)\\&=\frac{\partial^{2}f}{\partial u^{2}}+\frac{\partial^{2}f}{\partial v^{2}} \end{align*} $$
What am I missing here?
