Find that of $\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2}$ equals $= \frac{\partial^2 f}{\partial u^2} + ...$ Given:
$$\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = \frac{\partial^2 f}{\partial u^2} + \frac{\partial^2 f}{\partial v^2}$$
where
$$u = x \cos \theta + y \sin \theta$$
$$v = -x \sin \theta + y \cos \theta$$
I am trying to show how we can get:
$$ \frac{\partial f}{\partial y} = \frac{\partial f}{\partial u} \frac{\partial u}{\partial y} + \frac{\partial f}{\partial v} \frac{\partial v}{\partial y} = \frac{\partial f}{\partial u} \sin \theta+ \frac{\partial f}{\partial v} \cos \theta$$
Now, we can see that if we take the derivative of $u = x \cos \theta + y \sin \theta$ with respect to $y$ we can get why $\frac{\partial v}{\partial y} = sin(\theta)$ and same for $\frac{\partial u}{\partial y} = sin(\theta)$.
Question: how we can get that $ \frac{\partial f}{\partial y} = \frac{\partial f}{\partial u} \frac{\partial u}{\partial y} + \frac{\partial f}{\partial v} \frac{\partial v}{\partial y} $ in the first place please as if we summed $ \frac{\partial f}{\partial u} \frac{\partial u}{\partial y} + \frac{\partial f}{\partial v} \frac{\partial v}{\partial y} $ we will get $ 2 \frac{\partial f}{\partial y}  $ and not $ \frac{\partial f}{\partial y}  $?
 A: The keyword here is "chain of rule".
Consider a function $z=f(u,v)$ such that $\frac{\partial f}{\partial u}$ and $\frac{\partial f}{\partial v}$ there exists and $u=g(x,y)$ and $v=h(x,y)$ such that $\frac{\partial u}{\partial x},\frac{\partial u}{\partial y}, \frac{\partial v}{\partial x},\frac{\partial v}{\partial y}$ there exists. Then the 'chain of rule' states that
\begin{align}
\frac{\partial z}{\partial x}&=\frac{\partial z}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial z}{\partial v}\frac{\partial v}{\partial x}, \quad \tag{1}\\
\frac{\partial z}{\partial y}&=\frac{\partial z}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial z}{\partial v}\frac{\partial v}{\partial y}. \quad \tag{2}
\end{align}
The prove of this result is not "hard".
Then, setting $u=x\cos \theta+y\sin \theta$ and $v=-x\sin \theta+y\cos\theta$ we have by $(1)$ and $(2)$ that
$$\frac{\partial f}{\partial x}=\frac{\partial f}{\partial u}\cos \theta-\frac{\partial f}{\partial v}\sin \theta,$$
$$\frac{\partial f}{\partial y}=\frac{\partial f}{\partial u}\sin\theta +\frac{\partial f}{\partial v}\cos \theta.$$
Then, applying again the "chain of rule" in this case to $f_{x}\overset{\text{def}}{=}\frac{\partial f}{\partial x}$ and $f_{y}\overset{\text{def}}{=}\frac{\partial f}{\partial y}$ we have by $(1)$ and $(2)$ that
$$ \frac{\partial^{2}f}{\partial x^{2}}=\frac{\partial f_{x}}{\partial x}=\frac{\partial f_{x}}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial f_x}{\partial v}\frac{\partial v}{\partial x}=\frac{\partial f_{x}}{\partial u}\cos\theta -\frac{\partial f_x}{\partial v}\sin\theta,$$
$$ \frac{\partial^{2}f}{\partial y^{2}}=\frac{\partial f_{y}}{\partial y}=\frac{\partial f_{y}}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial f_y}{\partial v}\frac{\partial v}{\partial y}=\frac{\partial f_{y}}{\partial u}\sin\theta+\frac{\partial f_y}{\partial v}\cos\theta.$$
Solving and simplifying,
$$\frac{\partial^{2}f}{\partial x^{2}}+\frac{\partial^{2}f}{\partial y^{2}}=\frac{\partial}{\partial u}\left( \frac{\partial f}{\partial u}\cos \theta-\frac{\partial f}{\partial v}\sin \theta\right)\cos\theta-\frac{\partial }{\partial v}\left( 
\frac{\partial f}{\partial u}\cos \theta-\frac{\partial f}{\partial v}\sin \theta  \right)\sin\theta \\ +\frac{\partial}{\partial u}\left(\frac{\partial f}{\partial u}\sin\theta +\frac{\partial f}{\partial v}\cos \theta 
\right)\sin\theta\\ +\frac{\partial}{\partial v}\left(  \frac{\partial f}{\partial u}\sin\theta +\frac{\partial f}{\partial v}\cos \theta\right)\cos\theta\\ = \frac{\partial^{2}f}{\partial u^{2}}+\frac{\partial^{2}f}{\partial v^{2}},$$ as desired.
