Probably it is just a silly question.
I have an operator
$$L=(x-a)\frac{\partial}{\partial x}+(y-b)\frac{\partial}{\partial y}$$
with $a,b$ real numbers and $x,y$ real variables.
Let's say I need to change the variables (just a simple linear combination) as
$$z=\frac{x+y}{2}\\
t=x-y$$
Basically I would simply substitute $$\frac{\partial}{\partial x}=\frac{\partial}{\partial z}\frac{\partial z}{\partial x}+\frac{\partial}{\partial t}\frac{\partial t}{\partial x}=\frac{1}{2}\frac{\partial}{\partial z}+\frac{\partial}{\partial t}\\ \frac{\partial}{\partial y}=\frac{\partial}{\partial z}\frac{\partial z}{\partial y}+\frac{\partial}{\partial t}\frac{\partial t}{\partial y}=\frac{1}{2}\frac{\partial}{\partial z}-\frac{\partial}{\partial t}$$ then $$L=(z+\frac{1}{2}t-a)(\frac{1}{2}\frac{\partial}{\partial z}+\frac{\partial}{\partial t})+(z-\frac{1}{2}t-b)(\frac{1}{2}\frac{\partial}{\partial z}-\frac{\partial}{\partial t})$$
But I had the doubt, at some point, that I have to transform completely the $\nabla_{x,y}=\frac{\partial}{\partial x}\hat{e}_x+\frac{\partial}{\partial y}\hat{e}_y$ operator like $$\nabla_{z,t}=(\frac{\partial}{\partial z}\frac{\partial z}{\partial x}+\frac{\partial}{\partial t}\frac{\partial t}{\partial x})\hat{e}_x+(\frac{\partial}{\partial z}\frac{\partial z}{\partial y}+\frac{\partial}{\partial t}\frac{\partial t}{\partial y})\hat{e}_y=\cdots=\frac{\partial}{\partial z}\hat{e}_z+\frac{\partial}{\partial t}\hat{e}_t$$
This option clearly leads to a quite different result $$L=((x-a),0)\cdot\nabla_{x,y}+(0,(y-b))\cdot\nabla_{x,y} \to L = ((z+\frac{1}{2}t-a),0)\cdot\nabla_{z,t}+(0,(z-\frac{1}{2}t-b))\cdot\nabla_{z,t}$$
Assuming I can still do addition/multiplications, which one is wrong and why?
Thanks in advance.