Let $\Omega \subseteq \mathbb R^2$ be an open subset, and $L$ a (linear) differential operator of order 2 on $C^2(\Omega)$, given by
$$L u = \sum_{i, j = 1}^2 a_{i, j} \partial_i \partial_j u + \sum_{i=1}^2 b_i \partial_i u + c u$$
where $a_{i, j}, b_i, c \in C(\Omega, \mathbb R)$. Furthermore, let $U \subseteq \mathbb R^2$ be open, and $\phi: \Omega \to U$ an $C^2$-diffeomorphism.
I now want to show that the operator $\tilde L w := (L (w \circ \phi)) \circ \phi^{-1}$ is a linear differential operator on $C^2(U)$ of the same type as $L$ (parabolic, hyperbolic, ultra-hyperbolic etc.), and that the (total) symbol of $\tilde L$ is given by
$$\tilde L_p(y, \xi) = - \xi^T J_\phi(x) A(x) J_\phi^T (x) \xi \quad \text{ with } y = \phi(x), A = (a_{i, j})_{1 \leq i, j \leq 2}$$
where $J_\phi$ ist the Jacobiean matrix of $\phi$ (and where the symbol of $L$ is defined as $L(x, \xi) = - \xi^t A(x) \xi$ for $A = (a_{i, j}$)).
Now I tried to write out
$$(L(w \circ \phi)) \circ \phi^{-1} = \left( \sum_{i, j}^2 a_{i, j} \partial_i \partial_j (w \circ \phi) + \sum_{i = 1}^2 b_i \partial_i (w \circ \phi) + c (w \circ \phi) \right) \circ \phi^{-1}$$
but I'm no really sure how to proceed from there. I can't just "pull" the $\phi^{-1}$ into the individual sums, or can I? I think this would require some sort of distributivity which I don't see given here?
I know that to check the type of $\tilde L$, I'd need to check the eigenvalues of the coefficient matrix of the highest derivative of $\tilde L$ (for example if the matrix $A$ of $L$ is singular, then $L$ is parabolic, if $A$ has $n-1$ eigenvalues $> 0$ and one $< 0$ then $L$ is hyperbolic, and so on).
I presume this matrix for $\tilde L$ might end up being a product of the coefficient matrix $A$ of the original operator $L$ and the Jacobian matrix of $\phi$ because then it'd be plausible that I would arrive at the above formula for the symbol. But I'm not sure how to rewrite $\tilde L$ to get the coefficient matrix of its' 2nd-order derivatives to check the type of the operator and arrive at the the symbol formula.