Computing second order partial derivatives I am tasked with the following question:

Consider the wave equation for linearly $x$ polarised waves travelling in the $\pm z$ directions: 
  $$\frac{\partial^2 E_x}{\partial t^2}=c^2\frac{\partial^2 E_x}{\partial z^2}\tag{1}$$
  Transform Eq. $(1)$ to the independent variables $q = z − ct$ and $s = z + ct$ and show
  that
  $$\frac{\partial^2 E_x}{\partial s \partial q}=0$$

When I am confronted with questions of this type I always use the 'tree diagram for partial derivatives' method which is outlined in this video - Chain rule and tree diagrams of multivariable functions (KristaKingMath) by Krista King.
So writing 
$$E_{x}=E_{x}(q,s)$$
$$q=q(z,t) \qquad\text{and}\qquad s=s(z,t)$$
then
$$\frac{\partial E_x}{\partial z}=\frac{\partial E_x}{\partial s}\cdot\frac{\partial s}{\partial z}+\frac{\partial E_x}{\partial q}\cdot\frac{\partial q}{\partial z}$$
Now since $q = z − ct$ and $s = z + ct$
$$\frac{\partial s}{\partial z}=\frac{\partial q}{\partial z}=1$$
then
$$\frac{\partial E_x}{\partial z}=\frac{\partial E_x}{\partial s}+\frac{\partial E_x}{\partial q}\tag{a}$$
Now with respect to time:
$$\frac{\partial E_x}{\partial t}=\frac{\partial E_x}{\partial q}\cdot\frac{\partial q}{\partial t}+\frac{\partial E_x}{\partial s}\cdot\frac{\partial s}{\partial t}$$
Since,
$$-\frac{\partial q}{\partial z}=\frac{\partial s}{\partial z}=c$$
then
$$\frac{\partial E_x}{\partial t}=c\biggl(\frac{\partial E_x}{\partial s}-\frac{\partial E_x}{\partial q}\biggr)\tag{b}$$
From $(\mathrm{a})$ (since it holds for all functions)
$$\frac{\partial }{\partial z}=\frac{\partial }{\partial s}+\frac{\partial}{\partial q}\tag{2}$$
So differentiating $(\mathrm{b})$ wrt $z$ using $(2)$
$$\frac{\partial^2 E_x}{\partial z \partial t}=c\left(\frac{\partial }{\partial s}+\frac{\partial}{\partial q}\right)\left(\frac{\partial E_x}{\partial s}-\frac{\partial E_x}{\partial q}\right)=c\left(\frac{\partial^2 E_x}{\partial s^2}-\frac{\partial^2 E_x}{\partial s \partial q}+\frac{\partial^2 E_x}{\partial q \partial s}-\frac{\partial^2 E_x}{\partial q^2}\right)$$
$$=c\left(\frac{\partial^2 E_x}{\partial s^2}-\frac{\partial^2 E_x}{\partial q^2}\right)$$
So the expression that I wanted to show is zero,
$$\frac{\partial^2 E_x}{\partial s \partial q}$$ has cancelled out of the expression. Here I have assumed that $E_x$ is sufficiently 'well-behaved' such that $$\frac{\partial^2 E_x}{\partial s \partial q}=\frac{\partial^2 E_x}{\partial q \partial s}$$ 

This is the answer given by the author:


Since the solution given by the author doesn't explicitly show that 
$$\frac{\partial^2 E_x}{\partial s \partial q}=0,$$
could someone please explain how I can show that it is zero?
 A: To show that $\dfrac{\partial^2 E_x}{\partial s \partial q}=0$ compute the second partial derivatives in the original PDE, $\dfrac{\partial^2 E_x}{\partial t^2}-c^2\dfrac{\partial^2 E_x}{\partial z^2}=0$ in terms of the partial derivatives wrt to the new variables. In fact, you did it half the way as you did it for the first derivatives,
$$\frac{\partial E_x}{\partial z}=\frac{\partial E_x}{\partial s}+\frac{\partial E_x}{\partial q}$$
and
$$\frac{\partial E_x}{\partial t}=c\biggl(\frac{\partial E_x}{\partial s}-\frac{\partial E_x}{\partial q}\biggr)\tag{b}$$
So,
$$\frac{\partial^2 E_x}{\partial z^2}=\left(\frac{\partial }{\partial s}+\frac{\partial}{\partial q}\right)\left(\frac{\partial E_x}{\partial s}+\frac{\partial E_x}{\partial q}\right)=\left(\frac{\partial^2 E_x}{\partial s^2}+2\frac{\partial^2 E_x}{\partial q \partial s}+\frac{\partial^2 E_x}{\partial q^2}\right)$$
and
$$\frac{\partial^2 E_x}{\partial t^2}=c\left(\frac{\partial }{\partial s}-\frac{\partial}{\partial q}\right)c\left(\frac{\partial E_x}{\partial s}-\frac{\partial E_x}{\partial q}\right)=c^2\left(\frac{\partial^2 E_x}{\partial s^2}-2\frac{\partial^2 E_x}{\partial q \partial s}+\frac{\partial^2 E_x}{\partial q^2}\right)$$
Substituting:
$$0=\dfrac{\partial^2 E_x}{\partial t^2}-c^2\dfrac{\partial^2 E_x}{\partial z^2}=c^2\left(\frac{\partial^2 E_x}{\partial s^2}-2\frac{\partial^2 E_x}{\partial q \partial s}+\frac{\partial^2 E_x}{\partial q^2}\right)-c^2\left(\frac{\partial^2 E_x}{\partial s^2}+2\frac{\partial^2 E_x}{\partial q \partial s}+\frac{\partial^2 E_x}{\partial q^2}\right)=$$
$$=-4c^2\frac{\partial^2 E_x}{\partial q \partial s}$$
which means,
$$\frac{\partial^2 E_x}{\partial q \partial s}=0$$
A: Notation: let $E_x:=E$, $\partial_s:=\partial/\partial s$, $\partial_s^2:= \partial^2/\partial s^2$, etc. 
You already found the first derivatives in the new variables. Let's write this in a suggestive way of the new conjugate (up to $c$) differential operators. $$\begin{align*} \partial_z E = (\partial_s  + \partial_q) E, \\ \partial_t E = c(\partial_s  - \partial_q)E\end{align*} $$ 
Then, just like the familiar ring of polynomials, we have a (non-commutative) ring of differential operators with$$\begin{align*}\partial_z = \partial_s + \partial_q \Rightarrow \partial_z^2 &= \partial_s^2 + 2\partial_{qs}+\partial_q^2, \\ \partial_t = c(\partial_s - \partial_q)\Rightarrow\partial_t^2 &= c^2(\partial_s^2-2\partial_{qs}+\partial_q^2)\end{align*} $$
As $E(t,z)$ satisfies the wave equation, $\partial_{qs}E = 0.$
