A Lax pair for the Burgers equation $u_t+2 \, u \, u_x+ u_{xx} =0$ is,
$$L = \partial_x +u \text{ and } M=-\partial_{xx} -2 \, u \, \partial_{x}$$
To get the resulting differential equation from the Lax equation,
$$ \frac{dL}{dt}+[L,M] = 0$$
The solutions claim,
$$\frac{dL}{dt} = u_t $$
What I don't understand is why the partial derivative with respect to $x$ drops out, can someone please explain why?
Attempted Justification
We have $L \psi = \lambda \psi$ and $\psi_t=M \psi$, we then take the time derivative of the first equation to give,
$$ \frac{dL}{dt} \psi + L \psi_t= \lambda \psi_t \implies \frac{dL}{dt} \psi + L M \psi = M L\psi $$
Thus the lax equation seemingly already takes into account the time derivative which acts on $\psi$. But I am still not comfortable with dropping partial derivative with respect to $x$.
The differential $dL/dt$ is the time derivative of the operator $L$! As only the term $u(x,t)$ depends on $t$ the result $dL/dt = u_t$ directly follows.
Maybe you are confused because of the total derivative (and not a partial derivative). It turns out that this is nothing to worry as the operator $L(t)$ is only a function of $t$ as it acts as an operator in $x$ (so $x$ is like the matrix indices $i$ and $j$ in the discrete version $L(t)_{ij}$). I believe this should be made explicit in the notation as otherwise it might be indeed confusing.