Partial Derivatives involving Change of Variables I am currently reading Elementary Fluid Dynamics by Acheson and following along $3.9$. In this section, the author uses a change of variables: $z=3c-2c_0$ (where $c_0$ is a constant and $z,c$ are functions of $x,t$) to transform the equation:
$$ c_t + (2c_0-3c)c_x=0 \quad\implies\quad z_t-zz_x=0  $$
which is simply the advection equation with $z$ as the speed and hence the solution is:
$$ z = F(x+zt) $$
All of this I understand just fine. However, immediately after the author verifies this is the solution by actually differentiating and substituting this form for $z$ into the PDE. This is where I am a little confused. Why is it that:
$$ \frac{\partial}{\partial t}(F) = F'(\xi)\frac{\partial \xi}{\partial t} + F'(\xi)\frac{\partial \xi}{\partial z}\frac{\partial z}{\partial t} $$
where $\xi = x+zt$. It seems to me that either of these terms on their own would be equal to $\frac{\partial}{\partial t}(F)$ but I don't quite see why both are necessary. This is something that I keep running into and it always confuses me. Note my confusion is not with the chain rule, I understand that just fine and see how either term individually would be found, it's the fact that both terms are needed that confuses especially considering $F$ is a function of only one variable.
Example 2:
Another similar example that I can follow along but can't quite justify to myself comes from a fluid dynamics paper. In the paper, the film height $h=h(x,y,t)$ and the height-averaged per volume solute concentration $\phi = \phi(x,y,t)$ are reformulated as:
$$ \xi = h/h_* \qquad\text{and}\qquad \psi = h\phi $$
Later on in the paper, the author takes the partial derivative with respect to $h$ of a function given as $G=G(\xi,\phi)$. Hence, in this case it makes sense to me that this partial derivative will be of two terms, namely:
$$ \frac{\partial}{\partial h}G = \frac{\partial G}{\partial \xi}\frac{\partial\xi}{\partial h} + \frac{\partial G}{\partial \phi}\frac{\partial \phi}{\partial h} = \frac{1}{h_*}G_\xi -\frac{\psi}{h^2} G_\phi $$
but I can't quite reconcile these two examples. In the second case, $G$ is a function of two variables both of which depend on $h$. Hence, it seems intuitive that both terms will be needed. However, in the first case, we only have a function of one variable.
 A: For the implicit solution $$
z = F(\xi) ,\qquad \xi = x+zt  $$
where $z = z(x,t)$, we note that all the above quantities are functions of $(x,t)$. Thus, partial differentiation with respect to $t$ gives directly $$
\partial_t z = F'(\xi)\, \partial_t \xi , \qquad \partial_t \xi = 0 + (\partial_t z)\, t + z ,
$$
according to the chain rule and the product rule. This was the easy computation, which is pretty much straightforward to follow. Now, let us view $\xi$ as a function of $(x,t,z)$. It follows that time-domain partial differentiation takes the following form $$
\partial_t \xi = \partial_t\xi + \partial_z \xi\, \partial_t z = z + t\, \partial_t z ,
$$
which is a total derivative in time. Note that both approaches give the exact same result. Hope this observation / explanation helps.
NB. This identity is then used to write $\partial_t z = \frac{zF'(\xi)}{1-tF'(\xi)}$ and similar computations are performed for $\partial_x z$ to end the proof.
The second example involves the computation of a total derivative. In fact, we have set $G = G(\xi,\phi)$ where $\xi$ and $\phi$ may be expressed in terms of $h$. Similarly to above, $$
\partial_h G = \partial_\xi G\, \partial_h \xi + \partial_\phi G\, \partial_h \phi = \frac{1}{h_*}\partial_\xi G - \frac{\psi}{h^2} \partial_\phi G .
$$
