I have, e.g. a following PDE:
$\dfrac{\partial A}{\partial t} \:=\: -\,\dfrac{\partial\left(u\cdot A\right)}{\partial z}$
where
$A = \dfrac{B C}{D E}$,
where $B,\, C,\, D = f\left(t,\,z\right) \neq \rm const.$, but $E = \rm const.$, $u = f\left(t,\,z\right) \neq \rm const.$ is a velocity.
How do I substitute $A = \big(BC\big)\big/\big(DE\big)$ to the original PDE. I expect that (this is not a correct mathematical expression, but just to document my thinking):
$\partial A = \dfrac{1}{E} \cdot \partial\left(\dfrac{BC}{D}\right)$,
but I struggle to separate $B, C$ and $D$ to a single derivative terms.
$$\dfrac{\partial A}{\partial t} \:=\: -\,\dfrac{\partial\left(u A\right)}{\partial z}$$
$$\dfrac{\partial \dfrac{B C}{D E}}{\partial t} \:=\: -\,\dfrac{\partial\left( \dfrac{u B C}{D E}\right)}{\partial z}$$ Since $E$=constant : $$\dfrac{\partial \dfrac{B C}{D}}{\partial t} \:=\: -\,\dfrac{\partial\left( \dfrac{u B C}{D}\right)}{\partial z}$$
https://en.wikipedia.org/wiki/Chain_rule
$$\dfrac{C}{D}\dfrac{\partial B}{\partial t} + \dfrac{B}{D}\dfrac{\partial C}{\partial t} + BC\dfrac{\partial \frac{1}{D}}{\partial t}\:=-\dfrac{BC}{D}\dfrac{\partial u}{\partial z} -\dfrac{uC}{D}\dfrac{\partial B}{\partial z} - \dfrac{uB}{D}\dfrac{\partial C}{\partial z} - uBC\dfrac{\partial \frac{1}{D}}{\partial z}$$ $\frac{\partial \frac{1}{D}}{\partial t}=-\frac{1}{D^2}\frac{\partial D}{\partial t}$
$$\dfrac{C}{D}\dfrac{\partial B}{\partial t} + \dfrac{B}{D}\dfrac{\partial C}{\partial t} - \frac{BC}{D^2}\frac{\partial D}{\partial t}\:=-\dfrac{BC}{D}\dfrac{\partial u}{\partial z} -\dfrac{uC}{D}\dfrac{\partial B}{\partial z} - \dfrac{uB}{D}\dfrac{\partial C}{\partial z} + \frac{uBC}{D^2}\frac{\partial D}{\partial z}$$