Construction of derivative for a mixer I have two streams, $S_1$ and $S_2$, mixing together into one outlet stream $S_3$.  The flow rate of $S_1$ is $Q_1$, $S_2$ is $Q_2=kQ_1$, mixing into $S_3$ with flow rate $Q_3$.  Each stream has a property $G$ which mixes linearly with $Q$.  $G_1$ is a function of $Q_1$ and $G_2$ is a function of $Q_2$.  I am trying to find $\frac{dG_3}{dQ_3}$ as a function of $\frac{dG_1}{dQ_1}$ and $\frac{dG_2}{dQ_2}$.
Here is one approach I have tried:
$Q_3 = Q_1+Q_2 = Q_1(1+k)$
$G_3 = \frac{G_1Q_1+G_2Q_2}{Q_1+Q_2} = \frac{G_1+kG_2}{1+k}$
$\frac{dG_3}{dQ_3} = \frac{dG_3}{dQ_1}\frac{dQ_1}{dQ_3}$
$ = \left(\frac1{1+k}\frac{dG_1}{dQ_1}+\frac{k}{1+k}\frac{dG_2}{dQ_1}\right) \left(\frac1{1+k}\right)$
$ = \left(\frac1{1+k}\frac{dG_1}{dQ_1}+\frac{k^2}{1+k}\frac{dG_2}{dQ_2}\right) \left(\frac1{1+k}\right)$
$ = \frac1{(1+k)^2}\left(\frac{dG_1}{dQ_1}+k^2\frac{dG_2}{dQ_2}\right)$
This would be fine, except that there is one major issue it does not encapsulate:
If $G_1$ and $G_2$ are constant, the above formula gives $\frac{dG_3}{dQ_3} = 0$.  But this is not true as if I mix $S_1$ and $S_2$ in different proportions I get a different $G_3$, $\frac{dG_3}{dQ_3}$ is non-zero. Also I expect if $k=0$ (only $S_1$ feeds $S_3$), then $\frac{dG_3}{dQ_3}$ should be $0$, and also should be $0$ when only $S_2$ feeds $S_3$ (ie. $Q_1=0$).  But between those two extremes $\frac{dG_3}{dQ_3}$ should be non-zero. 
So somehow I have to build in the perturbation of $k$ into this formulation to get the correct formula for $\frac{dG_3}{dQ_3}$.
 A: This reminds me of similar problems I did in Differential Equations, and from looking at your math, it seems correct. 
As for the issues you bring up, in the case of: 

"If $G_1$ and $G_2$ are constant, the above formula gives $\frac{dG_3}{dQ_3} =0$... If I mix $S_1$ and $S_2$ in different proportions I get a different $G_3$, $\frac{dG_3}{dQ_3}$ is non-zero." 

Both of these statements being simultaneously true is not a contradiction. If $G_1$ and $G_2$ are constants, the derivative will be zero, but unless the streams stop all together, $G_1$ and $G_2$ cannot be constants, since they rely on the flow rate. 
The second statement is mostly true, but I think potentially you're getting the meaning of proportion and flow rate mixed up. $S_1$ and $S_2$ are being mixed at all, implies that both streams must have some flow rate (ie $Q_1\neq 0$ and $Q_2\neq 0$).Thus, the only time $\frac{dG_3}{dQ_3} =0$ is when at least one of the streams isn't flowing at all. 
Also notice that when $k=0, \frac{dG_3}{dQ_3} = \frac{dG_1}{dQ_1}$, which doesn't necessarily mean $\frac{dG_3}{dQ_3} = 0$.
Your math above your paragraph all looks right to me. Hope this helps some.
