Does the chain rule automatically imply the full derivative?

Question

Starting with this equation

$$\ln{q}=\frac{u}{1-u}(\ln{\beta}+2\ln{u})$$

Where

$$u=\frac{\partial \ln{q}}{\partial \ln{C}}$$

First I note that the partial derivative of the equation w.r.t. $\ln{\beta}$ is:

$$\frac{\partial \ln{q}}{\partial \ln{\beta}}=\frac{u}{1-u}$$

Now I want to partial differentiate the starting equation w.r.t. $\ln{u}$:

$$\frac{\partial \ln{q}}{\partial \ln{u}}=\frac{1}{1-u}(\ln{q}+2u)$$

But note

$$\frac{1}{1-u}=\frac{\partial \ln{C}}{\partial \ln{q}} \frac{\partial \ln{q}}{\partial \ln{\beta}}$$

So the fourth displayed equation can be rewritten:

$$\frac{\partial \ln{q}}{\partial \ln{u}}=\frac{\partial \ln{C}}{\partial \ln{q}} \frac{\partial \ln{q}}{\partial \ln{\beta}} (\ln{q}+2u)$$

I multiply through by $\frac{\partial \ln{\beta}}{\partial \ln{q}}$:

$$\frac{\partial \ln{\beta}}{\partial \ln{q}} \frac{\partial \ln{q}}{\partial \ln{u}}=\frac{\partial \ln{C}}{\partial \ln{q}} (\ln{q}+2u)$$

And then invoking the chain rule on the LHS I end up with:

$$\frac{\partial \ln{\beta}}{\partial \ln{u}}=\frac{1}{u} (\ln{q}+2u)$$

My question is the following -- By implication in the partial derivative of the original equation w.r.t. $\ln{\beta}$ (third displayed equation), $\frac{\partial \ln{u}}{\partial \ln{\beta}}=0$. Does this mean that the equation here at the end is equal to $0$ or maybe even undefined? Or does the expression $\frac{\partial \ln{\beta}}{\partial \ln{q}} \frac{\partial \ln{q}}{\partial \ln{u}}$ in fact imply the full derivative $\frac{d \ln{\beta}}{d \ln{q}} \frac{d \ln{q}}{d \ln{u}}$, in which case the final equation may very well not be equal to $0$.

Answer 1

Sorry, but the form of the chain rule you are using is for derivatives, not partial derivatives. $\frac{\partial \ln{\beta}}{\partial \ln{q}} \frac{\partial \ln{q}}{\partial \ln{u}} \ne \frac{\partial \ln{\beta}}{\partial \ln{u}}$.

Since $u$ and $\beta$ are independent variables, $\frac{\partial \ln{\beta}}{\partial \ln{u}} = \frac{\partial \ln{u}}{\partial \ln{\beta}} = 0$, but$$\frac{\partial \ln{\beta}}{\partial \ln{u}} \ne \frac{1}{u} (\ln{q}+2u)$$