I am given two terms and the solution of a total differential based on these terms with respect to two endogenous variables. However, I am not sure how they arrived at this result?
The setting (meaning of the variables shouldn't matter):
First Term: $u = \frac{\lambda}{\lambda + \theta q(\theta)}$
Second Term: $(1 - \beta)(p - z) - \beta c\theta - \frac{(r + \lambda)c}{q(\theta)} = 0$
The total differential of the SECOND TERM w.r.t. endogenous variables $p$ and $\theta$:
$\frac{\partial \theta}{\partial p} > 0$ and so $\frac{\partial u}{\partial p} <0$
How do I arrive at this result for the total differential (I know how to calculate the total differential but I don't know how they arrive at this solution)?
If you totally differentiate $$ (1-\beta)(p-z) - \beta c \theta - \frac{(r+\lambda)c}{q(\theta)} = 0 $$ you get $$ (1-\beta)\,dp - \beta c \, d\theta + (r+\lambda)c\frac{q'(\theta)}{[q(\theta)]^2} \,d\theta = 0 $$ $$ \frac{d\theta}{dp} = \frac{1-\beta}{c\left[\beta - (r+\lambda)\frac{q'(\theta)}{[q(\theta)]^2} \right]} $$
To show $\frac{d\theta}{dp} > 0$, there must be given some other information. For instance, assume $0<\beta<1$ and $c,r,\lambda$ are all strictly positive and $q'(\theta) < 0$. Then $\frac{d\theta}{dp} > 0$ holds.
For the other inequality, note that $$ \frac{du}{dp} = \frac{du}{d\theta} \frac{d\theta}{dp} $$
Since $\frac{d\theta}{dp} > 0$ is already shown, using the first expression, you need to show that $\frac{du}{d\theta} <0$. This second derivative is $$ \frac{du}{d\theta} = \frac{-\lambda \left[ q(\theta) + \theta q'(\theta)\right]}{\left[\lambda + \theta q(\theta)\right]^2} $$ Again, you need some conditions on $q(\theta)$ to ensure $q(\theta) + \theta q'(\theta) >0$ so that $\frac{du}{d\theta} <0$. For example, $q(\theta) = \theta^{-1/2}$ would work but from some other conditions in the model maybe you can show that $q(\theta) + \theta q'(\theta) >0$ directly.