Total derivative evaluates to zero: problem while doing comparative statics

Question

I am playing with a toy model version for my research and I have an implicit equation defining the equilibrium: $$\phi = \frac {S}{NF(\phi(u-\eta)+\eta - c)}$$

where $S,N,u,\eta,c$ are parameters and $F(.)$ is the cdf function. I have a cost minimizing objective function defined as $Min C_1S$, where $C_1$ is a separate constant cost coefficient.

I need to perform comparative statics analysis of the objective function wrt different parameters, so I substituted the value of $S$ from the implicit equation above and then proceeded to compute the total derivative of the objective function wrt the different parameters. My expressions evaluate to zero, so I am clearly doing something wrong and I can't understand what.

Here's my attempt: Say the objective function is OF. Then I need $\frac{dOF}{dN}$. According to total derivative formula, we should have: $\frac{dOF}{dN}=\frac{\partial OF}{\partial \phi}\frac{d\phi}{dN}+\frac{\partial OF}{\partial N}$. Now I wasn't sure of $\frac{d\phi}{dN}$, so I tried two approaches.

Firstly, I tried $\frac{\partial\phi}{\partial N}$ because $\phi$ is affected by all these parameters $N, S,$ etc. I used the implicit function theorem to arrive at $\frac{-\phi F(\phi(u-\eta)+\eta -c)}{NF(\phi(u-\eta)+\eta -c) + \phi N(u-\eta)f(\phi(u-\eta)+\eta -c)}$.

Multiplying this with $\frac{\partial OF}{\partial \phi}$ I finally get $-C_1\phi F(\phi(u-\eta)+\eta -c)$. But $\frac{\partial OF}{\partial N}$ evaluates to $C_1\phi F(\phi(u-\eta)+\eta -c)$, so the two terms cancel each other out.

Then I thought, perhaps I need to calculate $\frac{d\phi}{dN}$ since this is the equilibrium and we don't need the ceteris paribus condition. From the implicit function, I took derivative of the entire expression wrt N. This is what I get:

$\frac{\partial \phi}{\partial N}NF(\phi(u-\eta)+\eta -c)+\phi F(\phi(u-\eta)+\eta -c)+\phi Nf(\phi(u-\eta)+\eta -c)\frac{\partial \phi}{\partial N}(u-\eta)$, here $f()$ is the pdf.

I substitute the value of $\frac{\partial \phi}{\partial N}$ from the implicit function theorem and $\frac{d\phi}{dN}$ evaluates to zero. This also happens when I follow this procedure for other parameters, say $\eta$.

Answer 1

Consider the minimisation problem $$ OF(N,u,\eta,c) = \min_\phi C_1 S(\phi; N,u,\eta,c) $$ Here, I'm assuming that the minimisation is with respect to $\phi$. (don't know if this is correct as it is not specified in your model).

Let $\phi^\ast(N,u,\eta,c)$ be the (assumed unique) optimal solution. And assume that it is identified by the suitable first order condition.

Then we can use the enveloppe theorem to conclude that: $$ \frac{d OF(N,u,\eta,c)}{d N} = C_1 \frac{d S(\phi^\ast; N, u, \eta, c)}{d N} = C_1 \phi^\ast F(\phi^\ast(u-\eta) + \eta - c). $$

I'm mainly guesssing here as it is not very clear from you question which variable is endogenously set by the equilibrium condition and what variable you are minimising over.