Abraham (1987) Simple Job Market Matching Model I have a question about a derivation in Abraham (1987)'s simple job market matching model (equations 3 through 7):

She begins by writing down tautologies:

J - V = L - U = E

where J is the number of jobs, V is the number of vacancies, L is the total labor force, U is the total unemployed, and E is the total employed. Consider a probability function that a particular vacancy will be filled:

p(V/U) where $p_u &gt; 0, p_v &lt;0, p_{uu} &lt; 0, p_{vv} &gt; 0$.

Now, letting the total quits and firings from the current stock of employees be sE we must have:
\begin{equation}
 p(V/U)V - sE =0 
\end{equation}
Letting U/E = u, V/E = v, and dividing through by E we get:

\begin{equation}
p(v/u)v = s
\end{equation}

Now things get tricky. She writes "along the set of points satisfying (the above)" we have:

\begin{equation}
\frac{dv}{du} = - \frac{p_u}{p+p_v v} 
\end{equation}

My question is: how? Suppose I treat this as an implicit differentiation problem. Write $F(v,u) = p(v/u)v-s = 0$ then we know we know:

\begin{equation}
\frac{d v}{d u} = - \frac{\partial F}{du} / \frac{\partial F}{dv} 
\end{equation}

I obtain $F_v = p + (v/u)p_v $ and $F_u =  - (\frac{v}{u})^2 p_{u}$ which yields:

\begin{equation}
\frac{dv}{du} =  \frac{(\frac{v}{u})^2 p_{u}}{p + (v/u)p_v} 
\end{equation}

This doesn't look very close. Am I missing something? What's going on here?

EDIT:

Inspired by the answer below, I agree it is as simple as totally differentiating P(u,v) = p(v/u).

\begin{equation} 
p(v/u) = P(v,u) \rightarrow P(v,u)v - s = 0 
\end{equation}

Now, let's totally differentiate this expression with respect to v,u, and s.

\begin{equation} 
\left[P_v(v,u)v + P\right] dv + v P_u du - ds = 0
\end{equation}

Consider $ds = 0 $ e.g. no change in the separation rate. This yields:

\begin{equation} 
\frac{dv}{du} = - \frac{vP_u}{vP_v + P}
\end{equation}

Substituting $p(v/u)$ for $P(v,u)$ gives the expression off by a factor v.
Note that function $p$ actually depends upon a single variable $u/v$ and I prefer avoiding the abuse of notation 
(source of confusion) and write $p(u/v) = P(u,v)$. This implies:

\begin{equation}
p'(u/v) = P_u(u,v)v = P_v(u,v) \cdot (-v^2/u). 
\end{equation}

So if $P(u,v)v = s$ total differentiation along $ds=0$ yields

\begin{equation}
P_u(u,v)vdu + (P+P_v(u,v)v)dv = 0, 
\end{equation}

or equivalently 

\begin{equation}
\frac{dv}{du} = - \frac{P_uv}{P+P_v v}. 
\end{equation}

This result is still slighly different from those given in your question... Who can help further?