Linearization and the effect of a change There is the following system of four equations and four endogenous variables $(K,L,w,q)$. Assume $F$ is a concave function.
$\partial F(K,L)/\partial K = r + (1-p)$
$\partial F(K,L)/\partial L = w$
$F(K,L)-rK-wL-q-(1-p)K = 0$
$v(w,p)=\bar{v}$
It seems to linearize the system around the equilibrium values to get $\frac{\partial K}{\partial p}$ and $\frac{\partial L}{\partial p}$ and it seems that these equations become
$\frac{\partial K}{\partial p} = \frac{-F_{LL}-F_{KL}\partial w/\partial p}{(F_{KK}F_{LL}-F^2_{KL})}$
$\frac{\partial L}{\partial p} = \frac{F_{KK}\partial w/\partial p + F_{KL}}{(F_{KK}F_{LL}-F^2_{KL})}$
I tried to Taylar expansion but I could not get these equations. How do we get these last equations? These equations are from p.85 Kousky et al. (2006)
Start from the first order conditions:
$$
\begin{align*}
&amp;F_k = r + (1- p)\\
&amp;F_L = w\\
\end{align*}
$$
Now totally differentiate with respect to $p$:
$$
\begin{align*}
&amp;F_{KK} dK + F_{KL} dL = - dp\\
&amp;F_{LK} dK + F_{LL} dL = \frac{\partial w}{\partial p} dp\\
\end{align*}
$$
This is equivalent to:
$$
\begin{align*}
&amp;dK = \frac{- dp - F_{KL} dL}{F_{KK}}\\
&amp;dL = \frac{\frac{\partial w}{\partial p}dp - F_{LK} dK}{F_{LL}}
\end{align*}
$$
substituting the second into the first gives:
$$
\begin{align*}
&amp;dK = \frac{(- dp)F_{LL} - F_{KL}(\frac{\partial w}{\partial p} dp - F_{LK} dK)}{F_{KK} F_{LL}},\\
\to &amp;\left(1 - \frac{F_{KL} F_{LK}}{F_{LL} F_{KK}}\right) dK = \frac{-F_{LL}dp- F_{KL}\frac{\partial w}{\partial p} dp}{F_{KK} F_{LL}},\\
\to &amp;dK = \frac{-F_{LL} - F_{KL}\frac{\partial w}{\partial p}}{F_{LL} F_{KK} - (F_{LK})^2} dp
\end{align*}
$$
This gives:
$$
\frac{\partial K}{\partial p} = \frac{-F_{LL} - F_{KL}\frac{\partial w}{\partial p}}{F_{LL} F_{KK} - (F_{LK})^2}
$$
The expression for $\frac{\partial L}{\partial p}$ can be derived in a similar way.