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I have a system of three equations in three unknowns, $k$, $\theta$ and $w$. I am interested in the behavior of the variables when one parameter changes. I first specify the system:

 Clear[eq1, eq2, eq3, k, theta, w, a, eta, r, s, p, rho, beta, solk, 
 soltheta, solw];

 eta = 0.5; a = 1/3; r = 1/20; s = 1/10; p = 1/5; rho = 0.7;

 f = (a*k^rho + (1 - a))^(1/rho);

 eq1[beta_] := 
 p - ((1 - beta)*theta^(-eta)/(r + s + (1 - beta)*theta^(-eta))*
 D[f, k]/(r + s));

 eq2[beta_] := 
 p - ((1 - beta)*
 theta^(-eta)/(r + s + 
    beta*theta^(1 - eta) + (1 - beta)*theta^(-eta))*f/k*1/(r + s));

 eq3[beta_] := 
 w - (beta*( 
  r + s + theta^(1 - eta))*(a*k^rho + (1 - a))^(1/rho)/(r + s + 
    beta*theta^(1 - eta) + (1 - beta)*theta^(-eta)));

Then, I solve the system numerically and I plot the solution as functions on the parameter $\beta$:

solk[beta_?NumericQ] := {k} /. 
FindRoot[{eq1[beta] == 0, eq2[beta] == 0, 
 eq3[beta] == 0}, {{k, 0.00009}, {theta, 1000000}, {w, 0.6}}];

soltheta[beta_?NumericQ] := {theta} /. 
FindRoot[{eq1[beta] == 0, eq2[beta] == 0, 
 eq3[beta] == 0}, {{k, 0.00009}, {theta, 1000000}, {w, 0.6}}];

solw[beta_?NumericQ] := {w} /. 
FindRoot[{eq1[beta] == 0, eq2[beta] == 0, 
 eq3[beta] == 0}, {{k, 0.00009}, {theta, 1000000}, {w, 0.6}}];

Plot[solk[beta], {beta, 0.1, 0.9}]
Plot[soltheta[beta], {beta, 0.1, 0.9}]
Plot[solw[beta], {beta, 0.1, 0.9}]

Now, I would like to get some insight about why, in particular, $w$ behaves like it does.

For instance, I could compute the total derivative of eq3:

Dt[eq3[beta]] // FullSimplify

But the output is rather complicated and difficult to interpret. What I would like to obtain is an expression in which $\frac{\partial{w}}{\partial{\beta}}$ depends on $\frac{\partial{k}}{\partial{\beta}}$ and $\frac{\partial{\theta}}{\partial{\beta}}$. How can I do that?

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  • $\begingroup$ Do you need dependencies of derivatives on solutions of equations 1,2,3? $\endgroup$ – Alex Trounev Nov 5 '18 at 10:41
  • $\begingroup$ Maybe D[w /. First@Solve[eq3[beta] == 0, w] /. {k -> k[beta], theta -> theta[beta]}, beta] // Simplify? You could use Rationalize[eq3[beta]] to simplify the numbers, which seem to be simple fractions. Or if you like Dt, then perhaps Dt[w /. First@Solve[Rationalize@eq3[beta] == 0, w]] /. {Dt[beta] -> 1} // Simplify? $\endgroup$ – Michael E2 Nov 5 '18 at 12:48
  • $\begingroup$ @Alex I guess so. Since the system is block recursive I would like to know how the derivative of $w$ wrt $\beta$ depends on $k^*(\beta)$ and $\theta^*(\beta)$ where stars means the functions solves the system eq1 and eq2 $\endgroup$ – Giop Nov 6 '18 at 12:57
  • $\begingroup$ @MichaelE2, thank you. So the output of D[w /. First@Solve[eq3[beta] == 0, w] /. {k -> k[beta], theta -> theta[beta]}, beta] // Simplify gives me $w$ as a function of $k(\beta)$ and $\theta(\beta)$ and their derivatives. Unfortunately the expression is still too complicated to be studied analytically $\endgroup$ – Giop Nov 6 '18 at 14:15
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It is necessary to replace the variables in the equations for solutions, and then calculate the derivative, for example,

e3 = 
 eq3[beta] /. {w -> solw[beta], k -> solk[beta], 
   theta -> soltheta[beta]}

Out[]= -((
  beta (2/3 + solk[beta]^0.7/3)^1.42857 (3/20 + soltheta[beta]^0.5))/(
  3/20 + (1 - beta)/soltheta[beta]^0.5 + beta soltheta[beta]^0.5)) + 
 solw[beta]
e3D = D[e3, beta]

Out[]= -(((2/3 + solk[beta]^0.7/3)^1.42857 (3/20 + 
     soltheta[beta]^0.5))/(
  3/20 + (1 - beta)/soltheta[beta]^0.5 + beta soltheta[beta]^0.5)) - (
 0.333333 beta (2/3 + solk[beta]^0.7/3)^0.428571 (3/20 + 
    soltheta[beta]^0.5) Derivative[1][solk][beta])/(
 solk[beta]^0.3 (3/20 + (1 - beta)/soltheta[beta]^0.5 + 
    beta soltheta[beta]^0.5)) - (
 0.5 beta (2/3 + solk[beta]^0.7/3)^1.42857 Derivative[1][soltheta][
   beta])/((3/20 + (1 - beta)/soltheta[beta]^0.5 + 
    beta soltheta[beta]^0.5) soltheta[beta]^0.5) + (
 beta (2/3 + solk[beta]^0.7/3)^1.42857 (3/20 + 
    soltheta[beta]^0.5) (-(1/soltheta[beta]^0.5) + 
    soltheta[beta]^0.5 - (
    0.5 (1 - beta) Derivative[1][soltheta][beta])/
    soltheta[beta]^1.5 + (0.5 beta Derivative[1][soltheta][beta])/
    soltheta[beta]^0.5))/(3/20 + (1 - beta)/soltheta[beta]^0.5 + 
   beta soltheta[beta]^0.5)^2 + Derivative[1][solw][beta]

Other equations are calculated similarly.

e1 = eq1[beta] /. {w -> solw[beta], k -> solk[beta], 
   theta -> soltheta[beta]}; e1D = D[e1, beta]
 e2 = 
     eq2[beta] /. {w -> solw[beta], k -> solk[beta], 
       theta -> soltheta[beta]}; e2D = D[e2, beta]
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