# Partial derivative on a Solve solution

Trying to solve a demand-supply equilibrium here and them find a partial derivative of the equilibrium with respect to the parameters.

S[q_] := -Log[q]
d[Il_, beta_] := phi*v*zeta/(1 - beta - delta)*(A/(v*zeta*(1 + phi*Il)))^(v/(v - 1))
S1[Il_, beta_, ig_] := S[(delta + beta)*i  - beta*ig - (1 - beta - delta)*Il]


d is the demand and S1 is the supply. The equilibrium price is expressed in terms of the parameters beta and ig.

f[beta_, ig_] := x /. Solve[d[x, beta] == S1[x, beta, ig], x, Reals][]


Now I want to look at the partial derivative of the equilibrium wrt ig as a function of beta.

Eff[beta_] := D[f[beta, ig], ig] /. ig -> 0.03


But Eff() at any value is taking forever to compute. The function f() is well-defined and is giving me quick output when I plug in some arguments.

Is there anything wrong in the code or is it just computationally time-consuming?

Values of the symbols are as follows.

phi = 0.2
v = 1.5
zeta = 1
delta = 0.04
A = 1
i = 0.05

• The problem is not Eff[], but The Solve n the definition of f . With all the undefined symbols like: phi, v, zeta, delta, ig it is hard, to solve. Try giving numerical values to these symbols Apr 21 at 20:33
• Yes. I had assigned values to them, which i missed out on mentioning. Edited now Apr 21 at 20:38
• If I evaluate your code I get the error: Solve::nsmet: This system cannot be solved with the methods available to Solve. If you give the extra variables numeric values you might want to try NSolve instead of solve: the form of your equation is not one that can be inverted symbolically I think so you would have to resolve to numerical methods anyway
– Gert
Apr 21 at 20:48
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If you replace Solve with NSolve and calculate your derivative the old fashioned way you can quickly evaluate it for enough values to make a proper plot.

phi=0.2; v=1.5; zeta=1; delta=0.04; A=1; i=0.05;

d[Il_,beta_]:=1. (phi v zeta)/(1-beta-delta) ( A/(v zeta(1+phi Il)))^(v/(v-1));

S1[Il_,beta_,ig_]:=-1. Log[(delta+beta)*i-beta*ig-(1-beta-delta)*Il]

f[beta_,ig_] := Module[{x},
x/. NSolve[d[x,beta]==S1[x,beta,ig],x,Reals][]
];
Eff[beta_, dx_] := (f[beta,0.03+dx]-f[beta,0.03])/dx;
`

By setting dx to 0.0001 you can calculate the derivative for any beta pretty quickly. (Note that I also put factors 1. in front of your functions to let Mathematica evaluate them numerically from the beginning.)

• Thanks! That is indeed a quick fix to the problem Apr 22 at 4:19