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**Unformatted text preview: **e formula for d x*(α)/dα, totally differentiate this identity with respect to α :
2 ⋅ x*(α ) + 2 ⋅α ⋅ d x*(α )
d x*(α )
− e x * (α ) ⋅
dα
dα d x*(α )
dα and solve, to get ≡−
α ≡
α 0 2⋅ x*(α )
2⋅α − e x *(α ) Comparative Statics when there are Several Parameters Implicit differentiation also works when there is more than one parameter. Consider the problem max ƒ( x ;α , β )
x with first order condition
∂ ƒ( x*;α , β )
∂x =0 Since the solution function x* = x*(α,β ) satisfies this first order condition for all values of α and
β , we have the identity
∂ ƒ( x*(α , β );α , β )
≡0
α ,β
∂x
Note that the optimal value x* = x*(α,β ) is affected by both changes in α as well as changes in
β. To derive ∂ x*(α,β )/∂α, we totally differentiate the above identity with respect to α, and then
solve. If we want ∂ x*(α,β )/∂β, we totally differentiate the identity with respect to β, then solve.
For example, consider the maximization problem
max a ⋅ ln( x ) − β ⋅ x 2
x Since the first order condition is α⋅[x*]–1 – 2⋅β ⋅x* = 0 its solution function x*(α,β ) will satisfy the identity α⋅[x*(α,β )]–1 – 2⋅β ⋅x*(α,β ) α≡β 0
,
To get ∂ x*(α,β )/∂α, totally differentiate this identity with respect to α :
[ x*(α , β )]−1 − α ⋅ [ x*(α , β )]−2 ⋅ ∂x*(α , β )
∂x*(α , β )
− 2⋅ β ⋅
∂α
∂α ≡ α ,β 0 and solve to get:
Econ 100A 13 Mathematical Handout ∂x*(α , β )
∂α = [ x*(α , β )]−1
α ⋅ [ x*(α , β )]−2 + 2 ⋅ β On the other hand, to get ∂ x*(α,β )/∂β, totally differentiate the identity with respect to β :
−α ⋅ [ x*(α , β )]−2 ⋅ ∂x*(α , β )
∂x*(α , β )
− 2 ⋅ x*(α , β ) − 2 ⋅ β ⋅
∂β
∂β =0 and solve to get:
∂x*(α , β )
∂β = − 2 ⋅ x*(α , β )
α ⋅ [ x*(α , β )]−2 + 2 ⋅ β Comparative Statics when there are Several Control Variables Implicit differentiation also works when there is more than one control variable, and hence more
than one equation in the first order condition. Consider the example
max ƒ( x1 , x2 ;α )
x1 , x2 The first order conditions are that x1* and x2* solve the pair of equations
∗
∗
∗
∗
∂ ƒ( x1 , x2 ;α )
∂ ƒ( x1 , x2 ;α )
and
=0
=0
∂x1
∂x2
so the solution functions x1* = x1*(α) and x2* = x2*(α) satisfy the pair of identities
∗
∗
∂ ƒ( x1 (α ), x2 (α );α )
∂x1 ≡0
α and ∗
∗
∂ ƒ( x1 (α ), x2 (α );α )
∂x2 ≡0
α To get ∂x1*(α)/∂α and ∂x2*(α)/∂α, totally differentiate both of these identities respect to α, to get
∗
∗
∗
∗
∗
∗
∗
∗
∂ 2 ƒ ( x1 (α ), x2 (α );α ) ∂x1 (α )
∂ 2 ƒ ( x1 (α ), x2 (α );α ) ∂x2 (α )
∂ 2 ƒ ( x1 (α ), x2 (α );α )
⋅
+
⋅
+
∂ x12
∂α
∂ x1 ∂ x2
∂α
∂ x1 ∂α ≡0 ∗
∗
∗
∗
∗
∗
∗
∗
∂ 2 ƒ ( x1 (α ), x2 (α );α ) ∂x1 (α )
∂ 2 ƒ ( x1 (α ), x2 (α );α ) ∂x2 (α...

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